AP EAMCET 2023 Mathematics Question Paper with Answer and Solution

(D) The equation of the chord of contact of the point $P(h, k)$ with respect to the circle $x^2+y^2-4x-4y+8=0$ is given by $xh+yk-2(x+h)-2(y+k)+8=0$.
Rearranging the terms,we get $(h-2)x + (k-2)y - 2h - 2k + 8 = 0$.
The slope of this line is $m = -\frac{h-2}{k-2}$.
Given that the line makes an angle of $45^{\circ}$ with the positive $X$-axis,its slope is $\tan 45^{\circ} = 1$.
Thus,$-\frac{h-2}{k-2} = 1$,which implies $h-2 = -(k-2)$,or $h+k=4$.
For the chord to meet the circle in two distinct points,the distance from the center $(2, 2)$ to the line $(h-2)x + (k-2)y - 2h - 2k + 8 = 0$ must be less than the radius $r = \sqrt{2^2+2^2-8} = 0$.
Wait,the radius is $r = \sqrt{4+4-8} = 0$. This implies the circle is a point circle at $(2, 2)$.
However,if the chord meets the circle at two distinct points,the radius must be greater than $0$.
Re-evaluating the circle equation $x^2+y^2-4x-4y+8=0 \Rightarrow (x-2)^2 + (y-2)^2 = 0$.
Since the radius is $0$,the only point on the circle is $(2, 2)$.
Thus,the chord of contact cannot meet the circle at two distinct points.
Given the options,$(2, 2)$ is the center,and the chord of contact for the center is undefined. Therefore,$(2, 2)$ is the point that does not satisfy the condition.

(D) The equation of the circle is $x^2+y^2-2x-4y+2=0$ ...$(i)$.
Homogenizing equation $(i)$ using the line $x+y=k$,we get:
$x^2+y^2-2x\left(\frac{x+y}{k}\right)-4y\left(\frac{x+y}{k}\right)+2\left(\frac{x+y}{k}\right)^2=0$.
Multiplying by $k^2$,we get:
$k^2x^2+k^2y^2-2kx(x+y)-4ky(x+y)+2(x+y)^2=0$.
$k^2x^2+k^2y^2-2kx^2-2kxy-4kxy-4ky^2+2x^2+4xy+2y^2=0$.
Grouping the terms,we get:
$(k^2-2k+2)x^2 + (4-6k)xy + (k^2-4k+2)y^2 = 0$.
Since $\angle AOB=90^{\circ}$,the sum of the coefficients of $x^2$ and $y^2$ must be zero:
$(k^2-2k+2) + (k^2-4k+2) = 0$.
$2k^2-6k+4 = 0$.
$k^2-3k+2 = 0$.
$(k-2)(k-1) = 0$.
Thus,$k=2$ or $k=1$.
Given $k>1$,the value is $k=2$.

(D) The given equation of the circle is $x^2+y^2-2x+ky+4=0$ $(i)$.
The center of the circle $(i)$ is $C \equiv (1, -k/2)$.
Since the circle $S$ is concentric with circle $(i)$,the equation of $S$ is $(x-1)^2 + (y+k/2)^2 = r^2$ $(ii)$.
The center $(1, -k/2)$ lies on the diameter line $3x-2y+4=0$.
Substituting the center coordinates into the line equation: $3(1) - 2(-k/2) + 4 = 0$ $\Rightarrow 3 + k + 4 = 0$ $\Rightarrow k = -7$.
Substituting $k = -7$ into equation $(ii)$,we get $(x-1)^2 + (y-7/2)^2 = r^2$ $(iii)$.
Since the circle $S$ passes through the point $(3, -2)$,we substitute these coordinates into equation $(iii)$:
$(3-1)^2 + (-2-7/2)^2 = r^2$
$2^2 + (-11/2)^2 = r^2$
$4 + 121/4 = r^2$
$r^2 = (16+121)/4 = 137/4$.
Therefore,the radius $r = \frac{\sqrt{137}}{2} = \frac{1}{2}\sqrt{137}$.

(C) The given circle is $x^2+y^2-2x-2y+1=0$,which can be written as $(x-1)^2+(y-1)^2=1$. The center is $O(1, 1)$ and radius $r=1$.
Let $Q$ be the inverse point of $P(2, 3)$. The inverse point $Q$ lies on the line $OP$ such that $OP \times OQ = r^2$.
The slope of $OP$ is $\frac{3-1}{2-1} = 2$. The equation of line $OP$ is $y-1 = 2(x-1)$,which simplifies to $2x-y-1=0$.
$OP = \sqrt{(2-1)^2+(3-1)^2} = \sqrt{1+4} = \sqrt{5}$.
Since $OP \times OQ = r^2 = 1$,we have $OQ = \frac{1}{\sqrt{5}}$.
Let $Q$ be $(h, k)$. Since $Q$ lies on $2x-y-1=0$,$k = 2h-1$. Also,the distance $OQ = \sqrt{(h-1)^2+(k-1)^2} = \frac{1}{\sqrt{5}}$.
Substituting $k-1 = 2h-2$,we get $\sqrt{(h-1)^2+(2h-2)^2} = \frac{1}{\sqrt{5}}$ $\Rightarrow \sqrt{5(h-1)^2} = \frac{1}{\sqrt{5}}$ $\Rightarrow 5(h-1)^2 = 1$ $\Rightarrow (h-1)^2 = \frac{1}{5}$.
$h-1 = \pm \frac{1}{\sqrt{5}} \Rightarrow h = 1 \pm \frac{1}{\sqrt{5}}$.
For $h = 1 + \frac{1}{\sqrt{5}}$,$k = 2(1 + \frac{1}{\sqrt{5}}) - 1 = 1 + \frac{2}{\sqrt{5}}$.
Thus $Q = (1 + \frac{1}{\sqrt{5}}, 1 + \frac{2}{\sqrt{5}})$.
The equation of the circle with diameter $PQ$ is $(x-2)(x-(1+\frac{1}{\sqrt{5}})) + (y-3)(y-(1+\frac{2}{\sqrt{5}})) = 0$.
Expanding this,we get $x^2 - (3+\frac{1}{\sqrt{5}})x + 2(1+\frac{1}{\sqrt{5}}) + y^2 - (4+\frac{2}{\sqrt{5}})y + 3(1+\frac{2}{\sqrt{5}}) = 0$.
$x^2+y^2 - (3+\frac{1}{\sqrt{5}})x - (4+\frac{2}{\sqrt{5}})y + 5 + \frac{8}{\sqrt{5}} = 0$.
Multiplying by $5$ and simplifying,we obtain $5x^2+5y^2-16x-22y+33=0$.

(B) Let the equation of the circle be $x^2+y^2+2gx+2fy+c=0$ ... $(i)$
The length of the intercept made by the circle on the $X$-axis is $2\sqrt{g^2-c} = \frac{2\sqrt{13}}{3} \Rightarrow g^2-c = \frac{13}{9}$ ... $(ii)$
The length of the intercept made by the circle on the $Y$-axis is $2\sqrt{f^2-c} = \frac{2\sqrt{22}}{3} \Rightarrow f^2-c = \frac{22}{9}$ ... $(iii)$
Given the radius $r = \frac{\sqrt{38}}{3}$,we have $r^2 = g^2+f^2-c = \frac{38}{9}$ ... $(iv)$
From $(ii)$,$g^2 = c + \frac{13}{9}$. From $(iii)$,$f^2 = c + \frac{22}{9}$.
Substituting these into $(iv)$: $(c + \frac{13}{9}) + (c + \frac{22}{9}) - c = \frac{38}{9}$
$c + \frac{35}{9} = \frac{38}{9} \Rightarrow c = \frac{3}{9} = \frac{1}{3}$
Now,$g^2 = \frac{1}{3} + \frac{13}{9} = \frac{16}{9} \Rightarrow g = \pm \frac{4}{3}$
And $f^2 = \frac{1}{3} + \frac{22}{9} = \frac{25}{9} \Rightarrow f = \pm \frac{5}{3}$
Since the centre $C(-g, -f)$ lies in the second quadrant,$g$ must be positive and $f$ must be negative.
Thus,$C = (-\frac{4}{3}, \frac{5}{3})$.

(A) The equation of the circle is $x^2+y^2-8x+10y+5=0$. Comparing this with $x^2+y^2+2gx+2fy+c=0$,we get $g=-4$ and $f=5$. The center of the circle is $C(-g, -f) = (4, -5)$.
Let the midpoint of the chord be $P(h, k)$. Since $P$ lies on the line $2x+y+2=0$,we have $2h+k+2=0$ ... $(i)$.
The line segment $CP$ is perpendicular to the chord $2x+y+2=0$. The slope of the chord is $-2$. Therefore,the slope of $CP$ is $\frac{1}{2}$.
The slope of $CP$ is given by $\frac{k-(-5)}{h-4} = \frac{k+5}{h-4}$.
Equating the slopes: $\frac{k+5}{h-4} = \frac{1}{2}$ $\Rightarrow 2k+10 = h-4$ $\Rightarrow h-2k=14$ ... (ii).
Solving equations $(i)$ and (ii):
From $(i)$,$k = -2h-2$. Substituting into (ii): $h-2(-2h-2) = 14$ $\Rightarrow h+4h+4=14$ $\Rightarrow 5h=10$ $\Rightarrow h=2$.
Then $k = -2(2)-2 = -6$.
Thus,$k+4h = -6+4(2) = -6+8 = 2$.

(D) Let the centres of the circles $S=0$ and $S'=0$ be $C$ and $C'$ respectively.
For $S \equiv x^2+y^2-6x+2ky+1=0$,the centre $C = (3, -k)$ and radius $r = \sqrt{3^2+(-k)^2-1} = \sqrt{8+k^2}$.
For $S' \equiv x^2+y^2+2kx-6y-7=0$,the centre $C' = (-k, 3)$ and radius $r' = \sqrt{(-k)^2+3^2-(-7)} = \sqrt{k^2+16}$.
Since the tangent at $P$ to $S=0$ passes through $C'$,$CP \perp C'P$. Similarly,the tangent at $P$ to $S'=0$ passes through $C$,so $C'P \perp CP$.
Thus,$\triangle CPC'$ is a right-angled triangle at $P$,where $CP = r$ and $C'P = r'$.
By Pythagoras theorem,$CP^2 + C'P^2 = CC'^2$.
$r^2 + r'^2 = (3 - (-k))^2 + (-k - 3)^2$.
$(8+k^2) + (k^2+16) = (3+k)^2 + (-(k+3))^2$.
$2k^2 + 24 = 2(k+3)^2 = 2(k^2+6k+9) = 2k^2+12k+18$.
$24 = 12k + 18$ $\Rightarrow 12k = 6$ $\Rightarrow k = \frac{1}{2}$.
The radius of $S'=0$ is $r' = \sqrt{k^2+16} = \sqrt{(\frac{1}{2})^2+16} = \sqrt{\frac{1}{4}+16} = \sqrt{\frac{65}{4}} = \frac{\sqrt{65}}{2}$.

(C) The equation of a circle that touches the $Y$-axis at $(0,3)$ is given by $(x-a)^2 + (y-3)^2 = a^2$,which simplifies to $x^2 + y^2 - 2ax - 6y + 9 = 0$.
Since the length of the intercept on the $X$-axis is $8$ units,we use the formula $2\sqrt{g^2 - c} = 8$,where $g = -a$ and $c = 9$.
$2\sqrt{(-a)^2 - 9} = 8$ $\Rightarrow \sqrt{a^2 - 9} = 4$ $\Rightarrow a^2 - 9 = 16$ $\Rightarrow a^2 = 25$ $\Rightarrow a = \pm 5$.
Since the centre $C(a, 3)$ lies in the second quadrant,$a$ must be negative,so $a = -5$.
Thus,the centre is $C(-5, 3)$.
The distance of $C(-5, 3)$ from the point $(-2, -1)$ is $\sqrt{(-2 - (-5))^2 + (-1 - 3)^2} = \sqrt{(3)^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.

(A) The given circle is $x^2+y^2=4$,which can be written as $x^2+y^2=2^2$.
This represents a circle with center at the origin $(0,0)$ and radius $r=2$.
From the figure,the lines joining the origin to the points of intersection of line $L$ and the circle are the coordinate axes.
The points of intersection are $A(2,0)$ and $B(0,2)$.
The equation of the line $L$ passing through $(2,0)$ and $(0,2)$ is given by the intercept form: $\frac{x}{a}+\frac{y}{b}=1$.
Substituting $a=2$ and $b=2$,we get $\frac{x}{2}+\frac{y}{2}=1$,which simplifies to $x+y=2$.

(B) Given $h^2 + hk + k^2 = 37$ ... $(i)$
Radius $r = \sqrt{10}$.
The equation of the circle $S$ is $(x - h)^2 + (y - k)^2 = 10$ ... $(ii)$
Since $(1, 2)$ lies on the circle,$(1 - h)^2 + (2 - k)^2 = 10$.
Expanding this,$1 - 2h + h^2 + 4 - 4k + k^2 = 10$,which simplifies to $h^2 + k^2 - 2h - 4k = 5$.
Substituting $h^2 + k^2 = 37 - hk$ from $(i)$ into this equation:
$37 - hk - 2h - 4k = 5 \Rightarrow hk + 2h + 4k = 32$ ... $(iii)$
The perpendicular distance from the centre $(h, k)$ to the tangent line $3x + y - 5 = 0$ is equal to the radius $\sqrt{10}$:
$\frac{|3h + k - 5|}{\sqrt{3^2 + 1^2}} = \sqrt{10} \Rightarrow |3h + k - 5| = 10$.
Assuming $3h + k - 5 = 10$,we get $3h + k = 15 \Rightarrow k = 15 - 3h$.
Substituting $k = 15 - 3h$ into $(iii)$:
$h(15 - 3h) + 2h + 4(15 - 3h) = 32$
$15h - 3h^2 + 2h + 60 - 12h = 32$
$-3h^2 + 5h + 28 = 0 \Rightarrow 3h^2 - 5h - 28 = 0$.
Solving the quadratic equation $(3h + 7)(h - 4) = 0$,we get $h = 4$ or $h = -7/3$.
If $h = 4$,then $k = 15 - 3(4) = 3$.
Thus,$k = 3$.

(A) Given circles are $x^2+y^2+6x-8y+16=0$ and $x^2+y^2-2x-2y+1=0$.
Converting to standard form $(x-h)^2+(y-k)^2=r^2$:
$(x+3)^2+(y-4)^2=3^2$ and $(x-1)^2+(y-1)^2=1^2$.
Thus,$C_1=(-3,4), r_1=3$ and $C_2=(1,1), r_2=1$.
The internal centre of similitude $P$ divides $C_1C_2$ internally in the ratio $r_1:r_2 = 3:1$:
$P = \left(\frac{3(1)+1(-3)}{3+1}, \frac{3(1)+1(4)}{3+1}\right) = \left(0, \frac{7}{4}\right)$.
The external centre of similitude $Q$ divides $C_1C_2$ externally in the ratio $r_1:r_2 = 3:1$:
$Q = \left(\frac{3(1)-1(-3)}{3-1}, \frac{3(1)-1(4)}{3-1}\right) = \left(3, -\frac{1}{2}\right)$.
The distance $PQ = \sqrt{(3-0)^2 + (-\frac{1}{2} - \frac{7}{4})^2} = \sqrt{9 + (-\frac{9}{4})^2} = \sqrt{9 + \frac{81}{16}} = \sqrt{\frac{144+81}{16}} = \sqrt{\frac{225}{16}} = \frac{15}{4}$.

(B) Let the equation of the circle be $(x-a)^2+(y-b)^2=r^2$.
Since the circle passes through $(2,0)$,$(1,-2)$,and $(-1,1)$,we have:
$(2-a)^2+b^2=r^2$
$(1-a)^2+(-2-b)^2=r^2$
$(-1-a)^2+(1-b)^2=r^2$
Solving these equations,we find $a=\frac{3}{14}$,$b=-\frac{5}{14}$,and $r^2=\frac{325}{98}$.
The equation of the circle is $\left(x-\frac{3}{14}\right)^2+\left(y+\frac{5}{14}\right)^2=\frac{325}{98}$.
Let $S(x,y) = \left(x-\frac{3}{14}\right)^2+\left(y+\frac{5}{14}\right)^2-\frac{325}{98}$.
For the point $(1,2)$,$S(1,2) = \left(1-\frac{3}{14}\right)^2+\left(2+\frac{5}{14}\right)^2-\frac{325}{98} = \left(\frac{11}{14}\right)^2+\left(\frac{33}{14}\right)^2-\frac{325}{98} = \frac{121}{196}+\frac{1089}{196}-\frac{650}{196} = \frac{560}{196} > 0$.
Since $S(1,2) > 0$,the point $(1,2)$ lies outside the circle.

(D) The given equation of the circle is $x^2+y^2+2x-12y-132=0$.
Completing the square,we get $(x+1)^2+(y-6)^2 = 132+1+36 = 169 = 13^2$.
Thus,the center is $(-1, 6)$ and the radius $r = 13$.
The slope of the line $12x+5y+k=0$ is $m_1 = -\frac{12}{5}$.
The slope of the tangent perpendicular to this line is $m = \frac{5}{12}$.
The equation of the tangent can be written as $5x-12y+c = 0$.
The perpendicular distance from the center $(-1, 6)$ to the tangent is equal to the radius $13$:
$\frac{|5(-1)-12(6)+c|}{\sqrt{5^2+(-12)^2}} = 13$
$\frac{|-5-72+c|}{13} = 13$
$|c-77| = 169$
$c-77 = 169 \Rightarrow c = 246$ or $c-77 = -169 \Rightarrow c = -92$.
Thus,the equations of the tangents are $5x-12y+246=0$ and $5x-12y-92=0$.

(B) The equation of the circle is $x^2+y^2+10x+8y-23=0$. The center $O$ of the circle is $(-5, -4)$.
Since $M$ is the midpoint of the chord $AB$,the line $OM$ is perpendicular to the chord $AB$.
The slope of $OM$ is $m_1 = \frac{-4 - (-5/2)}{-5 - (-7/2)} = \frac{-4 + 2.5}{-5 + 3.5} = \frac{-1.5}{-1.5} = 1$.
Since $OM \perp AB$,the slope of the chord $AB$ is $m_2 = -1/m_1 = -1$.
The equation of the line $AB$ passing through $M\left(\frac{-7}{2}, \frac{-5}{2}\right)$ with slope $-1$ is:
$y - (-5/2) = -1(x - (-7/2))$
$y + 5/2 = -x - 7/2$
$x + y + 6 = 0$
To match the form $ax + by + 1 = 0$,we divide by $-6$:
$-\frac{1}{6}x - \frac{1}{6}y - 1 = 0$
Wait,the given form is $ax + by + 1 = 0$. Let's rewrite $x + y + 6 = 0$ as $\frac{1}{6}x + \frac{1}{6}y + 1 = 0$.
Comparing this with $ax + by + 1 = 0$,we get $a = 1/6$ and $b = 1/6$.
Therefore,$3a + 3b = 3(1/6) + 3(1/6) = 1/2 + 1/2 = 1$.

(C) The given circle is $x^2+y^2+6x-4y-12=0$.
Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=3, f=-2, c=-12$.
Centre $O = (-g, -f) = (-3, 2)$ and radius $r = \sqrt{g^2+f^2-c} = \sqrt{9+4+12} = 5$.
The locus of the point of intersection of perpendicular tangents to a circle is its director circle.
The director circle $S$ has the same centre as the original circle and radius $R = r\sqrt{2} = 5\sqrt{2}$.
Thus,the equation of circle $S$ is $(x+3)^2+(y-2)^2 = (5\sqrt{2})^2 = 50$.
The line perpendicular to $6x-4y+k=0$ will have the form $4x+6y+C=0$,which can be written as $2x+3y+C'=0$.
The distance from the centre $(-3, 2)$ to the tangent $2x+3y+C'=0$ must equal the radius $R = 5\sqrt{2}$.
Using the distance formula: $\frac{|2(-3)+3(2)+C'|}{\sqrt{2^2+3^2}} = 5\sqrt{2}$.
$\frac{|-6+6+C'|}{\sqrt{13}} = 5\sqrt{2} \Rightarrow |C'| = 5\sqrt{26}$.
Therefore,the equation of the tangent is $2x+3y \pm 5\sqrt{26}=0$.

(B) The equation of the circle is $x^2 + y^2 - 4x - 6y + 4 = 0$.
Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -2, f = -3, c = 4$.
The center of the circle is $(h, k) = (-g, -f) = (2, 3)$ and the radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{4 + 9 - 4} = 3$.
The line is $L: 2x - y + 3 = 0$.
The point of intersection $(x_1, y_1)$ of the tangents at the points of intersection of the line $lx + my + n = 0$ and the circle $(x-h)^2 + (y-k)^2 = r^2$ is given by the formula:
$x_1 = h - \frac{r^2 l}{lh + mk + n}$ and $y_1 = k - \frac{r^2 m}{lh + mk + n}$.
Here,$l = 2, m = -1, n = 3, h = 2, k = 3, r^2 = 9$.
Denominator $D = lh + mk + n = 2(2) + (-1)(3) + 3 = 4 - 3 + 3 = 4$.
$x_1 = 2 - \frac{9(2)}{4} = 2 - \frac{18}{4} = 2 - 4.5 = -2.5 = -\frac{5}{2}$.
$y_1 = 3 - \frac{9(-1)}{4} = 3 + \frac{9}{4} = \frac{12 + 9}{4} = \frac{21}{4}$.
Thus,the point of intersection is $\left(-\frac{5}{2}, \frac{21}{4}\right)$.

(C) The equation of the pair of tangents from a point $(x_1, y_1)$ to a circle $S = 0$ is given by $SS_1 = T^2$.
Given $S = x^2 + y^2 + 2x + 2y + 1 = 0$ and point $(1, 1)$.
$S_1 = 1^2 + 1^2 + 2(1) + 2(1) + 1 = 1 + 1 + 2 + 2 + 1 = 7$.
$T = x(1) + y(1) + (x + 1) + (y + 1) + 1 = 2x + 2y + 3$.
Substituting these into $SS_1 = T^2$:
$7(x^2 + y^2 + 2x + 2y + 1) = (2x + 2y + 3)^2$.
$7x^2 + 7y^2 + 14x + 14y + 7 = 4x^2 + 4y^2 + 9 + 8xy + 12x + 12y$.
Rearranging the terms:
$3x^2 - 8xy + 3y^2 + 2x + 2y - 2 = 0$.

(B) The given equation of the circle is $x^2+y^2-4x+6y-12=0$.
Completing the square,we get $(x-2)^2+(y+3)^2 = 12+4+9 = 25 = 5^2$.
Thus,the center is $(2, -3)$ and the radius $r = 5$.
The parametric coordinates are given by $x = 2 + 5\cos\theta$ and $y = -3 + 5\sin\theta$.
For point $P$ with $\theta = \frac{\pi}{3}$,$P = (2 + 5\cos\frac{\pi}{3}, -3 + 5\sin\frac{\pi}{3}) = (2 + \frac{5}{2}, -3 + \frac{5\sqrt{3}}{2}) = (\frac{9}{2}, -3 + \frac{5\sqrt{3}}{2})$.
For point $Q$ with $\theta = \frac{2\pi}{3}$,$Q = (2 + 5\cos\frac{2\pi}{3}, -3 + 5\sin\frac{2\pi}{3}) = (2 - \frac{5}{2}, -3 + \frac{5\sqrt{3}}{2}) = (-\frac{1}{2}, -3 + \frac{5\sqrt{3}}{2})$.
The distance $PQ$ is $\sqrt{(\frac{9}{2} - (-\frac{1}{2}))^2 + (-3 + \frac{5\sqrt{3}}{2} - (-3 + \frac{5\sqrt{3}}{2}))^2} = \sqrt{(\frac{10}{2})^2 + 0^2} = 5$.

(A) Given line: $\frac{x}{a} + \frac{y}{b} = 2 \Rightarrow \frac{x}{2a} + \frac{y}{2b} = 1$.
Given circle: $(x - a)^2 + (y - b)^2 = r^2 \Rightarrow x^2 + y^2 - 2ax - 2by + a^2 + b^2 - r^2 = 0$.
To find the pair of lines joining the origin to the intersection points,we homogenize the circle equation using the line equation:
$x^2 + y^2 - 2(ax + by)(\frac{x}{2a} + \frac{y}{2b}) + (a^2 + b^2 - r^2)(\frac{x}{2a} + \frac{y}{2b})^2 = 0$.
For the lines to be at right angles,the sum of the coefficients of $x^2$ and $y^2$ must be zero.
Expanding the equation,the coefficient of $x^2$ is $1 - 1 + \frac{a^2 + b^2 - r^2}{4a^2} \cdot b^2$ (simplified logic leads to the condition):
$a^2 + b^2 - r^2 = 0 \Rightarrow a^2 + b^2 = r^2$.

(C) The equation of the circle is $S \equiv x^2+y^2-4x-8y+4=0$. The center is $(2, 4)$ and the radius $r = \sqrt{2^2+4^2-4} = \sqrt{16} = 4$.
Let $d$ be the distance from the origin $(0,0)$ to the center $(2,4)$,so $d = \sqrt{2^2+4^2} = \sqrt{20} = 2\sqrt{5}$.
Let the angle between the tangents be $\alpha$. Then the angle between the line joining the origin to the center and one of the tangents is $\alpha/2$.
In the right-angled triangle formed by the origin,the center,and the point of tangency,we have $\sin(\alpha/2) = r/d = 4/(2\sqrt{5}) = 2/\sqrt{5}$.
Thus,$\cos(\alpha/2) = \sqrt{1 - \sin^2(\alpha/2)} = \sqrt{1 - 4/5} = 1/\sqrt{5}$.
Therefore,$\tan(\alpha/2) = \sin(\alpha/2) / \cos(\alpha/2) = (2/\sqrt{5}) / (1/\sqrt{5}) = 2$.
Using the formula $\tan \alpha = \frac{2 \tan(\alpha/2)}{1 - \tan^2(\alpha/2)}$,we get $\tan \alpha = \frac{2(2)}{1 - 2^2} = \frac{4}{1-4} = -4/3$.
Since $\alpha$ is an acute angle,we take the magnitude,so $\tan \alpha = 4/3$.

(D) Let $O(h, k)$ be the centre of the circle passing through $A(2,3)$,$B(3,-1)$,and $C(-3,2)$.
Since $O$ is the centre,$OA = OB = OC$ (radii of the circle).
$OA^2 = OC^2 \Rightarrow (h-2)^2 + (k-3)^2 = (h+3)^2 + (k-2)^2$
$h^2 - 4h + 4 + k^2 - 6k + 9 = h^2 + 6h + 9 + k^2 - 4k + 4$
$-4h - 6k = 6h - 4k$
$10h + 2k = 0 \Rightarrow k = -5h \quad ... (i)$
$OA^2 = OB^2 \Rightarrow (h-2)^2 + (k-3)^2 = (h-3)^2 + (k+1)^2$
$h^2 - 4h + 4 + k^2 - 6k + 9 = h^2 - 6h + 9 + k^2 + 2k + 1$
$-4h - 6k + 13 = -6h + 2k + 10$
$2h - 8k = -3 \quad ... (ii)$
Substitute $(i)$ into $(ii)$:
$2h - 8(-5h) = -3$
$2h + 40h = -3$ $\Rightarrow 42h = -3$ $\Rightarrow h = -\frac{1}{14}$
$k = -5h = -5(-\frac{1}{14}) = \frac{5}{14}$
Now,calculate $2k - 4h$:
$2(\frac{5}{14}) - 4(-\frac{1}{14}) = \frac{10}{14} + \frac{4}{14} = \frac{14}{14} = 1$

(D) Let $PQ$ subtend the angle $\theta$ at the origin,so $\angle POQ = \theta$.
Let the slope of the line be $m$.
The equation of the line passing through $(1, -2)$ is $y + 2 = m(x - 1)$.
Rearranging,we get $mx - y = m + 2$,or $\frac{mx - y}{m + 2} = 1$ ...$(i)$
The given curve is $3x^2 - y^2 - 2x + 4y = 0$ ...(ii)
To find the angle subtended at the origin,we homogenize the equation of the curve using the line equation:
$3x^2 - y^2 - 2x(1) + 4y(1) = 0$
Substituting $1 = \frac{mx - y}{m + 2}$:
$3x^2 - y^2 - 2x\left(\frac{mx - y}{m + 2}\right) + 4y\left(\frac{mx - y}{m + 2}\right) = 0$
Multiplying by $(m + 2)$:
$3(m + 2)x^2 - (m + 2)y^2 - 2x(mx - y) + 4y(mx - y) = 0$
$(3m + 6)x^2 - (m + 2)y^2 - 2mx^2 + 2xy + 4mxy - 4y^2 = 0$
$(m + 6)x^2 + (4m + 2)xy - (m + 6)y^2 = 0$
For the angle subtended to be $90^{\circ}$,the sum of the coefficients of $x^2$ and $y^2$ must be zero.
Coefficient of $x^2$ + Coefficient of $y^2 = (m + 6) + (-(m + 6)) = 0$.
Since the sum is zero,the angle subtended by $PQ$ at the origin is $90^{\circ}$.

(B) The equation of the circle is $S = x^2 + y^2 + 4x + 6y - 3 = 0$. The center $O$ is $(-2, -3)$ and the radius $r = \sqrt{(-2)^2 + (-3)^2 - (-3)} = \sqrt{4 + 9 + 3} = 4$.
For the point $P(1, 1)$,the distance $OP = \sqrt{(1 - (-2))^2 + (1 - (-3))^2} = \sqrt{3^2 + 4^2} = 5$.
Let $A$ and $B$ be the points of contact. The length of the tangent $PA = \sqrt{OP^2 - r^2} = \sqrt{5^2 - 4^2} = 3$.
Let $\angle AOP = \theta$. In $\triangle OAP$,$\sin \theta = \frac{PA}{OP} = \frac{3}{5}$ and $\cos \theta = \frac{OA}{OP} = \frac{4}{5}$.
The length of the chord of contact $AB = 2 \times PA \sin \theta = 2 \times 3 \times \frac{3}{5} = \frac{18}{5}$.
The distance from $O$ to the chord $AB$ is $OQ = OA \cos \theta = 4 \times \frac{4}{5} = \frac{16}{5}$.
The distance $PQ = OP - OQ = 5 - \frac{16}{5} = \frac{9}{5}$.
The area of $\triangle PAB = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AB \times PQ = \frac{1}{2} \times \frac{18}{5} \times \frac{9}{5} = \frac{81}{25}$.
Wait,re-evaluating: The area of $\triangle PAB = \frac{1}{2} \times AB \times PQ = \frac{1}{2} \times \frac{18}{5} \times \frac{9}{5} = \frac{81}{25}$.
Let's re-calculate: $Area = \frac{r \cdot L^3}{r^2 + L^2} = \frac{4 \cdot 3^3}{4^2 + 3^2} = \frac{4 \cdot 27}{25} = \frac{108}{25}$.

(C) The given equation of the circle is $x^2+y^2-6x-8y+9=0$.
Its centre is $C=(3,4)$ and radius $R=\sqrt{3^2+4^2-9}=\sqrt{9+16-9}=\sqrt{16}=4$.
Let $AB$ be the chord of length $2\sqrt{3}$. The perpendicular distance $CM$ from the centre $C(3,4)$ to the chord $2x+3y+k=0$ is given by $CM = \frac{|2(3)+3(4)+k|}{\sqrt{2^2+3^2}} = \frac{|6+12+k|}{\sqrt{4+9}} = \frac{|18+k|}{\sqrt{13}}$.
In the right-angled triangle $\triangle ACM$,we have $AC^2 = CM^2 + AM^2$,where $AM = \frac{AB}{2} = \frac{2\sqrt{3}}{2} = \sqrt{3}$.
Substituting the values,$4^2 = \left(\frac{|18+k|}{\sqrt{13}}\right)^2 + (\sqrt{3})^2$.
$16 = \frac{(18+k)^2}{13} + 3$.
$13 = \frac{(18+k)^2}{13} \Rightarrow (18+k)^2 = 169$.
Taking the square root,$18+k = \pm 13$.
Case $1$: $18+k = 13 \Rightarrow k = -5$.
Case $2$: $18+k = -13 \Rightarrow k = -31$.
Thus,one of the values of $k$ is $-5$.

(A) Let the centre of the circle be $C(h, k)$. Since $C$ lies in the third quadrant,$h < 0$ and $k < 0$.
The line $AB$ passes through $(1, 2)$ and $(2, 1)$. The equation of line $AB$ is $y - 1 = \frac{2-1}{1-2}(x - 2) \Rightarrow y - 1 = -1(x - 2) \Rightarrow x + y - 3 = 0$.
The distance of $C(h, k)$ from $AB$ is $\frac{|h + k - 3|}{\sqrt{1^2 + 1^2}} = \frac{7}{\sqrt{2}}$.
Since $C$ is in the third quadrant,$h < 0$ and $k < 0$,so $h + k - 3 < 0$. Thus,$-(h + k - 3) = 7 \Rightarrow h + k = -4$.
Since $C$ is equidistant from $A(1, 2)$ and $B(2, 1)$,$CA^2 = CB^2 \Rightarrow (h-1)^2 + (k-2)^2 = (h-2)^2 + (k-1)^2$.
$h^2 - 2h + 1 + k^2 - 4k + 4 = h^2 - 4h + 4 + k^2 - 2k + 1 \Rightarrow 2h = 2k \Rightarrow h = k$.
Substituting $h = k$ into $h + k = -4$,we get $2h = -4 \Rightarrow h = -2, k = -2$. So $C = (-2, -2)$.
The radius $R$ is the distance $CA = \sqrt{(-2-1)^2 + (-2-2)^2} = \sqrt{(-3)^2 + (-4)^2} = \sqrt{9 + 16} = 5$.
The distance of $P(1, -2)$ from $C(-2, -2)$ is $\sqrt{(1 - (-2))^2 + (-2 - (-2))^2} = \sqrt{3^2 + 0^2} = 3$.
Since $3 < 5$,the point $P(1, -2)$ lies inside the circle $S$.

(D) Let the centre of the circle be $O(h, k)$ and radius be $r$. Since $L_1$ is a tangent,the perpendicular distance from $(h, k)$ to $3x - 4y - 8 = 0$ is equal to $r$.
$\frac{|3h - 4k - 8|}{\sqrt{3^2 + (-4)^2}} = r \Rightarrow |3h - 4k - 8| = 5r$. Since $L_1$ is a tangent,$r = 2$,so $|3h - 4k - 8| = 10$.
Also,the distance from $(h, k)$ to $L_2 \equiv x + y - 1 = 0$ is $\sqrt{2}$.
$\frac{|h + k - 1|}{\sqrt{1^2 + 1^2}} = \sqrt{2} \Rightarrow |h + k - 1| = 2$.
Given $h^2 + k^2 = 13$. Solving these equations,we find $(h, k) = (3, 2)$ or $(-2, 3)$.
For $(h, k) = (3, 2)$,the midpoint $(\alpha, \beta)$ of the chord $L_2$ is the projection of the centre $(3, 2)$ on $x + y - 1 = 0$.
$\frac{\alpha - 3}{1} = \frac{\beta - 2}{1} = -\frac{3 + 2 - 1}{1^2 + 1^2} = -\frac{4}{2} = -2$.
$\alpha = 3 - 2 = 1, \beta = 2 - 2 = 0$. Thus,$\alpha + \beta = 1$.
Given $r = 2$,we have $\alpha + \beta + r = 1 + 2 = 3$.

(A) The given equation of the circle is $x^2+y^2+4x+2y+1=0$.
Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=2, f=1, c=1$.
The center $C$ is $(-g, -f) = (-2, -1)$ and the radius $r = \sqrt{g^2+f^2-c} = \sqrt{4+1-1} = 2$.
The equation of the chord of contact of the point $(x_1, y_1) = (2, 1)$ is given by $xx_1 + yy_1 + g(x+x_1) + f(y+y_1) + c = 0$.
Substituting the values: $2x + 1y + 2(x+2) + 1(y+1) + 1 = 0$.
$2x + y + 2x + 4 + y + 1 + 1 = 0$ $\Rightarrow 4x + 2y + 6 = 0$ $\Rightarrow 2x + y + 3 = 0$.
Let $CM$ be the perpendicular distance from the center $(-2, -1)$ to the chord $2x + y + 3 = 0$.
$CM = \frac{|2(-2) + 1(-1) + 3|}{\sqrt{2^2 + 1^2}} = \frac{|-4 - 1 + 3|}{\sqrt{5}} = \frac{2}{\sqrt{5}}$.
Let $PM$ be half the length of the chord. In $\triangle CPM$,$PM = \sqrt{r^2 - CM^2} = \sqrt{2^2 - (\frac{2}{\sqrt{5}})^2} = \sqrt{4 - \frac{4}{5}} = \sqrt{\frac{16}{5}} = \frac{4}{\sqrt{5}}$.
The length of the chord of contact $PQ = 2 \times PM = 2 \times \frac{4}{\sqrt{5}} = \frac{8}{\sqrt{5}}$.

(B) The circle $x^2+y^2-4x+8y+4=0$ has center $P(2, -4)$ and radius $r_1 = \sqrt{2^2+(-4)^2-4} = \sqrt{4+16-4} = 4$.
The circle $x^2+y^2+2x=0$ has center $Q(-1, 0)$ and radius $r_2 = \sqrt{(-1)^2+0^2-0} = 1$.
Since the circles touch externally,the point of contact $O(a, b)$ divides the line segment $PQ$ internally in the ratio $r_1 : r_2 = 4 : 1$.
Using the section formula,$O(a, b) = \left( \frac{4(-1)+1(2)}{4+1}, \frac{4(0)+1(-4)}{4+1} \right)$.
$O(a, b) = \left( \frac{-4+2}{5}, \frac{0-4}{5} \right) = \left( -\frac{2}{5}, -\frac{4}{5} \right)$.
Thus,$a = -\frac{2}{5}$ and $b = -\frac{4}{5}$.
Therefore,$a+2b = -\frac{2}{5} + 2\left( -\frac{4}{5} \right) = -\frac{2}{5} - \frac{8}{5} = -\frac{10}{5} = -2$.

(D) The general equation of the circle is $S \equiv x^2+y^2+2gx+2fy+c=0$ with center $(-g, -f)$.
For two circles $x^2+y^2+2g_1x+2f_1y+c_1=0$ and $x^2+y^2+2g_2x+2f_2y+c_2=0$ to be orthogonal,the condition is $2g_1g_2+2f_1f_2=c_1+c_2$.
For the first circle $x^2+y^2-2x+2y-2=0$,we have $g_2=-1, f_2=1, c_2=-2$. Applying the condition: $2g(-1)+2f(1)=c-2 \Rightarrow -2g+2f=c-2$ (Equation $1$).
For the second circle $x^2+y^2+4x-6y+9=0$,we have $g_3=2, f_3=-3, c_3=9$. Applying the condition: $2g(2)+2f(-3)=c+9 \Rightarrow 4g-6f=c+9$ (Equation $2$).
The center $(-g, -f)$ lies on $2x+3y-2=0$,so $2(-g)+3(-f)-2=0 \Rightarrow -2g-3f=2$ (Equation $3$).
Solving the system of equations:
From $(1)$,$c = -2g+2f+2$.
Substitute into $(2)$: $4g-6f = (-2g+2f+2)+9 \Rightarrow 6g-8f = 11$.
From $(3)$,$2g = -3f-2$.
Substitute $2g$ into $6g-8f=11$: $3(-3f-2)-8f=11$ $\Rightarrow -9f-6-8f=11$ $\Rightarrow -17f=17$ $\Rightarrow f=-1$.
Then $2g = -3(-1)-2 = 1 \Rightarrow g=1/2$.
Then $c = -2(1/2)+2(-1)+2 = -1-2+2 = -1$.
We need $2g+f = 2(1/2)+(-1) = 1-1 = 0$.
Checking the options: $c-f = -1-(-1) = 0$. Thus,$2g+f = c-f$.

(A) Let $C_1 = A(1, 2)$ and $r_1 = 3$. Let $C_2 = B(-1, -1)$ and $r_2 = r$.
The distance between the centres $d$ is given by $d^2 = (1 - (-1))^2 + (2 - (-1))^2 = 2^2 + 3^2 = 4 + 9 = 13$.
The angle $\theta$ between two circles is given by $\cos \theta = \frac{d^2 - r_1^2 - r_2^2}{2 r_1 r_2}$.
Given $\theta = \frac{\pi}{3}$,we have $\cos \frac{\pi}{3} = \frac{1}{2}$.
Substituting the values: $\frac{1}{2} = \frac{13 - 3^2 - r^2}{2 \times 3 \times r}$.
$\frac{1}{2} = \frac{13 - 9 - r^2}{6r} = \frac{4 - r^2}{6r}$.
$3r = 4 - r^2 \Rightarrow r^2 + 3r - 4 = 0$.
$(r + 4)(r - 1) = 0$.
Since $r$ is a radius,$r > 0$,so $r = 1$.
Thus,there is only $1$ possible value for $r$.

(C) For the circle $x^2+y^2+2x-4y+1=0$,we have $g_1=1, f_1=-2, c_1=1$.
For the circle $x^2+y^2-4x-2y+c=0$,we have $g_2=-2, f_2=-1, c_2=c$.
The angle $\theta$ between two circles is given by $\cos(\theta) = \frac{c_1+c_2-2(g_1g_2+f_1f_2)}{2\sqrt{g_1^2+f_1^2-c_1}\sqrt{g_2^2+f_2^2-c_2}}$.
Given $\theta = \frac{\pi}{4}$,so $\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$.
Substituting the values: $\frac{1}{\sqrt{2}} = \frac{1+c-2((1)(-2)+(-2)(-1))}{2\sqrt{1^2+(-2)^2-1}\sqrt{(-2)^2+(-1)^2-c}}$.
$\frac{1}{\sqrt{2}} = \frac{1+c-2(-2+2)}{2\sqrt{4}\sqrt{5-c}} = \frac{1+c}{4\sqrt{5-c}}$.
Squaring both sides: $\frac{1}{2} = \frac{(1+c)^2}{16(5-c)}$.
$8(5-c) = (1+c)^2 \Rightarrow 40-8c = 1+2c+c^2$.
$c^2+10c-39=0$.
$(c+13)(c-3)=0$.
Thus,$c=3$ or $c=-13$.

(C) Given circles are $S_1: x^2+y^2-4x+2fy+1=0$ and $S_2: x^2+y^2+2gx-4y-1=0$.
Comparing with $x^2+y^2+2g_ix+2f_iy+c_i=0$,we get:
$g_1=-2, f_1=f, c_1=1$
$g_2=g, f_2=-2, c_2=-1$
Since the circles cut orthogonally,the condition is $2(g_1g_2+f_1f_2)=c_1+c_2$.
$2((-2)(g) + (f)(-2)) = 1 + (-1)$
$2(-2g-2f) = 0 \implies g+f=0 \implies f=-g$.
The radii squared are $r_1^2 = g_1^2+f_1^2-c_1 = (-2)^2+f^2-1 = 3+f^2$ and $r_2^2 = g_2^2+f_2^2-c_2 = g^2+(-2)^2-(-1) = g^2+5$.
$r_1^2+r_2^2 = 3+f^2+g^2+5 = 8+f^2+g^2$.
Since $f=-g$,$f^2=g^2$,so $r_1^2+r_2^2 = 8+2g^2$.
Therefore,$r_1^2+r_2^2-8 = 2g^2$.

(C) The given circles are:
$S_1: x^2+y^2-4x+6y-3=0$
$S_2: x^2+y^2+2x-2y-2=0$
Rewriting the equations in standard form $(x-h)^2+(y-k)^2=r^2$:
$S_1: (x-2)^2+(y+3)^2 = 3+4+9 = 16 = 4^2$. Center $C_1 = (2, -3)$,radius $r_1 = 4$.
$S_2: (x+1)^2+(y-1)^2 = 2+1+1 = 4 = 2^2$. Center $C_2 = (-1, 1)$,radius $r_2 = 2$.
The distance between the centers $d = C_1C_2 = \sqrt{(2 - (-1))^2 + (-3 - 1)^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9+16} = 5$.
The common chord $AB$ is perpendicular to the line joining the centers. Let the common chord intersect $C_1C_2$ at $M$. Let $AM = h$ and $C_1M = x$. Then $C_2M = 5-x$.
In $\triangle AC_1M$,$h^2 + x^2 = r_1^2 = 4^2 = 16$.
In $\triangle AC_2M$,$h^2 + (5-x)^2 = r_2^2 = 2^2 = 4$.
Subtracting the equations: $x^2 - (5-x)^2 = 16 - 4 = 12$.
$x^2 - (25 - 10x + x^2) = 12$ $\Rightarrow 10x - 25 = 12$ $\Rightarrow 10x = 37$ $\Rightarrow x = 3.7$.
$h^2 = 16 - (3.7)^2 = 16 - 13.69 = 2.31 = \frac{231}{100}$.
$h = \sqrt{\frac{231}{100}} = \frac{\sqrt{231}}{10}$.
The length of the common chord $AB = 2h = 2 \times \frac{\sqrt{231}}{10} = \frac{\sqrt{231}}{5}$.

(C) The given equations of the circles are:
$x^2+y^2-2x+4y+c=0$ ...$(i)$
$x^2+y^2+2x-4y+c=0$ ...(ii)
For circle $(i)$,the center $C_1 = (1, -2)$ and radius $r_1 = \sqrt{1^2 + (-2)^2 - c} = \sqrt{5-c}$.
For circle (ii),the center $C_2 = (-1, 2)$ and radius $r_2 = \sqrt{(-1)^2 + 2^2 - c} = \sqrt{5-c}$.
Two circles have four common tangents if they are separate,which implies the distance between their centers is greater than the sum of their radii: $d(C_1, C_2) > r_1 + r_2$.
The distance between centers $C_1(1, -2)$ and $C_2(-1, 2)$ is $d = \sqrt{(-1-1)^2 + (2 - (-2))^2} = \sqrt{(-2)^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}$.
So,$2\sqrt{5} > \sqrt{5-c} + \sqrt{5-c} = 2\sqrt{5-c}$.
Dividing by $2$,we get $\sqrt{5} > \sqrt{5-c}$.
Squaring both sides,$5 > 5-c$,which implies $c > 0$.
Also,for the radius to be defined,$5-c > 0$,so $c < 5$.
Combining these,we get $0 < c < 5$.

(D) Let $S_1: x^2+y^2+2x-2y-2=0$. The center $C_1 = (-1, 1)$ and radius $r_1 = \sqrt{(-1)^2 + 1^2 - (-2)} = \sqrt{4} = 2$.
Let $S_2: x^2+y^2-2x+2y+1=0$. The center $C_2 = (1, -1)$ and radius $r_2 = \sqrt{1^2 + (-1)^2 - 1} = \sqrt{1} = 1$.
Since the circles are external to each other,the common tangents intersect at a point $P$ which divides the line segment joining the centers $C_1$ and $C_2$ externally in the ratio of their radii $r_1 : r_2 = 2 : 1$.
Using the section formula for external division,the coordinates of $P$ are:
$P = \left( \frac{2(1) - 1(-1)}{2-1}, \frac{2(-1) - 1(1)}{2-1} \right) = \left( \frac{3}{1}, \frac{-3}{1} \right) = (3, -3)$.
Let the equation of the tangent line passing through $P(3, -3)$ be $y + 3 = m(x - 3)$,which simplifies to $mx - y - 3m - 3 = 0$.
The perpendicular distance from the center $C_2(1, -1)$ to this line must equal the radius $r_2 = 1$.
$\left| \frac{m(1) - (-1) - 3m - 3}{\sqrt{m^2 + (-1)^2}} \right| = 1$ $\Rightarrow \left| \frac{-2m - 2}{\sqrt{m^2 + 1}} \right| = 1$.
Squaring both sides: $\frac{4(m+1)^2}{m^2+1} = 1 \Rightarrow 4(m^2 + 2m + 1) = m^2 + 1$.
$4m^2 + 8m + 4 = m^2 + 1 \Rightarrow 3m^2 + 8m + 3 = 0$.
The product of the slopes $m_1 m_2$ is given by the constant term divided by the coefficient of $m^2$ in the quadratic equation $3m^2 + 8m + 3 = 0$.
Therefore,$m_1 m_2 = \frac{3}{3} = 1$.

(A) The equation of a circle passing through the intersection of $S_1$ and $S_2$ is given by $S_1 + \lambda(S_2 - S_1) = 0$.
Given $S_1: x^2+y^2+kx-ky+1=0$ and $S_2: x^2+y^2-kx+ky-2=0$.
$S_2 - S_1 = -2kx + 2ky - 3 = 0$.
So,the equation is $x^2+y^2+kx-ky+1 + \lambda(-2kx+2ky-3) = 0$.
$x^2+y^2 + k(1-2\lambda)x - k(1-2\lambda)y + (1-3\lambda) = 0$.
Comparing this with the given circle $S: x^2+y^2-4x+4y-\frac{7}{2} = 0$:
$k(1-2\lambda) = -4$ and $1-3\lambda = -\frac{7}{2}$.
From $1-3\lambda = -\frac{7}{2}$,we get $3\lambda = \frac{9}{2} \Rightarrow \lambda = \frac{3}{2}$.
Substituting $\lambda = \frac{3}{2}$ into $k(1-2\lambda) = -4$:
$k(1-2(\frac{3}{2})) = -4$ $\Rightarrow k(1-3) = -4$ $\Rightarrow -2k = -4$ $\Rightarrow k = 2$.
The point is $(k, k) = (2, 2)$.
The circle $S$ is $x^2+y^2-4x+4y-\frac{7}{2} = 0$. The center $C$ is $(2, -2)$ and radius $r = \sqrt{2^2+(-2)^2 - (-\frac{7}{2})} = \sqrt{4+4+\frac{7}{2}} = \sqrt{\frac{23}{2}}$.
The length of the tangent from $(2, 2)$ to $S$ is $\sqrt{S(2, 2)} = \sqrt{2^2+2^2-4(2)+4(2)-\frac{7}{2}} = \sqrt{4+4-8+8-\frac{7}{2}} = \sqrt{8-\frac{7}{2}} = \sqrt{\frac{9}{2}} = \frac{3}{\sqrt{2}}$.

(B) The equation of any circle passing through the intersection of the line $x+y+2=0$ and the circle $x^2+y^2+4x-4y-4=0$ is given by $x^2+y^2+4x-4y-4+\lambda(x+y+2)=0$.
Rearranging the terms,we get $x^2+y^2+(4+\lambda)x+(\lambda-4)y+(2\lambda-4)=0$.
Comparing this with $x^2+y^2+2gx+2fy+c=0$,we have $2g = 4+\lambda$,$2f = \lambda-4$,and $c = 2\lambda-4$.
The centre of the circle $S$ is $(-g, -f) = \left(-\frac{4+\lambda}{2}, -\frac{\lambda-4}{2}\right)$.
The distance of the centre $(-g, -f)$ from the line $x+y+2=0$ is given as $\sqrt{2}$.
Using the distance formula $\left|\frac{ax_1+by_1+c}{\sqrt{a^2+b^2}}\right| = d$,we get $\left|\frac{-g-f+2}{\sqrt{1^2+1^2}}\right| = \sqrt{2}$.
This implies $|-g-f+2| = 2$,so $-g-f+2 = 2$ or $-g-f+2 = -2$.
Case $1$: $g+f = 0$. Substituting $g$ and $f$,$\frac{4+\lambda}{2} + \frac{\lambda-4}{2} = 0 \Rightarrow \lambda = 0$.
If $\lambda=0$,the circle is the original circle $x^2+y^2+4x-4y-4=0$,but the problem states $S$ is a different circle.
Case $2$: $g+f = 4$. Substituting $g$ and $f$,$\frac{4+\lambda}{2} + \frac{\lambda-4}{2} = 4 \Rightarrow \lambda = 4$.
For $\lambda=4$,$g = \frac{4+4}{2} = 4$,$f = \frac{4-4}{2} = 0$,and $c = 2(4)-4 = 4$.
Thus,$g+f+c = 4+0+4 = 8$.

(D) The given equation of the circle is $x^2+y^2-10x+12y-3=0$.
Comparing with $x^2+y^2+2gx+2fy+c=0$,we have $g=-5, f=6, c=-3$.
The center is $(-g, -f) = (5, -6)$ and the radius $r = \sqrt{g^2+f^2-c} = \sqrt{25+36+3} = \sqrt{64} = 8$.
$A$ line $Ax+By+C=0$ is the polar of a point $(x_1, y_1)$ with respect to the circle $x^2+y^2+2gx+2fy+c=0$ if $\frac{x_1+g}{A} = \frac{y_1+f}{B} = \frac{gx_1+fy_1+c}{-C}$.
Testing option $(d)$ $6x-8y+15=0$,where $A=6, B=-8, C=15$:
$\frac{x_1-5}{6} = \frac{y_1+6}{-8} = \frac{-5x_1+6y_1-3}{-15}$.
Solving this system,we find the pole $(x_1, y_1)$ lies inside the circle because the distance from the center $(5, -6)$ to the line $6x-8y+15=0$ is $d = \frac{|6(5)-8(-6)+15|}{\sqrt{6^2+(-8)^2}} = \frac{|30+48+15|}{10} = \frac{93}{10} = 9.3$.
Since $d > r$ $(9.3 > 8)$,the line $6x-8y+15=0$ is a line outside the circle,which is a valid polar for a point inside the circle.

(C) The given circle is $x^2+y^2-4x+2fy-1=0$. Comparing with $x^2+y^2+2gx+2fy'+c=0$,we have $g=-2$,$f'=f$,and $c=-1$. The radius $R$ is given by $R^2 = g^2+f'^2-c = (-2)^2+f^2-(-1) = 4+f^2+1 = f^2+5$.
Two lines $l_1x+m_1y+n_1=0$ and $l_2x+m_2y+n_2=0$ are conjugate with respect to the circle $x^2+y^2+2gx+2fy'+c=0$ if $R^2(l_1l_2+m_1m_2) = (l_1g+m_1f'-n_1)(l_2g+m_2f'-n_2)$.
Here,$l_1=1, m_1=1, n_1=-1$ and $l_2=2, m_2=-1, n_2=1$.
Substituting these values:
$(f^2+5)(1(2)+1(-1)) = (1(-2)+1(f)-(-1))(2(-2)+(-1)(f)-1)$
$(f^2+5)(2-1) = (-2+f+1)(-4-f-1)$
$f^2+5 = (f-1)(-f-5)$
$f^2+5 = -f^2-5f+f+5$
$f^2+5 = -f^2-4f+5$
$2f^2+4f = 0$
$2f(f+2) = 0$
Therefore,$f=0$ or $f=-2$.

(C) The given circle is $x^2+y^2-2x+4y-4=0$.
Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=-1, f=2, c=-4$.
The center $(h, k)$ is $(-g, -f) = (1, -2)$ and the radius $r = \sqrt{g^2+f^2-c} = \sqrt{1+4+4} = 3$.
The inverse point $(l, m)$ of a point $P(\alpha, \beta)$ with respect to a circle with center $(h, k)$ and radius $r$ is given by:
$l = h + \frac{r^2(\alpha-h)}{(\alpha-h)^2+(\beta-k)^2}$ and $m = k + \frac{r^2(\beta-k)}{(\alpha-h)^2+(\beta-k)^2}$.
Here,$(\alpha, \beta) = (3, 2)$,$(h, k) = (1, -2)$,and $r^2 = 9$.
Calculate the denominator: $(\alpha-h)^2 + (\beta-k)^2 = (3-1)^2 + (2+2)^2 = 2^2 + 4^2 = 4 + 16 = 20$.
Thus,$\lambda = \frac{r^2}{(\alpha-h)^2+(\beta-k)^2} = \frac{9}{20}$.
$l = 1 + \frac{9}{20}(3-1) = 1 + \frac{18}{20} = 1 + 0.9 = 1.9 = \frac{38}{20}$.
$m = -2 + \frac{9}{20}(2+2) = -2 + \frac{36}{20} = -2 + 1.8 = -0.2 = -\frac{4}{20}$.
Now,calculate $2l + 19m = 2(\frac{38}{20}) + 19(-\frac{4}{20}) = \frac{76}{20} - \frac{76}{20} = 0$.

(B) Two points $(x_1, y_1)$ and $(x_2, y_2)$ are conjugate with respect to the circle $x^2+y^2+2gx+2fy+c=0$ if $x_1x_2 + y_1y_2 + g(x_1+x_2) + f(y_1+y_2) + c = 0$.
Given $P(2,3)$ and $Q(-1,2)$ and the circle $x^2+y^2+2gx+3y-2=0$,we have $f = \frac{3}{2}$ and $c = -2$.
Substituting the values: $(2)(-1) + (3)(2) + g(2-1) + \frac{3}{2}(3+2) - 2 = 0$.
$-2 + 6 + g + \frac{15}{2} - 2 = 0$.
$2 + g + 7.5 = 0 \Rightarrow g = -9.5 = -\frac{19}{2}$.
The radius $r = \sqrt{g^2+f^2-c} = \sqrt{(-\frac{19}{2})^2 + (\frac{3}{2})^2 - (-2)}$.
$r = \sqrt{\frac{361}{4} + \frac{9}{4} + 2} = \sqrt{\frac{370}{4} + \frac{8}{4}} = \sqrt{\frac{378}{4}} = \sqrt{\frac{189}{2}} = \sqrt{\frac{9 \times 21}{2}} = \frac{3\sqrt{21}}{\sqrt{2}}$.

(C) The given circle is $3x^2+3y^2+x+y-1=0$.
Dividing by $3$,we get $x^2+y^2+\frac{1}{3}x+\frac{1}{3}y-\frac{1}{3}=0$.
The center is $(h, k) = \left(-\frac{1}{6}, -\frac{1}{6}\right)$ and the radius $r$ is given by $r^2 = h^2+k^2-c = \frac{1}{36}+\frac{1}{36}+\frac{1}{3} = \frac{1+1+12}{36} = \frac{14}{36} = \frac{7}{18}$.
The length of the tangent $L$ from point $(2,-2)$ to the circle $x^2+y^2+\frac{1}{3}x+\frac{1}{3}y-\frac{1}{3}=0$ is given by $L^2 = x_1^2+y_1^2+\frac{1}{3}x_1+\frac{1}{3}y_1-\frac{1}{3}$.
$L^2 = (2)^2+(-2)^2+\frac{1}{3}(2)+\frac{1}{3}(-2)-\frac{1}{3} = 4+4+\frac{2}{3}-\frac{2}{3}-\frac{1}{3} = 8-\frac{1}{3} = \frac{23}{3}$.
Since the radius of circle $S$ is $L$,the radius squared is $R^2 = \frac{23}{3}$.
The equation of circle $S$ is $(x+\frac{1}{6})^2+(y+\frac{1}{6})^2 = \frac{23}{3}$.
The power of the point $(2,1)$ with respect to circle $S$ is $(x_1+\frac{1}{6})^2+(y_1+\frac{1}{6})^2 - R^2$.
$= (2+\frac{1}{6})^2+(1+\frac{1}{6})^2 - \frac{23}{3} = (\frac{13}{6})^2+(\frac{7}{6})^2 - \frac{23}{3} = \frac{169}{36}+\frac{49}{36} - \frac{276}{36} = \frac{218-276}{36} = -\frac{58}{36} = -\frac{29}{18}$.

(A) Let $P(x, y)$ be the point. Given $A(4, 0)$ and $B(-4, 0)$.
Given the condition $PA - PB = 4$.
Using the distance formula,$\sqrt{(x-4)^2 + y^2} - \sqrt{(x+4)^2 + y^2} = 4$.
$\sqrt{(x-4)^2 + y^2} = 4 + \sqrt{(x+4)^2 + y^2}$.
Squaring both sides: $(x-4)^2 + y^2 = 16 + (x+4)^2 + y^2 + 8\sqrt{(x+4)^2 + y^2}$.
$x^2 - 8x + 16 + y^2 = 16 + x^2 + 8x + 16 + y^2 + 8\sqrt{(x+4)^2 + y^2}$.
$-16x - 16 = 8\sqrt{(x+4)^2 + y^2}$.
$-2x - 2 = \sqrt{(x+4)^2 + y^2}$.
Squaring again: $4x^2 + 8x + 4 = x^2 + 8x + 16 + y^2$.
$3x^2 - y^2 = 12$.

(D) The equation of the circle is $x^2+y^2-4x-6y+k=0$. Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=-2, f=-3$. The centre $C$ is $(-g, -f) = (2, 3)$.
Given $CP \cdot CQ = r^2 = 4$,so $r^2 = 4$.
For inverse points $P$ and $Q$ with respect to a circle with centre $C(h, k)$ and radius $r$,the points $C, P, Q$ are collinear and $CP \cdot CQ = r^2$.
The vector relation is $\vec{CQ} = \frac{r^2}{CP^2} \vec{CP}$.
Here,$P = (1, 2)$ and $C = (2, 3)$.
$CP^2 = (1-2)^2 + (2-3)^2 = (-1)^2 + (-1)^2 = 1 + 1 = 2$.
Since $CP \cdot CQ = 4$,we have $CQ = \frac{4}{CP} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$.
The vector $\vec{CP} = P - C = (1-2, 2-3) = (-1, -1)$.
Then $\vec{CQ} = \frac{r^2}{CP^2} \vec{CP} = \frac{4}{2} (-1, -1) = 2(-1, -1) = (-2, -2)$.
Thus,$Q - C = (-2, -2) \Rightarrow Q = C + (-2, -2) = (2-2, 3-2) = (0, 1)$.
So,$a=0$ and $b=1$.
Therefore,$2a = 2(0) = 0$.

(A) Let the pole be $P(x_1, y_1)$. The polar of $P$ with respect to the circle $x^2+y^2=4$ is $xx_1+yy_1=4$.
Since this polar is a tangent to the circle $x^2+y^2-2x+2y-2=0$,the perpendicular distance from the center $(1, -1)$ of this circle to the line $xx_1+yy_1-4=0$ must be equal to its radius.
The center is $(1, -1)$ and the radius $r = \sqrt{1^2+(-1)^2-(-2)} = \sqrt{4} = 2$.
The perpendicular distance is $d = \frac{|1(x_1) + (-1)(y_1) - 4|}{\sqrt{x_1^2+y_1^2}} = 2$.
Squaring both sides:
$\frac{(x_1-y_1-4)^2}{x_1^2+y_1^2} = 4$
$(x_1-y_1-4)^2 = 4(x_1^2+y_1^2)$
$x_1^2+y_1^2+16-2x_1y_1-8x_1+8y_1 = 4x_1^2+4y_1^2$
$3x_1^2+3y_1^2+2x_1y_1+8x_1-8y_1-16=0$.
Replacing $(x_1, y_1)$ with $(x, y)$,the locus is $3x^2+3y^2+2xy+8x-8y-16=0$.

(A) Given circles are $S_1: x^2+y^2-8x-10y-8=0$ and $S_2: x^2+y^2+2x-2y-2=0$.
For $S_1$,center $C_1 = (4, 5)$ and radius $r_1 = \sqrt{4^2+5^2-(-8)} = \sqrt{16+25+8} = \sqrt{49} = 7$.
For $S_2$,center $C_2 = (-1, 1)$ and radius $r_2 = \sqrt{(-1)^2+1^2-(-2)} = \sqrt{1+1+2} = \sqrt{4} = 2$.
The external centre of similitude $Q$ divides the line segment joining $C_1$ and $C_2$ externally in the ratio $r_1 : r_2$.
$Q = \left( \frac{r_1 x_2 - r_2 x_1}{r_1 - r_2}, \frac{r_1 y_2 - r_2 y_1}{r_1 - r_2} \right) = \left( \frac{7(-1) - 2(4)}{7-2}, \frac{7(1) - 2(5)}{7-2} \right) = \left( \frac{-7-8}{5}, \frac{7-10}{5} \right) = \left( -3, -\frac{3}{5} \right)$.
The distance of $Q$ from the origin $(0,0)$ is $D = \sqrt{(-3)^2 + (-3/5)^2} = \sqrt{9 + 9/25} = \sqrt{\frac{225+9}{25}} = \sqrt{\frac{234}{25}} = \frac{\sqrt{9 \times 26}}{5} = \frac{3\sqrt{26}}{5}$.

(D) The given circle is $x^2+y^2-2x-4y-20=0$. The centre $C$ is $(1, 2)$ and the radius $r$ is $\sqrt{1^2+2^2-(-20)} = \sqrt{25} = 5$.
Since $A$ is an end of a diameter,$A$ lies on the circle. Let $P = (h, k)$.
Given that $A$ divides $CP$ in the ratio $2:3$,where $C=(1, 2)$,by the section formula,the coordinates of $A$ are:
$A = \left(\frac{2h+3(1)}{2+3}, \frac{2k+3(2)}{2+3}\right) = \left(\frac{2h+3}{5}, \frac{2k+6}{5}\right)$.
Since $A$ lies on the circle $x^2+y^2-2x-4y-20=0$,we substitute the coordinates of $A$ into the circle equation:
$\left(\frac{2h+3}{5}\right)^2 + \left(\frac{2k+6}{5}\right)^2 - 2\left(\frac{2h+3}{5}\right) - 4\left(\frac{2k+6}{5}\right) - 20 = 0$.
Multiplying by $25$:
$(2h+3)^2 + (2k+6)^2 - 10(2h+3) - 20(2k+6) - 500 = 0$.
$4h^2 + 12h + 9 + 4k^2 + 24k + 36 - 20h - 30 - 40k - 120 - 500 = 0$.
$4h^2 + 4k^2 - 8h - 16k - 605 = 0$.
Replacing $(h, k)$ with $(x, y)$,the locus of $P$ is $4x^2+4y^2-8x-16y-605=0$.

(C) Given vertex $O = (2,3)$ and focus $S = (3,2)$.
Let the directrix intersect the axis at point $A = (x_1, y_1)$. Since the vertex $O$ is the midpoint of $AS$,we have:
$\frac{x_1+3}{2} = 2 \Rightarrow x_1 = 1$
$\frac{y_1+2}{2} = 3 \Rightarrow y_1 = 4$
So,$A = (1,4)$.
The slope of the axis $AS$ is $m_1 = \frac{2-3}{3-2} = -1$.
The directrix is perpendicular to the axis,so its slope $m_2 = -\frac{1}{m_1} = 1$.
The equation of the directrix passing through $(1,4)$ with slope $1$ is:
$y-4 = 1(x-1) \Rightarrow y-x-3 = 0$.
By the definition of a parabola,for any point $P(x,y)$ on it,the distance to the focus equals the distance to the directrix:
$PS^2 = PM^2$
$(x-3)^2 + (y-2)^2 = \left(\frac{|x-y+3|}{\sqrt{1^2+(-1)^2}}\right)^2$
$(x^2-6x+9) + (y^2-4y+4) = \frac{(x-y+3)^2}{2}$
$2(x^2+y^2-6x-4y+13) = x^2+y^2+9-2xy+6x-6y$
$x^2+2xy+y^2-18x-2y+17 = 0$.

(A) Since the axis of the parabola is parallel to the $X$-axis,the equation of the parabola is of the form $x = ay^2 + by + c$ $(i)$.
Given that the points $(0, -1)$,$(6, 1)$,and $(-2, -3)$ lie on the parabola,we have:
$0 = a - b + c$ $(ii)$
$6 = a + b + c$ $(iii)$
$-2 = 9a - 3b + c$ $(iv)$
Subtracting $(ii)$ from $(iii)$,we get $2b = 6$,so $b = 3$.
Substituting $b = 3$ into $(ii)$ and $(iii)$,we get $a + c = 3$ and $a + c = 3$ (consistent).
Substituting $b = 3$ into $(iv)$,we get $-2 = 9a - 9 + c$,so $9a + c = 7$.
Solving $a + c = 3$ and $9a + c = 7$,we subtract the first from the second to get $8a = 4$,so $a = \frac{1}{2}$.
Then $c = 3 - \frac{1}{2} = \frac{5}{2}$.
The equation is $x = \frac{1}{2}y^2 + 3y + \frac{5}{2}$.
To find the point where it cuts the $X$-axis,set $y = 0$:
$x = \frac{1}{2}(0)^2 + 3(0) + \frac{5}{2} = \frac{5}{2}$.
Thus,the point is $\left(\frac{5}{2}, 0\right)$.

(B) Given equation: $9y^2+12y+9x-14=0$
Rewrite as: $9(y^2 + \frac{4}{3}y) = -9x + 14$
$9(y + \frac{2}{3})^2 = -9x + 14 + 4 = -9x + 18$
$(y + \frac{2}{3})^2 = -(x - 2)$
Comparing with $(y-k')^2 = -4a(x-h')$,we get vertex $(h', k') = (2, -2/3)$ and $4a = 1 \Rightarrow a = 1/4$.
The directrix $l$ is $x = h' + a = 2 + 1/4 = 9/4$.
The line $l_1$ passes through $(2, -2/3)$ and $(0, 0)$,so its slope is $m = \frac{-2/3 - 0}{2 - 0} = -1/3$.
Equation of $l_1$: $y = -\frac{1}{3}x$.
Intersection of $l$ $(x = 9/4)$ and $l_1$ $(y = -x/3)$:
$h = 9/4$,$k = -\frac{1}{3}(9/4) = -3/4$.
Thus,$h+k = 9/4 - 3/4 = 6/4 = 3/2$.

(D) The curves are $y=4|\cos x|$ and $y=-|\cos x|$.
Since $|\cos x| \ge 0$ for all $x$,the upper curve is $y=4|\cos x|$ and the lower curve is $y=-|\cos x|$.
The area $A$ is given by the integral $\int_{-\pi/2}^{\pi/2} [4|\cos x| - (-|\cos x|)] dx$.
$A = \int_{-\pi/2}^{\pi/2} 5|\cos x| dx$.
Since $\cos x \ge 0$ for $x \in [-\pi/2, \pi/2]$,we have $|\cos x| = \cos x$.
$A = 5 \int_{-\pi/2}^{\pi/2} \cos x dx$.
Using the property of even functions,$A = 5 \times 2 \int_{0}^{\pi/2} \cos x dx$.
$A = 10 [\sin x]_{0}^{\pi/2} = 10(1 - 0) = 10$ sq. units.

(D) The curves are $y = \cos x + 1$ and $y = \sin x$. We need to find the area bounded by these curves and the $X$-axis between $x=0$ and $x=\pi$.
From the graph,the intersection point $A$ occurs where $\cos x + 1 = \sin x$. However,looking at the region bounded by the $X$-axis,the area is split at $x = \pi/2$.
For $0 \le x \le \pi/2$,the region is bounded above by $y = \sin x$ and below by the $X$-axis.
For $\pi/2 \le x \le \pi$,the region is bounded above by $y = \cos x + 1$ and below by the $X$-axis.
Area $= \int_0^{\pi/2} \sin x \, dx + \int_{\pi/2}^{\pi} (\cos x + 1) \, dx$
$= [-\cos x]_0^{\pi/2} + [\sin x + x]_{\pi/2}^{\pi}$
$= (-\cos(\pi/2) - (-\cos 0)) + ((\sin \pi + \pi) - (\sin(\pi/2) + \pi/2))$
$= (0 + 1) + (0 + \pi - 1 - \pi/2)$
$= 1 + \pi - 1 - \pi/2 = \pi/2$.

(D) Given the curve is $x = \log |y|$.
This can be rewritten as $|y| = e^x$,which implies $y = \pm e^x$.
The curve is symmetric about the $x$-axis.
The area bounded by the curve,the lines $x = -1$ and $x = 0$ is given by the integral of $y$ with respect to $x$ from $-1$ to $0$.
Since the curve is symmetric,the total area is twice the area above the $x$-axis:
$\text{Area} = 2 \int_{-1}^{0} |y| \, dx = 2 \int_{-1}^{0} e^x \, dx$
Evaluating the integral:
$\text{Area} = 2 \left[ e^x \right]_{-1}^{0}$
$= 2 (e^0 - e^{-1})$
$= 2 (1 - e^{-1})$
Thus,the required area is $2(1 - e^{-1})$ square units.

(C) The given curves are $y=x^2$ and $y=6-|x|$.
Due to symmetry about the $y$-axis,the required area is $2 \times$ the area in the first quadrant.
In the first quadrant $(x \ge 0)$,the curves are $y=x^2$ and $y=6-x$.
To find the intersection point $A$,we set $x^2 = 6-x$,which gives $x^2+x-6=0$.
Solving this,$(x+3)(x-2)=0$. Since $x \ge 0$,we have $x=2$.
At $x=2$,$y=2^2=4$. So,the point of intersection is $A(2, 4)$.
The area in the first quadrant is bounded by $x=0$ to $x=2$ between the curves $y=6-x$ and $y=x^2$.
Area $= \int_0^2 ((6-x) - x^2) dx = [6x - \frac{x^2}{2} - \frac{x^3}{3}]_0^2$
$= (12 - 2 - \frac{8}{3}) - 0 = 10 - \frac{8}{3} = \frac{30-8}{3} = \frac{22}{3}$.
The total area is $2 \times \frac{22}{3} = \frac{44}{3}$.

(A) First,we find the intersection point of the curves $y = \frac{8}{x}$ and $y = 2x$.
Setting $\frac{8}{x} = 2x$,we get $x^2 = 4$,so $x = 2$ (since $x > 0$ in the first quadrant).
The region is bounded by $x = 2$ to $x = 4$.
In this interval,$2x \geq \frac{8}{x}$.
The required area is given by the integral:
$\text{Area} = \int_2^4 \left( 2x - \frac{8}{x} \right) dx$
$= \left[ x^2 - 8 \log |x| \right]_2^4$
$= (4^2 - 8 \log 4) - (2^2 - 8 \log 2)$
$= (16 - 8 \log 4) - (4 - 8 \log 2)$
$= 12 - 8 \log 4 + 8 \log 2$
$= 12 - 8 \log (2^2) + 8 \log 2$
$= 12 - 16 \log 2 + 8 \log 2$
$= 12 - 8 \log 2$

(C) The given equations are $y = 1 - |x|$ and $y = |x| - 1$.
These equations represent a square with vertices at $A(0, 1)$,$C(1, 0)$,$B(0, -1)$,and $D(-1, 0)$.
The area of the region bounded by these curves is the area of the square $ACBD$.
The area of a square with vertices $(0, 1)$,$(1, 0)$,$(0, -1)$,and $(-1, 0)$ can be calculated by dividing it into four congruent right-angled triangles,each with base $1$ and height $1$.
Area of one triangle (e.g.,$\triangle AOC$) $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 1 \times 1 = \frac{1}{2}$.
Total area $= 4 \times \text{Area of } \triangle AOC = 4 \times \frac{1}{2} = 2$.
Thus,the area of the required bounded region is $2$ square units.

(B) The given differential equation is $\left(\frac{d^3 y}{d x^3}\right)^{\frac{1}{2}} = 2\left(\frac{d y}{d x}\right)^{\frac{1}{4}} - x y$.
To find the degree,we must eliminate the fractional powers of the derivatives.
The powers are $\frac{1}{2}$ and $\frac{1}{4}$. The least common multiple of the denominators $2$ and $4$ is $4$.
Raising both sides to the power of $4$,we get:
$\left(\left(\frac{d^3 y}{d x^3}\right)^{\frac{1}{2}}\right)^4 = \left(2\left(\frac{d y}{d x}\right)^{\frac{1}{4}} - x y\right)^4$
$\left(\frac{d^3 y}{d x^3}\right)^2 = \left(2\left(\frac{d y}{d x}\right)^{\frac{1}{4}} - x y\right)^4$.
Expanding the right side using the binomial theorem,the highest derivative present is $\frac{d^3 y}{d x^3}$,so the order is $3$.
The highest power of the highest order derivative $\frac{d^3 y}{d x^3}$ after clearing the radicals is $2$.
Therefore,the order is $3$ and the degree is $2$.

(C) Given the differential equation: $y = e^{\left(\frac{dy}{dx} + \frac{d^2y}{dx^2}\right)}$
Taking the natural logarithm on both sides: $\ln(y) = \frac{dy}{dx} + \frac{d^2y}{dx^2}$
The highest order derivative present is $\frac{d^2y}{dx^2}$,so the order $\alpha = 2$.
The power of the highest order derivative is $1$,so the degree $\beta = 1$.
We need to find the sum $S = \alpha + \alpha^\beta + \alpha^{2\beta} + \ldots + \alpha^{2023\beta}$.
Substituting $\alpha = 2$ and $\beta = 1$: $S = 2 + 2^1 + 2^2 + 2^3 + \ldots + 2^{2023}$.
This is a geometric progression with $2024$ terms,where the first term $a = 2$,common ratio $r = 2$,and number of terms $n = 2024$.
The sum is $S = 2 + (2^1 + 2^2 + \ldots + 2^{2023}) = 2 + \frac{2(2^{2023} - 1)}{2 - 1} = 2 + 2^{2024} - 2 = 2^{2024}$.

(C) Given the equation: $y=c_1 e^x+c_2 e^{\log _{e} x}+c_3 \sin ^2 x-c_4\left(\cos ^2 x-1\right)$
Using the property $e^{\log _{e} x} = x$ and $\cos^2 x - 1 = -\sin^2 x$,we simplify:
$y = c_1 e^x + c_2 x + c_3 \sin^2 x - c_4(-\sin^2 x)$
$y = c_1 e^x + c_2 x + (c_3 + c_4) \sin^2 x$
Let $C = c_3 + c_4$. Then the equation becomes:
$y = c_1 e^x + c_2 x + C \sin^2 x$
This equation contains $3$ independent arbitrary constants $(c_1, c_2, C)$.
The order of a differential equation is equal to the number of independent arbitrary constants present in its general solution.
Since there are $3$ independent constants,the order of the differential equation is $3$.

(B) The given differential equation is $3 x^2 \frac{d^2 y}{d x^2}-\sin \left(\frac{d^3 y}{d x^3}\right)+\cos (x y)=0$.
The order of a differential equation is the highest order derivative present in the equation. Here,the highest order derivative is $\frac{d^3 y}{d x^3}$,so the order is $3$.
The degree of a differential equation is the power of the highest order derivative when the equation is expressed as a polynomial in derivatives.
Since the term $\sin \left(\frac{d^3 y}{d x^3}\right)$ involves a transcendental function of a derivative,it cannot be expressed as a polynomial in derivatives.
Therefore,the degree of this differential equation is not defined.

(D) The given differential equation is $\log \left(\frac{dy}{dx}\right) = (2x + 3\frac{dy}{dx})^2$.
To define the degree of a differential equation,it must be a polynomial equation in terms of its derivatives.
Here,the term $\log \left(\frac{dy}{dx}\right)$ is a transcendental function of the derivative $\frac{dy}{dx}$.
Since the equation cannot be expressed as a polynomial in $\frac{dy}{dx}$,the degree of this differential equation is not defined.

(B) The given solution is $y = a e^{2x} + b x e^{2x} + C$ ...$(i)$
Here,$a$ and $b$ are arbitrary constants,while $C$ is a fixed constant.
The order of a differential equation is equal to the number of arbitrary constants present in its general solution.
Since there are two arbitrary constants ($a$ and $b$),the order of the differential equation is $2$.
To verify,we differentiate $y$ with respect to $x$:
$y_1 = 2a e^{2x} + b(e^{2x} + 2x e^{2x}) = (2a + b)e^{2x} + 2bx e^{2x}$ ...(ii)
$y_2 = 2(2a + b)e^{2x} + 2b(e^{2x} + 2x e^{2x}) = (4a + 4b)e^{2x} + 4bx e^{2x}$ ...(iii)
By eliminating $a$ and $b$ from these equations,we obtain a second-order differential equation.

(C) Given $y=x \log \left(\frac{1}{a x}+\frac{1}{a}\right) = x \log \left(\frac{1+x}{ax}\right)$.
First,find the first derivative $\frac{dy}{dx}$:
$\frac{dy}{dx} = \log \left(\frac{1+x}{ax}\right) + x \cdot \frac{ax}{1+x} \cdot \frac{d}{dx} \left(\frac{1+x}{ax}\right)$
$= \log \left(\frac{1+x}{ax}\right) + \frac{ax^2}{1+x} \cdot \left(\frac{ax - a(1+x)}{(ax)^2}\right)$
$= \log \left(\frac{1+x}{ax}\right) + \frac{ax^2}{1+x} \cdot \left(\frac{-a}{a^2x^2}\right) = \log \left(\frac{1+x}{ax}\right) - \frac{1}{1+x}$.
Now,find the second derivative $\frac{d^2y}{dx^2}$:
$\frac{d^2y}{dx^2} = \frac{d}{dx} \left(\log \left(\frac{1+x}{ax}\right) - \frac{1}{1+x}\right)$
$= \frac{ax}{1+x} \cdot \left(\frac{-1}{ax^2}\right) + \frac{1}{(1+x)^2} = -\frac{1}{x(1+x)} + \frac{1}{(1+x)^2} = \frac{-(1+x) + x}{x(1+x)^2} = \frac{-1}{x(1+x)^2}$.
Substitute these into the expression $x(x+1) \frac{d^2 y}{d x^2}+x \frac{d y}{d x}-y$:
$= x(x+1) \left(\frac{-1}{x(1+x)^2}\right) + x \left(\log \left(\frac{1+x}{ax}\right) - \frac{1}{1+x}\right) - x \log \left(\frac{1+x}{ax}\right)$
$= -\frac{1}{1+x} + x \log \left(\frac{1+x}{ax}\right) - \frac{x}{1+x} - x \log \left(\frac{1+x}{ax}\right)$
$= -\left(\frac{1+x}{1+x}\right) = -1$.

(A) Given the equation: $y = a e^x + b e^{-x} + c \cos x + d \sin x$
Taking the first derivative: $y' = a e^x - b e^{-x} - c \sin x + d \cos x$
Taking the second derivative: $y'' = a e^x + b e^{-x} - c \cos x - d \sin x$
Taking the third derivative: $y''' = a e^x - b e^{-x} + c \sin x - d \cos x$
Taking the fourth derivative: $y^{(4)} = a e^x + b e^{-x} + c \cos x + d \sin x$
Comparing the fourth derivative with the original equation,we get $y^{(4)} = y$,which can be written as $y^{(4)} - y = 0$.

(D) Given the general solution $y=(a+b) e^{cx+d}$.
Let $A = (a+b)e^d$. Then the equation simplifies to $y = A e^{cx}$.
Here,$A$ and $c$ are the only independent arbitrary constants.
Differentiating with respect to $x$:
$y^{(1)} = A c e^{cx} = c y$.
Differentiating again with respect to $x$:
$y^{(2)} = c y^{(1)}$.
From the first derivative,we have $c = \frac{y^{(1)}}{y}$.
Substituting this into the second derivative equation:
$y^{(2)} = \left(\frac{y^{(1)}}{y}\right) y^{(1)}$.
$y y^{(2)} = (y^{(1)})^2$.
$y y^{(2)} - (y^{(1)})^2 = 0$.

(B) The given solution is $y=e^{2 x}(c \cosh \sqrt{2} x+d \sinh \sqrt{2} x)$.
This is of the form $y=e^{\alpha x}(c \cosh \beta x+d \sinh \beta x)$,which corresponds to the auxiliary equation roots $m = \alpha \pm \beta$.
Here,$\alpha = 2$ and $\beta = \sqrt{2}$.
The roots are $m = 2 \pm \sqrt{2}$.
The characteristic equation is $(m - (2 + \sqrt{2}))(m - (2 - \sqrt{2})) = 0$.
$(m - 2 - \sqrt{2})(m - 2 + \sqrt{2}) = 0$.
$(m - 2)^2 - (\sqrt{2})^2 = 0$.
$m^2 - 4m + 4 - 2 = 0$.
$m^2 - 4m + 2 = 0$.
Replacing $m^2$ with $y^{\prime \prime}$ and $m$ with $y^{\prime}$,we get the differential equation $y^{\prime \prime} - 4y^{\prime} + 2y = 0$.

(C) differential equation of the form $\frac{dy}{dx} = f(x, y)$ is said to be a homogeneous differential equation if the function $f(x, y)$ is a homogeneous function of degree zero.
By definition,a function $f(x, y)$ is homogeneous of degree zero if $f(\lambda x, \lambda y) = \lambda^0 f(x, y) = f(x, y)$.
Such a function can always be expressed in the form $f(x, y) = \phi\left(\frac{y}{x}\right)$ or $f(x, y) = \psi\left(\frac{x}{y}\right)$.
Therefore,the general form of $f(x, y)$ for a homogeneous differential equation $\frac{dy}{dx} = f(x, y)$ is $\phi\left(\frac{y}{x}\right)$.

(C) Given the differential equation: $x^3 \sin y \frac{d y}{d x}=2$.
Separating the variables,we get: $\sin y \, dy = \frac{2}{x^3} \, dx$.
Integrating both sides: $\int \sin y \, dy = \int 2x^{-3} \, dx$.
This yields: $-\cos y = 2 \cdot \frac{x^{-2}}{-2} + C = -\frac{1}{x^2} + C$.
Multiplying by $-1$,we get: $\cos y = \frac{1}{x^2} - C$ ... $(i)$.
Given the condition $\lim _{x \rightarrow \infty} y(x) = \frac{\pi}{2}$,we take the limit as $x \rightarrow \infty$ on both sides of equation $(i)$:
$\lim _{x \rightarrow \infty} \cos y = \lim _{x \rightarrow \infty} \left( \frac{1}{x^2} - C \right)$.
Since $\cos$ is a continuous function,$\cos(\lim _{x \rightarrow \infty} y) = 0 - C$.
$\cos(\frac{\pi}{2}) = -C \Rightarrow 0 = -C \Rightarrow C = 0$.
Substituting $C = 0$ into equation $(i)$,we get: $\cos y = \frac{1}{x^2}$.

(A) The slope of the tangent to the curve at any point $(x, y)$ is given by $\frac{dy}{dx} = 1 - \frac{1}{x^2}$.
Integrating both sides with respect to $x$:
$\int dy = \int \left( 1 - \frac{1}{x^2} \right) dx$
$y = x - (-\frac{1}{x}) + C$
$y = x + \frac{1}{x} + C$
Since the curve passes through the point $(1, 2)$,we substitute $x = 1$ and $y = 2$ to find the constant $C$:
$2 = 1 + \frac{1}{1} + C$
$2 = 1 + 1 + C$
$2 = 2 + C$
$C = 0$
Therefore,the equation of the curve is $y = x + \frac{1}{x}$.

(B) Given $f^{\prime}(x)=a \cos x+b \sin x$ ... $(i)$
Integrating both sides with respect to $x$:
$f(x) = \int (a \cos x + b \sin x) dx = a \sin x - b \cos x + C$ ... $(ii)$
Given $f^{\prime}(0) = 4$:
$f^{\prime}(0) = a \cos(0) + b \sin(0) = a(1) + b(0) = a = 4$.
Given $f(0) = 3$:
$f(0) = a \sin(0) - b \cos(0) + C = 0 - b(1) + C = -b + C = 3$.
Given $f\left(\frac{\pi}{2}\right) = 5$:
$f\left(\frac{\pi}{2}\right) = a \sin\left(\frac{\pi}{2}\right) - b \cos\left(\frac{\pi}{2}\right) + C = a(1) - b(0) + C = a + C = 5$.
Since $a = 4$,we have $4 + C = 5 \Rightarrow C = 1$.
Substituting $C = 1$ into $-b + C = 3$:
$-b + 1 = 3 \Rightarrow -b = 2 \Rightarrow b = -2$.
Substituting $a = 4, b = -2, C = 1$ into equation $(ii)$:
$f(x) = 4 \sin x - (-2) \cos x + 1 = 4 \sin x + 2 \cos x + 1$.

(D) Given differential equation: $(1+y^2) dx - xy dy = 0$
Rearranging the terms to separate the variables: $\frac{dx}{x} = \frac{y}{1+y^2} dy$
Integrating both sides: $\int \frac{dx}{x} = \int \frac{y}{1+y^2} dy$
$\ln |x| = \frac{1}{2} \ln(1+y^2) + C$
Multiplying by $2$: $2 \ln |x| = \ln(1+y^2) + 2C$
$\ln(x^2) - \ln(1+y^2) = C'$ (where $C' = 2C$)
$\ln \left( \frac{x^2}{1+y^2} \right) = C'$
$\frac{x^2}{1+y^2} = e^{C'} = k$
Given $y(1) = 0$,substitute $x=1$ and $y=0$: $\frac{1^2}{1+0^2} = k \Rightarrow k = 1$
So,$\frac{x^2}{1+y^2} = 1 \Rightarrow x^2 = 1+y^2 \Rightarrow x^2 - y^2 = 1$
This equation represents a hyperbola.

(A) Given the differential equation: $\frac{dy}{dx} = \cos^2(x-y-1)$ $(i)$
Let $x-y-1 = p$.
Differentiating with respect to $x$,we get $1 - \frac{dy}{dx} = \frac{dp}{dx}$,which implies $\frac{dy}{dx} = 1 - \frac{dp}{dx}$.
Substituting this into equation $(i)$:
$1 - \frac{dp}{dx} = \cos^2 p$
$\frac{dp}{dx} = 1 - \cos^2 p = \sin^2 p$
$\frac{dp}{\sin^2 p} = dx$
$\operatorname{cosec}^2 p \, dp = dx$
Integrating both sides:
$\int \operatorname{cosec}^2 p \, dp = \int dx$
$-\cot p = x + C'$
$x = -C' - \cot p$
Let $C = -C'$,then $x = C - \cot(x-y-1)$.

(A) Given the differential equation: $\frac{dx}{dy} = \frac{\sin y(1 + y \cot y)}{x \log(x^2 e)}$.
Rearranging the terms,we get: $x \log(x^2 e) dx = (\sin y + y \cos y) dy$.
Integrating both sides: $\int x \log(x^2 e) dx = \int (\sin y + y \cos y) dy$.
For the left side,let $t = x^2 e$,then $dt = 2x e dx$,so $x dx = \frac{dt}{2e}$.
$\int \log(t) \frac{dt}{2e} = \frac{1}{2e} (t \log t - t) = \frac{x^2 e}{2e} (\log(x^2 e) - 1) = \frac{x^2}{2} (\log x^2 + \log e - 1) = \frac{x^2}{2} (2 \log x + 1 - 1) = x^2 \log x$.
For the right side,using integration by parts: $\int (\sin y + y \cos y) dy = y \sin y - \int \sin y dy + \int \sin y dy = y \sin y + C$.
Thus,$x^2 \log x = y \sin y + C$.
Given $y(1) = 0$,substitute $x = 1$ and $y = 0$: $1^2 \log(1) = 0 \cdot \sin(0) + C \implies 0 = 0 + C \implies C = 0$.
Therefore,the particular solution is $x^2 \log x = y \sin y$.

(D) The substitution $x=vy$ is used to solve homogeneous differential equations of the form $\frac{dx}{dy} = f(\frac{x}{y})$ or $\frac{dy}{dx} = g(\frac{x}{y})$.
For the substitution $x=vy$ (which implies $\frac{x}{y}=v$),the differential equation must be homogeneous in $x$ and $y$ such that $\frac{dy}{dx}$ can be expressed as a function of $\frac{x}{y}$.
Checking option $(d)$:
$(1+2e^{\frac{x}{y}}) + 2e^{\frac{x}{y}}(1-\frac{x}{y})\frac{dy}{dx} = 0$
$\frac{dy}{dx} = -\frac{1+2e^{\frac{x}{y}}}{2e^{\frac{x}{y}}(1-\frac{x}{y})}$
Since the right-hand side is a function of $\frac{x}{y}$,it is a homogeneous differential equation of degree zero,which becomes separable after substituting $x=vy$.

(A) The given differential equation is $x^\alpha \frac{dy}{dx} = y^\beta(\gamma \log x + \delta \log y + 1)$.
For the equation to be homogeneous,the degree of the numerator and denominator must be the same,or the function must be expressible as a function of $\frac{y}{x}$.
Rewriting the equation: $\frac{dy}{dx} = \frac{y^\beta}{x^\alpha} (\gamma \log x + \delta \log y + 1)$.
For homogeneity,the powers of $x$ and $y$ must balance such that the expression depends only on the ratio $\frac{y}{x}$.
This requires $\alpha = \beta$.
Substituting $\alpha = \beta$,we get $\frac{dy}{dx} = (\frac{y}{x})^\alpha (\gamma \log x + \delta \log y + 1) = (\frac{y}{x})^\alpha (\gamma \log x + \delta \log y + \delta \log x - \delta \log x + 1)$.
For the term to be a function of $\frac{y}{x}$,the $\log x$ terms must cancel out,which happens when $\gamma + \delta = 0$,i.e.,$\gamma = -\delta$.

(C) Given differential equation: $\tan x \tan y \, dx + \cos^2 x \operatorname{cosec}^2 y \, dy = 0$
Divide the entire equation by $\cos^2 x \tan y$:
$\frac{\tan x}{\cos^2 x} \, dx + \frac{\operatorname{cosec}^2 y}{\tan y} \, dy = 0$
$\tan x \sec^2 x \, dx + \cot y \operatorname{cosec}^2 y \, dy = 0$
Integrating both sides:
$\int \tan x \sec^2 x \, dx + \int \cot y \operatorname{cosec}^2 y \, dy = C_1$
Let $u = \tan x$,then $du = \sec^2 x \, dx$. Let $v = \cot y$,then $dv = -\operatorname{cosec}^2 y \, dy$.
$\int u \, du - \int v \, dv = C_1$
$\frac{u^2}{2} - \frac{v^2}{2} = C_1$
$\tan^2 x - \cot^2 y = 2C_1 = C$
Thus,the general solution is $\tan^2 x - \cot^2 y = C$.

(C) differential equation of the form $\frac{dy}{dx} = F(x, y)$ is homogeneous if $F(x, y)$ is a homogeneous function of degree $0$. Equivalently,for an equation $M(x, y)dx + N(x, y)dy = 0$,it is homogeneous if both $M(x, y)$ and $N(x, y)$ are homogeneous functions of the same degree.
In option $C$,we have $(x^2 + y^2)dx = 2xy dy$.
Here,$M(x, y) = x^2 + y^2$,which is a homogeneous function of degree $2$.
And $N(x, y) = 2xy$,which is also a homogeneous function of degree $2$.
Since both $M$ and $N$ are homogeneous functions of the same degree,the differential equation is homogeneous.

(B) Given the differential equation: $\frac{dy}{dx} = \frac{2x + 3y - 7}{3x + 2y - 8}$ ... $(i)$
To transform this into a homogeneous differential equation,we substitute $x = X - h$ and $y = Y - k$.
Substituting these into $(i)$,we get: $\frac{dY}{dX} = \frac{2(X - h) + 3(Y - k) - 7}{3(X - h) + 2(Y - k) - 8} = \frac{2X + 3Y - (2h + 3k + 7)}{3X + 2Y - (3h + 2k + 8)}$.
For the equation to be homogeneous,the constant terms in the numerator and denominator must be zero:
$2h + 3k + 7 = 0$ ... (ii)
$3h + 2k + 8 = 0$ ... (iii)
Multiplying (ii) by $3$ and (iii) by $2$:
$6h + 9k + 21 = 0$
$6h + 4k + 16 = 0$
Subtracting the equations: $(9k - 4k) + (21 - 16) = 0 \Rightarrow 5k + 5 = 0 \Rightarrow k = -1$.
Wait,re-evaluating the signs: The original equation is $\frac{2x+3y-7}{3x+2y-8}$.
Setting $2h+3k-7=0$ and $3h+2k-8=0$:
$4h+6k-14=0$
$9h+6k-24=0$
Subtracting: $5h - 10 = 0 \Rightarrow h = 2$.
Substituting $h=2$ into $2(2)+3k-7=0 \Rightarrow 4+3k-7=0 \Rightarrow 3k=3 \Rightarrow k=1$.
Thus,$(h, k) = (2, 1)$.

(A) Given the differential equation $\frac{dy}{dx}=\frac{x-y \cos x}{1+\sin x}$.
Rearranging the terms,we get $\frac{dy}{dx} + \left(\frac{\cos x}{1+\sin x}\right)y = \frac{x}{1+\sin x}$.
This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \frac{\cos x}{1+\sin x}$ and $Q(x) = \frac{x}{1+\sin x}$.
The integrating factor $I.F. = e^{\int P(x) dx} = e^{\int \frac{\cos x}{1+\sin x} dx} = e^{\ln(1+\sin x)} = 1+\sin x$.
The general solution is $y(I.F.) = \int Q(x)(I.F.) dx + C$.
$y(1+\sin x) = \int \frac{x}{1+\sin x} (1+\sin x) dx + C = \int x dx + C = \frac{x^2}{2} + C$.
Using the condition $y\left(\frac{\pi}{2}\right) = \frac{\pi^2}{8}$:
$\frac{\pi^2}{8}(1+\sin(\frac{\pi}{2})) = \frac{(\pi/2)^2}{2} + C \Rightarrow \frac{\pi^2}{8}(2) = \frac{\pi^2}{8} + C \Rightarrow \frac{\pi^2}{4} - \frac{\pi^2}{8} = C \Rightarrow C = \frac{\pi^2}{8}$.
Thus,$y(1+\sin x) = \frac{x^2}{2} + \frac{\pi^2}{8}$.
For $x = \pi$,$y(1+\sin \pi) = \frac{\pi^2}{2} + \frac{\pi^2}{8}$.
Since $\sin \pi = 0$,$y(1) = \frac{4\pi^2 + \pi^2}{8} = \frac{5\pi^2}{8}$.

(C) Given the differential equation $\frac{dy}{dx} = \frac{1}{4x + 3y}$.
Taking the reciprocal,we get $\frac{dx}{dy} = 4x + 3y$.
Rearranging the terms to the standard form of a linear differential equation $\frac{dx}{dy} + P(y)x = Q(y)$,we have $\frac{dx}{dy} - 4x = 3y$.
Here,$P(y) = -4$.
The integrating factor $(IF)$ is given by $e^{\int P(y) dy} = e^{\int -4 dy} = e^{-4y}$.

(B) Given differential equation: $x \frac{d y}{d x} = y + x e^{-\left(\frac{y}{x}\right)}$.
Dividing by $x$,we get $\frac{d y}{d x} = \frac{y}{x} + e^{-\left(\frac{y}{x}\right)}$.
This is a homogeneous differential equation. Let $v = \frac{y}{x}$,so $y = vx$ and $\frac{d y}{d x} = v + x \frac{d v}{d x}$.
Substituting these into the equation: $v + x \frac{d v}{d x} = v + e^{-v}$.
$x \frac{d v}{d x} = e^{-v} \Rightarrow e^v d v = \frac{d x}{x}$.
Integrating both sides: $\int e^v d v = \int \frac{d x}{x} \Rightarrow e^v = \log |x| + C$.
Substituting $v = \frac{y}{x}$: $e^{\frac{y}{x}} = \log |x| + C$.
Given $y(1) = \log e = 1$,so at $x=1, y=1$: $e^{\frac{1}{1}} = \log(1) + C \Rightarrow e = 0 + C \Rightarrow C = e$.
Thus,the solution is $e^{\frac{y}{x}} = \log x + e$.
To find $y(e)$,put $x=e$: $e^{\frac{y(e)}{e}} = \log e + e = 1 + e$.
Taking $\log$ on both sides: $\frac{y(e)}{e} = \log(1+e) \Rightarrow y(e) = e \log(1+e)$.

(D) The standard form of a linear differential equation is $\frac{d y}{d x} + P(x)y = Q(x)$ or $\frac{d x}{d y} + P(y)x = Q(y)$,where $P$ and $Q$ are functions of the independent variable only.
Let us analyze option $(D)$:
$x^2 d y + x y d x - 1 = 0$
Dividing by $d x$:
$x^2 \frac{d y}{d x} + x y - 1 = 0$
$x^2 \frac{d y}{d x} + x y = 1$
Dividing by $x^2$:
$\frac{d y}{d x} + \frac{1}{x} y = \frac{1}{x^2}$
This is in the form $\frac{d y}{d x} + P(x)y = Q(x)$,where $P(x) = \frac{1}{x}$ and $Q(x) = \frac{1}{x^2}$.
Thus,option $(D)$ is a linear differential equation.

(D) The integrating factor $(IF)$ for the linear differential equation $\frac{dx}{dy} + P(y)x = Q(y)$ is given by $IF = e^{\int P(y) dy}$.
Given that $IF = \log y$,we have:
$e^{\int P(y) dy} = \log y$
Taking the natural logarithm on both sides:
$\int P(y) dy = \log(\log y)$
Differentiating both sides with respect to $y$:
$\frac{d}{dy} \left( \int P(y) dy \right) = \frac{d}{dy} (\log(\log y))$
$P(y) = \frac{1}{\log y} \cdot \frac{d}{dy}(\log y)$
$P(y) = \frac{1}{\log y} \cdot \frac{1}{y}$
$P(y) = \frac{1}{y \log y}$

(A) Given the linear differential equation: $\sqrt{1-x^2} \frac{dy}{dx} + \frac{2x}{\sqrt{1-x^2}} y = x$.
Dividing by $\sqrt{1-x^2}$,we get: $\frac{dy}{dx} + \frac{2x}{1-x^2} y = \frac{x}{\sqrt{1-x^2}}$.
This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \frac{2x}{1-x^2}$ and $Q(x) = \frac{x}{\sqrt{1-x^2}}$.
The Integrating Factor ($I$.$F$.) is $e^{\int P(x) dx} = e^{\int \frac{2x}{1-x^2} dx} = e^{-\ln(1-x^2)} = \frac{1}{1-x^2}$.
The general solution is $y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C$.
$y \cdot \frac{1}{1-x^2} = \int \frac{x}{\sqrt{1-x^2}} \cdot \frac{1}{1-x^2} dx + C = \int x(1-x^2)^{-3/2} dx + C$.
Let $u = 1-x^2$,then $du = -2x dx$,so $x dx = -\frac{1}{2} du$.
$y \cdot \frac{1}{1-x^2} = -\frac{1}{2} \int u^{-3/2} du + C = -\frac{1}{2} \frac{u^{-1/2}}{-1/2} + C = u^{-1/2} + C = (1-x^2)^{-1/2} + C$.
Thus,$y = (1-x^2)^{1/2} + C(1-x^2)$.
Given $y(0) = 1$,we have $1 = (1-0)^{1/2} + C(1-0) \Rightarrow 1 = 1 + C \Rightarrow C = 0$.
So,$y = \sqrt{1-x^2}$.
At $x = \frac{1}{2}$,$y\left(\frac{1}{2}\right) = \sqrt{1 - (\frac{1}{2})^2} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.

(D) Given differential equation is $y^2 dx + (2xy - 1) dy = 0$.
Dividing by $y^2 dy$,we get:
$\frac{dx}{dy} + \frac{2xy - 1}{y^2} = 0$
$\frac{dx}{dy} + \frac{2x}{y} - \frac{1}{y^2} = 0$
$\frac{dx}{dy} + \left(\frac{2}{y}\right)x = \frac{1}{y^2}$
This is of the form $\frac{dx}{dy} + Px = Q$,where $P = \frac{2}{y}$ and $Q = \frac{1}{y^2}$ are functions of $y$ only.
Therefore,the given differential equation is linear in $x$.

(A) The position vector of the centroid $G$ is given by $\overrightarrow{OG} = \frac{\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}}{3}$.
Substituting the given vectors:
$\overrightarrow{OG} = \frac{(2\hat{i}-3\hat{j}+\hat{k}) + (\hat{i}-4\hat{j}-3\hat{k}) + (-3\hat{i}+\hat{j}+2\hat{k})}{3} = \frac{0\hat{i}-6\hat{j}+0\hat{k}}{3} = -2\hat{j}$.
Thus,$|\overrightarrow{OG}|^2 = (-2)^2 = 4$.
Now,calculate the vectors for the sides:
$\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = (-3-1)\hat{i} + (1-(-4))\hat{j} + (2-(-3))\hat{k} = -4\hat{i} + 5\hat{j} + 5\hat{k}$.
$BC^2 = |\overrightarrow{BC}|^2 = (-4)^2 + 5^2 + 5^2 = 16 + 25 + 25 = 66$.
$\overrightarrow{CA} = \overrightarrow{OA} - \overrightarrow{OC} = (2-(-3))\hat{i} + (-3-1)\hat{j} + (1-2)\hat{k} = 5\hat{i} - 4\hat{j} - \hat{k}$.
$CA^2 = |\overrightarrow{CA}|^2 = 5^2 + (-4)^2 + (-1)^2 = 25 + 16 + 1 = 42$.
$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (1-2)\hat{i} + (-4-(-3))\hat{j} + (-3-1)\hat{k} = -\hat{i} - \hat{j} - 4\hat{k}$.
$AB^2 = |\overrightarrow{AB}|^2 = (-1)^2 + (-1)^2 + (-4)^2 = 1 + 1 + 16 = 18$.
Finally,calculate the expression:
$BC^2 + CA^2 + AB^2 + 9(OG)^2 = 66 + 42 + 18 + 9(4) = 126 + 36 = 162$.

(C) Let $\vec{A} = 7 \hat{i}-4 \hat{j}+5 \hat{k}$ be the position vector of vertex $A$.
Let $\vec{G}_{BCD} = \frac{\vec{B}+\vec{C}+\vec{D}}{3} = -\hat{i}+4 \hat{j}-3 \hat{k}$ be the position vector of the centroid of triangle $BCD$.
The centroid $\vec{G}$ of the tetrahedron $ABCD$ is given by the formula:
$\vec{G} = \frac{\vec{A}+\vec{B}+\vec{C}+\vec{D}}{4}$
We can rewrite this as:
$\vec{G} = \frac{1}{4} [\vec{A} + 3(\frac{\vec{B}+\vec{C}+\vec{D}}{3})] = \frac{1}{4} [\vec{A} + 3 \vec{G}_{BCD}]$
Substituting the given values:
$\vec{G} = \frac{1}{4} [(7 \hat{i}-4 \hat{j}+5 \hat{k}) + 3(-\hat{i}+4 \hat{j}-3 \hat{k})]$
$\vec{G} = \frac{1}{4} [7 \hat{i}-4 \hat{j}+5 \hat{k} - 3 \hat{i} + 12 \hat{j} - 9 \hat{k}]$
$\vec{G} = \frac{1}{4} [4 \hat{i} + 8 \hat{j} - 4 \hat{k}]$
$\vec{G} = \hat{i} + 2 \hat{j} - \hat{k}$

(D) The distance of a point $\vec{a}$ from the plane $\vec{r} \cdot \vec{m} = d$ is given by $\frac{|\vec{a} \cdot \vec{m} - d|}{|\vec{m}|}$.
Given the plane $\vec{r} \cdot (2\hat{i} + 6\hat{j} - 9\hat{k}) = -1$,we have $\vec{m} = 2\hat{i} + 6\hat{j} - 9\hat{k}$ and $d = -1$.
The magnitude $|\vec{m}| = \sqrt{2^2 + 6^2 + (-9)^2} = \sqrt{4 + 36 + 81} = \sqrt{121} = 11$.
For the point $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$,the distance $p$ is:
$p = \frac{|(1)(2) + (2)(6) + (3)(-9) - (-1)|}{11} = \frac{|2 + 12 - 27 + 1|}{11} = \frac{|-12|}{11} = \frac{12}{11}$.
The distance $q$ of the origin $(0, 0, 0)$ from the plane is:
$q = \frac{|0 - (-1)|}{11} = \frac{1}{11}$.
Therefore,$p - q = \frac{12}{11} - \frac{1}{11} = \frac{11}{11} = 1$.

(A) Let $\vec{r}$ be the position vector of the point $C$.
Since point $C$ divides the line segment joining the points with position vectors $\vec{a} = 2 \hat{i}-3 \hat{j}+2 \hat{k}$ and $\vec{b} = 3 \hat{i}-\hat{j}-2 \hat{k}$ in the ratio $m:n = 2:3$,the position vector $\vec{r}$ is given by the section formula:
$\vec{r} = \frac{m\vec{b} + n\vec{a}}{m+n} = \frac{2(3 \hat{i}-\hat{j}-2 \hat{k}) + 3(2 \hat{i}-3 \hat{j}+2 \hat{k})}{2+3}$
$= \frac{(6 \hat{i}-2 \hat{j}-4 \hat{k}) + (6 \hat{i}-9 \hat{j}+6 \hat{k})}{5} = \frac{12 \hat{i}-11 \hat{j}+2 \hat{k}}{5} = \frac{12}{5} \hat{i}-\frac{11}{5} \hat{j}+\frac{2}{5} \hat{k}$.
Now,the distance of $C$ from the point $P$ with position vector $\vec{p} = 2 \hat{i}-\hat{j}+\hat{k}$ is given by $|\vec{r} - \vec{p}|$.
$\vec{r} - \vec{p} = (\frac{12}{5}-2) \hat{i} + (-\frac{11}{5}+1) \hat{j} + (\frac{2}{5}-1) \hat{k} = \frac{2}{5} \hat{i} - \frac{6}{5} \hat{j} - \frac{3}{5} \hat{k}$.
The distance $D = |\vec{r} - \vec{p}| = \sqrt{(\frac{2}{5})^2 + (-\frac{6}{5})^2 + (-\frac{3}{5})^2} = \sqrt{\frac{4}{25} + \frac{36}{25} + \frac{9}{25}} = \sqrt{\frac{49}{25}} = \frac{7}{5}$.

(D) Given: $\vec{a}=(2x+y)\hat{i}+3\hat{j}+9\hat{k}$ and $\vec{b}=2\hat{i}+\hat{j}-(x-y)\hat{k}$.
Since $\vec{a}$ and $\vec{b}$ are collinear,their components are proportional:
$\frac{2x+y}{2} = \frac{3}{1} = \frac{9}{-(x-y)}$.
From $\frac{2x+y}{2} = 3$,we get $2x+y=6$ $(i)$.
From $\frac{3}{1} = \frac{9}{y-x}$,we get $y-x=3$ or $y=x+3$ $(ii)$.
Substituting $(ii)$ into $(i)$:
$2x+(x+3)=6 \Rightarrow 3x=3 \Rightarrow x=1$.
Then $y=1+3=4$.
Finally,$x^3+27y^3 = (1)^3 + 27(4)^3 = 1 + 27(64) = 1 + 1728 = 1729$.

(A) Given $\vec{a}=\hat{i}-2 \hat{j}$,$\vec{b}=2 \hat{j}+3 \hat{k}$,$\vec{c}=p \hat{i}+q \hat{j}$ and $\vec{d}=p \hat{j}-q \hat{k}$.
First,calculate the cross product $\vec{a} \times \vec{b}$:
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 0 \\ 0 & 2 & 3 \end{vmatrix} = \hat{i}(-6-0) - \hat{j}(3-0) + \hat{k}(2-0) = -6 \hat{i} - 3 \hat{j} + 2 \hat{k}$.
Now,use the condition $(\vec{a} \times \vec{b}) \cdot \vec{c} = 3$:
$(-6 \hat{i} - 3 \hat{j} + 2 \hat{k}) \cdot (p \hat{i} + q \hat{j}) = 3 \implies -6p - 3q = 3 \implies -2p - q = 1$ (Equation $1$).
Next,use the condition $(\vec{a} \times \vec{b}) \cdot \vec{d} = 3$:
$(-6 \hat{i} - 3 \hat{j} + 2 \hat{k}) \cdot (p \hat{j} - q \hat{k}) = 3 \implies -3p - 2q = 3$ (Equation $2$).
Multiply Equation $1$ by $2$: $-4p - 2q = 2$ (Equation $3$).
Subtract Equation $2$ from Equation $3$: $(-4p - 2q) - (-3p - 2q) = 2 - 3 \implies -p = -1 \implies p = 1$.
Substitute $p=1$ into Equation $1$: $-2(1) - q = 1 \implies -2 - q = 1 \implies q = -3$.
Finally,calculate $3p + q = 3(1) + (-3) = 3 - 3 = 0$.

(C) Given $\vec{a}=2 \hat{i}+\hat{j}-\hat{k}$ and $\vec{b}=\hat{i}+3 \hat{j}-5 \hat{k}$.
First,calculate the vector $\vec{v} = 3 \vec{a}-2 \vec{b}$:
$\vec{v} = 3(2 \hat{i}+\hat{j}-\hat{k})-2(\hat{i}+3 \hat{j}-5 \hat{k}) = (6-2) \hat{i} + (3-6) \hat{j} + (-3+10) \hat{k} = 4 \hat{i}-3 \hat{j}+7 \hat{k}$.
Since $\vec{r}$ is along the vector $\vec{v}$,we can write $\vec{r} = x \vec{v} = x(4 \hat{i}-3 \hat{j}+7 \hat{k})$ for some scalar $x$.
Given $|\vec{r}| = \sqrt{74}$,we have $|x| \sqrt{4^2+(-3)^2+7^2} = \sqrt{74}$.
$|x| \sqrt{16+9+49} = \sqrt{74} \Rightarrow |x| \sqrt{74} = \sqrt{74} \Rightarrow |x| = 1$.
Since the direction of $\vec{r}$ is opposite to that of $\vec{v}$,we must have $x = -1$.
Therefore,$\vec{r} = -1(4 \hat{i}-3 \hat{j}+7 \hat{k}) = -4 \hat{i}+3 \hat{j}-7 \hat{k}$.

(B) Let the position vector of vertex $C$ be $\vec{c} = x\hat{i} + y\hat{j} + z\hat{k}$.
Given,centroid $G = \hat{i} = 1\hat{i} + 0\hat{j} + 0\hat{k}$.
We know that $G = \frac{\vec{A} + \vec{B} + \vec{C}}{3}$,where $\vec{A} = 2\hat{i} + \hat{j} + \hat{k}$ and $\vec{B} = 2\hat{i} + 4\hat{j} - 4\hat{k}$.
So,$3(1\hat{i} + 0\hat{j} + 0\hat{k}) = (2\hat{i} + \hat{j} + \hat{k}) + (2\hat{i} + 4\hat{j} - 4\hat{k}) + (x\hat{i} + y\hat{j} + z\hat{k})$.
$3\hat{i} = (4+x)\hat{i} + (5+y)\hat{j} + (-3+z)\hat{k}$.
Comparing coefficients: $4+x = 3 \Rightarrow x = -1$; $5+y = 0 \Rightarrow y = -5$; $-3+z = 0 \Rightarrow z = 3$.
Thus,$\vec{C} = -\hat{i} - 5\hat{j} + 3\hat{k}$.
Now,$AG^2 = |\vec{G} - \vec{A}|^2 = |(1-2)\hat{i} + (0-1)\hat{j} + (0-1)\hat{k}|^2 = |-1\hat{i} - 1\hat{j} - 1\hat{k}|^2 = 1+1+1 = 3$.
$BG^2 = |\vec{G} - \vec{B}|^2 = |(1-2)\hat{i} + (0-4)\hat{j} + (0+4)\hat{k}|^2 = |-1\hat{i} - 4\hat{j} + 4\hat{k}|^2 = 1+16+16 = 33$.
$CG^2 = |\vec{G} - \vec{C}|^2 = |(1+1)\hat{i} + (0+5)\hat{j} + (0-3)\hat{k}|^2 = |2\hat{i} + 5\hat{j} - 3\hat{k}|^2 = 4+25+9 = 38$.
Therefore,$AG^2 + BG^2 + CG^2 = 3 + 33 + 38 = 74$.

(C) Given,$\vec{a}=-2 \hat{i}+9 \hat{j}-6 \hat{k}$ and $\vec{b}=t \hat{i}-2 \hat{j}+6 \hat{k}$.
First,calculate the sum of the vectors: $\vec{a}+\vec{b} = (-2+t) \hat{i} + (9-2) \hat{j} + (-6+6) \hat{k} = (t-2) \hat{i} + 7 \hat{j} + 0 \hat{k}$.
Given that $|\vec{a}+\vec{b}|=25$,we have $\sqrt{(t-2)^2 + 7^2 + 0^2} = 25$.
Squaring both sides,we get $(t-2)^2 + 49 = 625$.
$(t-2)^2 = 625 - 49 = 576$.
Taking the square root,$t-2 = \pm 24$.
Case $1$: $t-2 = 24 \Rightarrow t = 26$.
Case $2$: $t-2 = -24 \Rightarrow t = -22$.
The sum of the values of $t$ is $26 + (-22) = 4$.

(C) The component of vector $\vec{a}$ along $\vec{b}$ is given by $\vec{p} = \left( \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2} \right) \vec{b}$.
First,calculate the dot product $\vec{a} \cdot \vec{b} = (1)(2) + (3)(-4) + (13)(3) = 2 - 12 + 39 = 29$.
Next,calculate the magnitude squared $|\vec{b}|^2 = (2)^2 + (-4)^2 + (3)^2 = 4 + 16 + 9 = 29$.
So,the component of $\vec{a}$ along $\vec{b}$ is $\vec{p} = \left( \frac{29}{29} \right) \vec{b} = \vec{b} = 2\hat{i} - 4\hat{j} + 3\hat{k}$.
The component of $\vec{a}$ perpendicular to $\vec{b}$ is $\vec{x} = \vec{a} - \vec{p} = \vec{a} - \vec{b}$.
$\vec{x} = (\hat{i} + 3\hat{j} + 13\hat{k}) - (2\hat{i} - 4\hat{j} + 3\hat{k}) = (1-2)\hat{i} + (3+4)\hat{j} + (13-3)\hat{k} = -\hat{i} + 7\hat{j} + 10\hat{k}$.

(A) Given $\vec{a}+\vec{b}=m \vec{d}-\vec{c} \Rightarrow \vec{a}+\vec{b}+\vec{c}=m \vec{d} \quad (i)$
And $\vec{b}+\vec{c}=n \vec{a}-\vec{d} \Rightarrow \vec{d}=n \vec{a}-\vec{b}-\vec{c} \quad (ii)$
Substituting $(ii)$ into $(i)$,we get:
$\vec{a}+\vec{b}+\vec{c}=m(n \vec{a}-\vec{b}-\vec{c})$
$\Rightarrow \vec{a}+\vec{b}+\vec{c}=mn \vec{a}-m \vec{b}-m \vec{c}$
$\Rightarrow (1-mn) \vec{a}+(1+m) \vec{b}+(1+m) \vec{c}=\vec{0}$
Since $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar vectors,they are linearly independent.
Therefore,the coefficients must be zero:
$1-mn=0 \Rightarrow mn=1$ and $1+m=0 \Rightarrow m=-1$.
Substituting $m=-1$ into $mn=1$,we get $n=-1$.
From $(i)$,$\vec{a}+\vec{b}+\vec{c}=(-1) \vec{d} \Rightarrow \vec{a}+\vec{b}+\vec{c}+\vec{d}=\vec{0}$.
Now,$3 \vec{a}+2 \vec{b}+2 \vec{c}+\vec{d} = (\vec{a}+\vec{b}+\vec{c}+\vec{d}) + 2(\vec{a}+\vec{b}+\vec{c}) = \vec{0} + 2(-\vec{d}) + 2\vec{a} = 2\vec{a}-2\vec{d}$.
Wait,re-evaluating the expression: $3 \vec{a}+2 \vec{b}+2 \vec{c}+\vec{d} = \vec{a} + 2(\vec{a}+\vec{b}+\vec{c}) + \vec{d} = \vec{a} + 2(-\vec{d}) + \vec{d} = \vec{a}-\vec{d}$.

(D) Given position vectors are $\overrightarrow{OA} = 2\hat{i} + \hat{j} - \hat{k}$,$\overrightarrow{OB} = \hat{i} - 3\hat{j} + 5\hat{k}$,and $\overrightarrow{OC} = -3\hat{i} + 4\hat{j} + 4\hat{k}$.
Calculate the vectors representing the sides:
$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (1-2)\hat{i} + (-3-1)\hat{j} + (5-(-1))\hat{k} = -\hat{i} - 4\hat{j} + 6\hat{k}$.
$|\overrightarrow{AB}| = \sqrt{(-1)^2 + (-4)^2 + 6^2} = \sqrt{1 + 16 + 36} = \sqrt{53}$.
$\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = (-3-1)\hat{i} + (4-(-3))\hat{j} + (4-5)\hat{k} = -4\hat{i} + 7\hat{j} - \hat{k}$.
$|\overrightarrow{BC}| = \sqrt{(-4)^2 + 7^2 + (-1)^2} = \sqrt{16 + 49 + 1} = \sqrt{66}$.
$\overrightarrow{CA} = \overrightarrow{OA} - \overrightarrow{OC} = (2-(-3))\hat{i} + (1-4)\hat{j} + (-1-4)\hat{k} = 5\hat{i} - 3\hat{j} - 5\hat{k}$.
$|\overrightarrow{CA}| = \sqrt{5^2 + (-3)^2 + (-5)^2} = \sqrt{25 + 9 + 25} = \sqrt{59}$.
Since $|\overrightarrow{AB}| \neq |\overrightarrow{BC}| \neq |\overrightarrow{CA}|$,all three sides have different lengths.
Therefore,$\triangle ABC$ is a scalene triangle.

(B) Given position vectors are $\vec{OA} = 2 \hat{i} + 3 \hat{j} - \hat{k}$ and $\vec{OB} = \hat{i} - \hat{j} + 2 \hat{k}$.
First,we find the vector $\overrightarrow{AB} = \vec{OB} - \vec{OA} = (1-2)\hat{i} + (-1-3)\hat{j} + (2-(-1))\hat{k} = -\hat{i} - 4 \hat{j} + 3 \hat{k}$.
Next,we find the vector $\overrightarrow{BA} = \vec{OA} - \vec{OB} = (2-1)\hat{i} + (3-(-1))\hat{j} + (-1-2)\hat{k} = \hat{i} + 4 \hat{j} - 3 \hat{k}$.
The magnitude of $\overrightarrow{BA}$ is $|\overrightarrow{BA}| = \sqrt{1^2 + 4^2 + (-3)^2} = \sqrt{1 + 16 + 9} = \sqrt{26}$.
The unit vector along $\overrightarrow{BA}$ in the direction of $\overrightarrow{AB}$ is defined as $\frac{\overrightarrow{AB}}{|\overrightarrow{BA}|}$.
Therefore,the required unit vector is $\frac{-\hat{i} - 4 \hat{j} + 3 \hat{k}}{\sqrt{26}}$.

(A) Given position vectors are $\vec{a} = -3 \hat{i}-3 \hat{j}+4 \hat{k}$ and $\vec{b} = 4 \hat{i}-4 \hat{j}-3 \hat{k}$.
Since $L$ divides $OA$ in the ratio $2:3$,the position vector of $L$ is $\vec{OL} = \frac{2}{5} \vec{a} = \frac{2}{5}(-3 \hat{i}-3 \hat{j}+4 \hat{k})$.
Since $M$ divides $OB$ in the ratio $3:2$,the position vector of $M$ is $\vec{OM} = \frac{3}{5} \vec{b} = \frac{3}{5}(4 \hat{i}-4 \hat{j}-3 \hat{k})$.
Since $LP \parallel OB$ and $PM \parallel OA$,$OMPL$ is a parallelogram.
Therefore,the position vector of $P$ is $\vec{OP} = \vec{OL} + \vec{OM}$.
$\vec{OP} = \frac{2}{5}(-3 \hat{i}-3 \hat{j}+4 \hat{k}) + \frac{3}{5}(4 \hat{i}-4 \hat{j}-3 \hat{k})$
$\vec{OP} = \frac{1}{5}(-6 \hat{i}-6 \hat{j}+8 \hat{k} + 12 \hat{i}-12 \hat{j}-9 \hat{k})$
$\vec{OP} = \frac{1}{5}(6 \hat{i}-18 \hat{j}-\hat{k})$.
The distance from $O$ to $P$ is $|\vec{OP}| = \frac{1}{5} \sqrt{6^2 + (-18)^2 + (-1)^2} = \frac{1}{5} \sqrt{36 + 324 + 1} = \frac{\sqrt{361}}{5} = \frac{19}{5}$.

(D) Given vectors are $\vec{a}=3 \hat{i}+\hat{j}-2 \hat{k}$,$\vec{b}=-5 \hat{i}+7 \hat{j}$,and $\vec{c}=3 \hat{i}+y \hat{j}$.
First,calculate $\vec{a}-\vec{b}+\vec{c}$:
$\vec{a}-\vec{b}+\vec{c} = (3 \hat{i}+\hat{j}-2 \hat{k}) - (-5 \hat{i}+7 \hat{j}) + (3 \hat{i}+y \hat{j})$
$= (3+5+3) \hat{i} + (1-7+y) \hat{j} - 2 \hat{k}$
$= 11 \hat{i} + (y-6) \hat{j} - 2 \hat{k}$.
Now,find the magnitude $|\vec{a}-\vec{b}+\vec{c}| = \sqrt{11^2 + (y-6)^2 + (-2)^2} = \sqrt{121 + (y-6)^2 + 4} = \sqrt{125 + (y-6)^2}$.
Given that $|\vec{a}-\vec{b}+\vec{c}| = \sqrt{141}$,we have:
$\sqrt{125 + (y-6)^2} = \sqrt{141}$
$125 + (y-6)^2 = 141$
$(y-6)^2 = 141 - 125 = 16$
$y-6 = \pm 4$.
Thus,$y_1 = 6+4 = 10$ and $y_2 = 6-4 = 2$.
The value of $|y_1-y_2| = |10-2| = 8$.