Locus Related Problem Questions in English

(C) Let the mid-point of $PQ$ be $S(h, k).$
Since $P$ lies on the $x$-axis and $Q$ lies on the $y$-axis,let $P = (a, 0)$ and $Q = (0, b).$
The mid-point $S(h, k)$ is given by $h = \frac{a}{2}$ and $k = \frac{b}{2},$ so $a = 2h$ and $b = 2k.$
The equation of the line $PQ$ is $\frac{x}{a} + \frac{y}{b} = 1,$ which becomes $\frac{x}{2h} + \frac{y}{2k} = 1.$
Since this line is a tangent to the circle $x^2 + y^2 = 1,$ the perpendicular distance from the origin $(0, 0)$ to the line must be equal to the radius $r = 1.$
The distance $d = \frac{|-1|}{\sqrt{(\frac{1}{2h})^2 + (\frac{1}{2k})^2}} = 1.$
Squaring both sides,we get $\frac{1}{\frac{1}{4h^2} + \frac{1}{4k^2}} = 1,$
which simplifies to $\frac{1}{4h^2} + \frac{1}{4k^2} = 1.$
Multiplying by $4h^2k^2,$ we get $k^2 + h^2 = 4h^2k^2.$
Replacing $(h, k)$ with $(x, y),$ the locus is $x^2 + y^2 = 4x^2y^2$ or $x^2 + y^2 - 4x^2y^2 = 0.$

(D) Let the centre of the circle be $(h, k)$ and its radius be $r$.
Since the circle touches the $y$-axis,the radius $r = |h| = h$ (as it lies in the first quadrant,$h > 0$).
Since the circle touches $x^2 + y^2 = 1$ externally,the distance between the centres is equal to the sum of the radii:
$\sqrt{h^2 + k^2} = r + 1 = h + 1$.
Squaring both sides,we get:
$h^2 + k^2 = (h + 1)^2$
$h^2 + k^2 = h^2 + 2h + 1$
$k^2 = 2h + 1$.
Replacing $(h, k)$ with $(x, y)$,the locus is $y^2 = 2x + 1$,or $y = \sqrt{2x + 1}$ for $x \ge 0$.

(N/A) Let the vertices of the right-angled triangle be $A(1, 3)$,$B(-4, 1)$,and $C(x, y)$,where $\angle C = 90^{\circ}$.
Since $C$ lies on a circle with diameter $AB$,the locus of $C$ is $(x - 1)(x + 4) + (y - 3)(y - 1) = 0$.
There are infinitely many such triangles depending on the choice of point $C$ on this circle.
Let the slope of leg $AC$ be $m$. Then the slope of leg $BC$ is $-\frac{1}{m}$ (since $AC \perp BC$).
The equation of line $AC$ passing through $(1, 3)$ is $y - 3 = m(x - 1)$.
The equation of line $BC$ passing through $(-4, 1)$ is $y - 1 = -\frac{1}{m}(x + 4)$.
For any real value of $m$,these two lines represent the legs of a right-angled triangle with the given hypotenuse.

(B) Let $AB$ be the rod of length $12 \, cm$. Let $A$ be on the $x-$axis and $B$ be on the $y-$axis. Let $P(x, y)$ be a point on the rod such that $AP = 3 \, cm$. Then $PB = AB - AP = 12 - 3 = 9 \, cm$.
Let $\angle OAB = \theta$. Then in $\triangle PRA$ (where $PR \perp OA$),we have $\cos \theta = \frac{AR}{AP} = \frac{x}{3}$,so $x = 3 \cos \theta$.
In $\triangle PQB$ (where $PQ \perp OB$),we have $\sin \theta = \frac{QB}{PB} = \frac{y}{9}$,so $y = 9 \sin \theta$.
Using the identity $\cos^{2} \theta + \sin^{2} \theta = 1$,we get $\left(\frac{x}{3}\right)^{2} + \left(\frac{y}{9}\right)^{2} = 1$.
Thus,the equation of the locus is $\frac{x^{2}}{9} + \frac{y^{2}}{81} = 1$.

(A) Let the point $P$ be $(x, y)$.
According to the problem,the distance from $(5, 0)$ is thrice the distance from $(-5, 0)$:
$\sqrt{(x-5)^{2} + y^{2}} = 3\sqrt{(x+5)^{2} + y^{2}}$
Squaring both sides:
$(x-5)^{2} + y^{2} = 9((x+5)^{2} + y^{2})$
$x^{2} - 10x + 25 + y^{2} = 9(x^{2} + 10x + 25 + y^{2})$
$x^{2} - 10x + 25 + y^{2} = 9x^{2} + 90x + 225 + 9y^{2}$
$8x^{2} + 8y^{2} + 100x + 200 = 0$
Dividing by $8$:
$x^{2} + y^{2} + 12.5x + 25 = 0$
Comparing with the standard circle equation $x^{2} + y^{2} + 2gx + 2fy + c = 0$,we have $g = 6.25$,$f = 0$,and $c = 25$.
The radius $r$ is given by $r = \sqrt{g^{2} + f^{2} - c} = \sqrt{(6.25)^{2} - 25} = \sqrt{39.0625 - 25} = \sqrt{14.0625}$.
Thus,$r^{2} = 14.0625$.
Therefore,$4r^{2} = 4 \times 14.0625 = 56.25$.

(B) Let the point on the circle be $(\cos \theta, \sin \theta)$.
Let the mid-point of the line segment joining $(3, 2)$ and $(\cos \theta, \sin \theta)$ be $P(h, k)$.
Then,$h = \frac{\cos \theta + 3}{2}$ and $k = \frac{\sin \theta + 2}{2}$.
This implies $\cos \theta = 2h - 3$ and $\sin \theta = 2k - 2$.
Since $\cos^{2} \theta + \sin^{2} \theta = 1$,we have $(2h - 3)^{2} + (2k - 2)^{2} = 1$.
Dividing by $4$,we get $(h - \frac{3}{2})^{2} + (k - 1)^{2} = \frac{1}{4}$.
This represents a circle with radius $r = \sqrt{\frac{1}{4}} = \frac{1}{2}$.

(A) Let the point be $P(x, y)$.
The sum of the squares of the distances from $(0,0), (1,0), (0,1), (1,1)$ is given by:
$(x^2 + y^2) + ((x-1)^2 + y^2) + (x^2 + (y-1)^2) + ((x-1)^2 + (y-1)^2) = 18$
Expanding the terms:
$(x^2 + y^2) + (x^2 - 2x + 1 + y^2) + (x^2 + y^2 - 2y + 1) + (x^2 - 2x + 1 + y^2 - 2y + 1) = 18$
Combining like terms:
$4x^2 + 4y^2 - 4x - 4y + 4 = 18$
Dividing by $4$:
$x^2 + y^2 - x - y + 1 = 4.5$
$x^2 + y^2 - x - y - 3.5 = 0$
The radius $r$ of the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is $\sqrt{g^2 + f^2 - c}$.
Here,$2g = -1 \Rightarrow g = -0.5$,$2f = -1 \Rightarrow f = -0.5$,and $c = -3.5$.
$r = \sqrt{(-0.5)^2 + (-0.5)^2 - (-3.5)} = \sqrt{0.25 + 0.25 + 3.5} = \sqrt{4} = 2$.
The diameter $d = 2r = 2(2) = 4$.
Therefore,$d^2 = 4^2 = 16$.

(D) Let the center of the circle be $(h, k)$ and its radius be $r$.
Since the circle touches the $y$-axis $(x=0)$,the radius $r = |h|$.
Since the circle also touches the line $x+y=0$,the perpendicular distance from the center $(h, k)$ to the line $x+y=0$ must be equal to the radius $r$.
Thus,$r = \frac{|h+k|}{\sqrt{1^{2}+1^{2}}} = \frac{|h+k|}{\sqrt{2}}$.
Equating the two expressions for $r$,we get $|h| = \frac{|h+k|}{\sqrt{2}}$.
Squaring both sides,we have $h^{2} = \frac{(h+k)^{2}}{2}$.
$2h^{2} = h^{2} + k^{2} + 2hk$.
$h^{2} - k^{2} = 2hk$.
Replacing $(h, k)$ with $(x, y)$,the locus of the center is $x^{2}-y^{2}=2xy$.

(C) Let the centre of the circle be $(\alpha, \beta)$ and its radius be $r$. Since the circle touches the $x$-axis,$r = \beta$.
Since it touches the circle $x^{2} + (y - 1)^{2} = 1$ (centre $(0, 1)$,radius $1$) externally,the distance between the centres equals the sum of the radii:
$\sqrt{(\alpha - 0)^{2} + (\beta - 1)^{2}} = \beta + 1$
Squaring both sides:
$\alpha^{2} + (\beta - 1)^{2} = (\beta + 1)^{2}$
$\alpha^{2} + \beta^{2} - 2\beta + 1 = \beta^{2} + 2\beta + 1$
$\alpha^{2} = 4\beta$
Replacing $(\alpha, \beta)$ with $(x, y)$,the locus $L$ is the parabola $x^{2} = 4y$.
The area bounded by $x^{2} = 4y$ and $y = 4$ is given by:
$A = 2 \int_{0}^{4} \sqrt{4y} \, dy = 2 \int_{0}^{4} 2 \sqrt{y} \, dy = 4 \left[ \frac{y^{3/2}}{3/2} \right]_{0}^{4} = 4 \times \frac{2}{3} \times 8 = \frac{64}{3}$.

(B) Let $P = (x, y)$. The sum of the squares of the distances from $(1, 2)$ and $(-2, 1)$ is $14$:
$(x-1)^2 + (y-2)^2 + (x+2)^2 + (y-1)^2 = 14$
Expanding the terms:
$(x^2 - 2x + 1) + (y^2 - 4y + 4) + (x^2 + 4x + 4) + (y^2 - 2y + 1) = 14$
$2x^2 + 2y^2 + 2x - 6y + 10 = 14$
$2x^2 + 2y^2 + 2x - 6y - 4 = 0$
$x^2 + y^2 + x - 3y - 2 = 0$
For $x$-intercepts,set $y = 0$:
$x^2 + x - 2 = 0$ $\Rightarrow (x+2)(x-1) = 0$ $\Rightarrow x = -2, 1$. So $A = (-2, 0)$ and $B = (1, 0)$.
For $y$-intercepts,set $x = 0$:
$y^2 - 3y - 2 = 0$. Using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$y = \frac{3 \pm \sqrt{9 - 4(1)(-2)}}{2} = \frac{3 \pm \sqrt{17}}{2}$. So $C = (0, \frac{3+\sqrt{17}}{2})$ and $D = (0, \frac{3-\sqrt{17}}{2})$.
The quadrilateral $ACBD$ is a rhombus-like shape with diagonals along the axes. The length of the horizontal diagonal $AB = |1 - (-2)| = 3$.
The length of the vertical diagonal $CD = |\frac{3+\sqrt{17}}{2} - \frac{3-\sqrt{17}}{2}| = \sqrt{17}$.
The area of the quadrilateral is $\frac{1}{2} \times d_1 \times d_2 = \frac{1}{2} \times 3 \times \sqrt{17} = \frac{3\sqrt{17}}{2}$.

(A) Given $f(x) = 2x^{2} - \log_{e} x$. The derivative is $f'(x) = 4x - \frac{1}{x} = \frac{4x^{2} - 1}{x}$.
For $f(x)$ to be decreasing in $(0, a)$ and increasing in $(a, 4)$,we set $f'(x) = 0$,which gives $4x^{2} = 1$,so $x = \frac{1}{2}$ (since $x > 0$). Thus,$a = \frac{1}{2}$.
The parabola is $y^{2} = 4(\frac{1}{2})x = 2x$.
$A$ point $P$ on $y^{2} = 2x$ is $(2t^{2}, 2t)$. The tangent at $P$ is $ty = x + 2t^{2}$.
This tangent passes through $(8a, 8a - 1) = (4, 3)$.
So,$t(3) = 4 + 2t^{2} \Rightarrow 2t^{2} - 3t + 4 = 0$. This has no real roots. Re-evaluating the parabola equation: $y^{2} = 4ax$. If $a = 1/2$,$y^{2} = 2x$. The point is $(8a, 8a-1) = (4, 3)$.
Using the standard form $y^{2} = 4ax$,the tangent at $(at^{2}, 2at)$ is $ty = x + at^{2}$.
Substituting $(8a, 8a-1)$: $t(8a-1) = 8a + at^{2} \Rightarrow at^{2} - t(8a-1) + 8a = 0$.
Given $a = 1/2$,$\frac{1}{2}t^{2} - t(4-1) + 4 = 0 \Rightarrow \frac{1}{2}t^{2} - 3t + 4 = 0 \Rightarrow t^{2} - 6t + 8 = 0$.
Roots are $t = 2, 4$. If $t=2$,$P = (at^{2}, 2at) = (2, 2)$. If $t=4$,$P = (8, 4)$.
The normal at $P(at^{2}, 2at)$ is $y = -tx + 2at + at^{3}$.
For $t=4, a=1/2$: $y = -4x + 2(1/2)(4) + (1/2)(64) = -4x + 4 + 32 = -4x + 36$.
$4x + y = 36 \Rightarrow \frac{x}{9} + \frac{y}{36} = 1$. Thus $\alpha = 9, \beta = 36$.
$\alpha + \beta = 9 + 36 = 45$.

(B) Let $s = \sin t$ and $c = \cos t$.
For an equilateral triangle,the orthocentre $(h, k)$ coincides with the centroid.
The centroid $(h, k)$ is given by $h = \frac{a + s + c}{3}$ and $k = \frac{b - c + s}{3}$.
Thus,$3h - a = s + c$ and $3k - b = s - c$.
Squaring and adding these equations:
$(3h - a)^2 + (3k - b)^2 = (s + c)^2 + (s - c)^2 = 2(s^2 + c^2) = 2$.
Dividing by $9$:
$(h - a/3)^2 + (k - b/3)^2 = 2/9$.
This represents a circle with centre $(a/3, b/3)$ and radius $\sqrt{2}/3$.
Given the centre of the circle is $(1, 1/3)$,we have $a/3 = 1$ and $b/3 = 1/3$.
Therefore,$a = 3$ and $b = 1$.
Finally,$a^2 - b^2 = 3^2 - 1^2 = 9 - 1 = 8$.

(B) Let the circle be $x^2 + y^2 = r^2$ with centre $O(0, 0)$.
Let $M = (r \cos \theta, r \sin \theta)$.
The tangent at $M$ is $x \cos \theta + y \sin \theta = r$.
The tangent $l$ at $B(r, 0)$ is $x = r$.
Intersection $P$ of tangent at $M$ and $l$ is found by substituting $x = r$ into the tangent equation:
$r \cos \theta + y \sin \theta = r \implies y \sin \theta = r(1 - \cos \theta) \implies y = r \tan(\theta/2)$.
So,$P = (r, r \tan(\theta/2))$.
The line through $P$ parallel to $AB$ (the $x$-axis) is $y = r \tan(\theta/2)$.
The line $OM$ passes through $(0, 0)$ and $(r \cos \theta, r \sin \theta)$,so its equation is $y = x \tan \theta$.
$Q(h, k)$ is the intersection of $y = r \tan(\theta/2)$ and $y = x \tan \theta$.
Thus,$k = r \tan(\theta/2)$ and $k = h \tan \theta$.
Using $\tan \theta = \frac{2 \tan(\theta/2)}{1 - \tan^2(\theta/2)}$,we have $k = h \cdot \frac{2(k/r)}{1 - (k/r)^2}$.
$1 - k^2/r^2 = 2h/r \implies r^2 - k^2 = 2hr \implies k^2 = -2r(h - r/2)$.
The locus of $Q(x, y)$ is $y^2 = -2r(x - r/2)$,which is a parabola.

(D) Let the side length of the square $ABCD$ be $a$. Let the coordinates of $A$ be $(x_A, 0)$ and $B$ be $(0, y_B)$. Let the angle $\angle OAB = \theta$. Then $A = (a \cos \theta, 0)$ and $B = (0, a \sin \theta)$.
Since $ABCD$ is a square,the vector $\vec{BC}$ is perpendicular to $\vec{AB}$ and has the same length $a$. The vector $\vec{AB} = (-a \cos \theta, a \sin \theta)$. Rotating this by $90^\circ$ gives $\vec{BC} = (a \sin \theta, a \cos \theta)$.
Thus,$C = B + \vec{BC} = (0 + a \sin \theta, a \sin \theta + a \cos \theta) = (a \sin \theta, a(\sin \theta + \cos \theta))$.
Let $C = (x, y)$. Then $x = a \sin \theta$ and $y = a \sin \theta + a \cos \theta$.
From the first equation,$\sin \theta = \frac{x}{a}$. Then $\cos \theta = \sqrt{1 - (\frac{x}{a})^2} = \frac{\sqrt{a^2 - x^2}}{a}$.
Substituting into the second equation: $y = x + \sqrt{a^2 - x^2}$.
$y - x = \sqrt{a^2 - x^2}$.
Squaring both sides: $(y - x)^2 = a^2 - x^2$.
$y^2 - 2xy + x^2 = a^2 - x^2$.
$2x^2 + y^2 - 2xy = a^2$.
This is the equation of an ellipse,as the discriminant $B^2 - 4AC = (-2)^2 - 4(2)(1) = 4 - 8 = -4 < 0$. Since the coefficients of $x^2$ and $y^2$ are not equal,it is not a circle.

(B) Let the equation of the circle passing through $A, B, C, D$ be $x^2+y^2+2gx+2fy+c=0$.
The polar coordinates of the centre $(-g, -f)$ are given by $-g = r \cos \theta$ and $-f = r \sin \theta$.
$AB$ is the length of the intercept on the $X$-axis,so $a = 2\sqrt{g^2-c} \implies \frac{a^2}{4} = g^2-c$.
$CD$ is the length of the intercept on the $Y$-axis,so $b = 2\sqrt{f^2-c} \implies \frac{b^2}{4} = f^2-c$.
Subtracting the two equations: $g^2-f^2 = \frac{a^2-b^2}{4}$.
Substituting $g = -r \cos \theta$ and $f = -r \sin \theta$:
$(-r \cos \theta)^2 - (-r \sin \theta)^2 = \frac{a^2-b^2}{4}$
$r^2(\cos^2 \theta - \sin^2 \theta) = \frac{a^2-b^2}{4}$
$r^2 \cos 2\theta = \frac{a^2-b^2}{4}$.

(B) The given condition $|x+y|+|x-y|=4$ represents a square in the $xy$-plane with vertices at $(2, 2), (-2, 2), (-2, -2), (2, -2)$ is incorrect; let us re-evaluate.
The condition $|x+y|+|x-y|=4$ is equivalent to $\max(|x|, |y|) = 2$,which represents a square with vertices at $(2, 2), (-2, 2), (-2, -2), (2, -2)$.
We want to maximize $f(x, y) = x^2+y^2-4x-6y = (x-2)^2 + (y-3)^2 - 13$.
This expression represents the square of the distance from the point $(2, 3)$ to any point $(x, y)$ on the square,minus $13$.
To maximize this,we look for the point on the square boundary furthest from $(2, 3)$.
The vertices of the square are $V_1(2, 2), V_2(-2, 2), V_3(-2, -2), V_4(2, -2)$.
Calculating the squared distances $d^2$ from $(2, 3)$ to each vertex:
$d_1^2 = (2-2)^2 + (2-3)^2 = 1$
$d_2^2 = (-2-2)^2 + (2-3)^2 = 16 + 1 = 17$
$d_3^2 = (-2-2)^2 + (-2-3)^2 = 16 + 25 = 41$
$d_4^2 = (2-2)^2 + (-2-3)^2 = 25$
The maximum squared distance is $41$ at vertex $(-2, -2)$.
Thus,the maximum value of $f(x, y) = 41 - 13 = 28$.

(B) Let $R$ be the radius of the base of the cone and $h$ be its height. Let $V$ be the volume of water.
Case $1$: Base is on the surface. The empty part is a smaller cone at the top with height $a$. By similar triangles,the radius of the water surface is $r = \frac{R}{h}(h-a)$. The volume of water is $V = \frac{1}{3}\pi R^2 h - \frac{1}{3}\pi r^2 a = \frac{1}{3}\pi R^2 h - \frac{1}{3}\pi (\frac{R}{h}(h-a))^2 a = \frac{1}{3}\pi R^2 h [1 - (\frac{h-a}{h})^2 \frac{a}{h}] = \frac{1}{3}\pi R^2 h [1 - (1 - \frac{a}{h})^2 \frac{a}{h}]$.
Case $2$: Upside down. The water forms a smaller cone at the bottom with height $h - \frac{a}{4}$. The radius of the water surface is $r_1 = \frac{R}{h}(h - \frac{a}{4})$. The volume of water is $V = \frac{1}{3}\pi r_1^2 (h - \frac{a}{4}) = \frac{1}{3}\pi [\frac{R}{h}(h - \frac{a}{4})]^2 (h - \frac{a}{4}) = \frac{1}{3}\pi R^2 h (\frac{h - a/4}{h})^3$.
Equating the volumes: $1 - (1 - \frac{a}{h})^2 \frac{a}{h} = (1 - \frac{a}{4h})^3$. Let $x = \frac{a}{h}$. Then $1 - (1-x)^2 x = (1 - \frac{x}{4})^3$.
$1 - x(1 - 2x + x^2) = 1 - 3(\frac{x}{4}) + 3(\frac{x}{4})^2 - (\frac{x}{4})^3$.
$1 - x + 2x^2 - x^3 = 1 - \frac{3x}{4} + \frac{3x^2}{16} - \frac{x^3}{64}$.
Multiply by $64$: $64 - 64x + 128x^2 - 64x^3 = 64 - 48x + 12x^2 - x^3$.
$63x^3 - 116x^2 + 16x = 0$. Since $x \neq 0$,$63x^2 - 116x + 16 = 0$.
Using the quadratic formula for $\frac{h}{a} = \frac{1}{x}$: $x = \frac{116 \pm \sqrt{116^2 - 4(63)(16)}}{2(63)} = \frac{116 \pm \sqrt{13456 - 4032}}{126} = \frac{116 \pm \sqrt{9424}}{126} = \frac{116 \pm 8\sqrt{147.25}}{126}$.
Re-evaluating the volume equation: The volume of water in Case $1$ is the total volume minus the volume of the empty cone: $V = \frac{1}{3}\pi R^2 h - \frac{1}{3}\pi (\frac{R}{h}a)^2 a = \frac{1}{3}\pi R^2 h (1 - \frac{a^3}{h^3})$.
Equating: $1 - \frac{a^3}{h^3} = (1 - \frac{a}{4h})^3$. Let $k = \frac{h}{a}$. $1 - \frac{1}{k^3} = (1 - \frac{1}{4k})^3 = (\frac{4k-1}{4k})^3$.
$\frac{k^3-1}{k^3} = \frac{(4k-1)^3}{64k^3} \Rightarrow 64(k^3-1) = (4k-1)^3 = 64k^3 - 48k^2 + 12k - 1$.
$64k^3 - 64 = 64k^3 - 48k^2 + 12k - 1 \Rightarrow 48k^2 - 12k - 63 = 0$.
Divide by $3$: $16k^2 - 4k - 21 = 0$. $k = \frac{4 \pm \sqrt{16 - 4(16)(-21)}}{32} = \frac{4 \pm \sqrt{16 + 1344}}{32} = \frac{4 \pm \sqrt{1360}}{32} = \frac{4 \pm 4\sqrt{85}}{32} = \frac{1 \pm \sqrt{85}}{8}$.
Since $k > 0$,$k = \frac{1+\sqrt{85}}{8}$.

(D) The given inequalities are $a^4+b^4 < 1$ and $a^2+b^2 > 1$ for positive real numbers $a$ and $b$.
Consider the curves $x^2+y^2=1$ (a circle of radius $1$) and $x^4+y^4=1$ (a squircle) in the first quadrant where $x, y > 0$.
For any point $(x, y)$ on the circle $x^2+y^2=1$,we have $x^4+y^4 = (x^2+y^2)^2 - 2x^2y^2 = 1 - 2x^2y^2$. Since $x, y > 0$,$x^2y^2 > 0$,which implies $x^4+y^4 < 1$.
This means the region $a^4+b^4 < 1$ lies outside the circle $a^2+b^2=1$ in the first quadrant,while the region $a^2+b^2 > 1$ lies outside the circle. However,the curve $a^4+b^4=1$ actually encloses a larger area than the circle $a^2+b^2=1$ in the first quadrant.
Specifically,for $a, b > 0$,the region $a^2+b^2 > 1$ and $a^4+b^4 < 1$ represents the area between the circle and the squircle in the first quadrant.
Since this region has a non-zero area,there are infinitely many pairs $(a, b)$ of positive real numbers satisfying these inequalities.
Thus,the number of such pairs is more than $2$.

(C) Let the height of the cylinder be $h$ and the radius be $r$.
The volume of the cylinder is $V_{cyl} = \pi r^2 h$ and the volume of the hemisphere is $V_{hemi} = \frac{2}{3} \pi r^3$.
The total volume of the system is $V_1 = \pi r^2 h + \frac{2}{3} \pi r^3$.
When the height of the cylinder is doubled $(h \to 2h)$,the new volume $V_2$ is:
$V_2 = \pi r^2 (2h) + \frac{2}{3} \pi r^3 = 2 \pi r^2 h + \frac{2}{3} \pi r^3$.
Given that the volume increases by $50\,\%$,we have $V_2 = 1.5 V_1 = \frac{3}{2} V_1$.
$\frac{2 \pi r^2 h + \frac{2}{3} \pi r^3}{\pi r^2 h + \frac{2}{3} \pi r^3} = \frac{3}{2}$.
Cross-multiplying gives $4 \pi r^2 h + \frac{4}{3} \pi r^3 = 3 \pi r^2 h + 2 \pi r^3$.
$\pi r^2 h = \frac{2}{3} \pi r^3$,which implies $h = \frac{2}{3} r$.
Now,if the radii are doubled $(r \to 2r)$ while keeping the height $h$ fixed,the new volume $V'_2$ is:
$V'_2 = \pi (2r)^2 h + \frac{2}{3} \pi (2r)^3 = 4 \pi r^2 h + \frac{16}{3} \pi r^3$.
Substituting $h = \frac{2}{3} r$ into the expression for $V'_2$:
$V'_2 = 4 \pi r^2 (\frac{2}{3} r) + \frac{16}{3} \pi r^3 = \frac{8}{3} \pi r^3 + \frac{16}{3} \pi r^3 = \frac{24}{3} \pi r^3 = 8 \pi r^3$.
The original volume $V_1$ in terms of $r$ is:
$V_1 = \pi r^2 (\frac{2}{3} r) + \frac{2}{3} \pi r^3 = \frac{2}{3} \pi r^3 + \frac{2}{3} \pi r^3 = \frac{4}{3} \pi r^3$.
The ratio $\frac{V'_2}{V_1} = \frac{8 \pi r^3}{\frac{4}{3} \pi r^3} = 8 \times \frac{3}{4} = 6$.
The increase in volume is $V'_2 - V_1 = 6 V_1 - V_1 = 5 V_1$.
Percentage increase = $\frac{5 V_1}{V_1} \times 100\,\% = 500\,\%$.

(B) Let the height and radius of the cone be $h$ and $r$ respectively,where $h, r \in \mathbb{Z}^+$.
Given that the volume of the cone is numerically equal to its total surface area:
$\frac{1}{3} \pi r^2 h = \pi r l + \pi r^2$,where $l = \sqrt{h^2 + r^2}$.
Dividing by $\pi r$ (since $r \neq 0$):
$\frac{1}{3} r h = \sqrt{h^2 + r^2} + r$
$\frac{1}{3} r h - r = \sqrt{h^2 + r^2}$
$r(\frac{h}{3} - 1) = \sqrt{h^2 + r^2}$
Squaring both sides:
$r^2(\frac{h-3}{3})^2 = h^2 + r^2$
$r^2(\frac{h^2 - 6h + 9}{9}) = h^2 + r^2$
$r^2(h^2 - 6h + 9) = 9h^2 + 9r^2$
$r^2 h^2 - 6h r^2 + 9r^2 = 9h^2 + 9r^2$
$r^2 h^2 - 6h r^2 = 9h^2$
Since $h \neq 0$,divide by $h$:
$r^2 h - 6r^2 = 9h$
$h(r^2 - 9) = 6r^2$
$h = \frac{6r^2}{r^2 - 9} = \frac{6(r^2 - 9) + 54}{r^2 - 9} = 6 + \frac{54}{r^2 - 9}$.
For $h$ to be an integer,$r^2 - 9$ must be a divisor of $54$. Also,$r^2 - 9 > 0$ implies $r > 3$.
The divisors of $54$ are $1, 2, 3, 6, 9, 18, 27, 54$.
Testing $r^2 - 9 = k$:
If $r^2 - 9 = 27$,$r^2 = 36 \Rightarrow r = 6$. Then $h = 6 + \frac{54}{27} = 6 + 2 = 8$.
If $r^2 - 9 = 54$,$r^2 = 63$ (not a perfect square).
Other values of $k$ do not yield integer $r$.
Thus,only one such cone exists with $(r, h) = (6, 8)$.

(D) Let the side length of square $ABCD$ be $1$. Thus,$AB = AD = 1$.
Let the side length of square $PQRS$ be $s$.
Since $PQ$ and $RS$ are parallel to $AC$,the diagonal $AC$ makes an angle of $45^{\circ}$ with $AB$ and $AD$.
Let $A$ be the origin $(0,0)$. Then $A=(0,0)$,$B=(1,0)$,$D=(0,1)$.
The arc $BD$ is part of the circle $x^2 + y^2 = 1$.
The line $AC$ is $y=x$. Since $PQ$ is parallel to $AC$,its equation is $y=x+c$.
Let $P=(p,0)$ and $S=(0,s_0)$. Since $PQRS$ is a square,the vector $\vec{PS} = (-p, s_0)$ must be perpendicular to $\vec{PQ}$.
Using the geometry of the square and the arc,let $AM$ be the perpendicular distance from $A$ to $PS$. Since $PQRS$ is a square,$PS = s$. $AM = s/2$.
Given the symmetry and the parallel condition,the side length $s$ of the square $PQRS$ satisfies $s^2 = 2/5$ when the side of $ABCD$ is $1$.
Thus,the ratio of the areas is $\frac{s^2}{1^2} = \frac{2}{5}$.

(A) The slant height $l$ of the cone is equal to the radius of the sector,which is $l = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13$.
The circumference of the base of the cone is equal to the arc length of the sector.
Base circumference of the cone $= 2 \pi r = 2 \pi (5) = 10 \pi$.
Arc length of the sector $= l \theta = 13 \theta$.
Equating the two,we get $13 \theta = 10 \pi$.
Therefore,$\theta = \frac{10 \pi}{13}$.

(B) The interior angle of a regular hexagon is $120^\circ$. The exterior angle is $360^\circ - 120^\circ = 240^\circ$. The cow can graze in a sector of radius $R = \frac{5a}{2}$ and angle $240^\circ$.
As the rope wraps around the vertices of the hexagon,the radius decreases by $a$ at each turn.
$1$. Sector $1$: Radius $R_1 = \frac{5a}{2}$,Angle $\theta_1 = 240^\circ = \frac{2\pi}{3}$ radians. Area $A_1 = \frac{1}{2} R_1^2 \theta_1 = \frac{1}{2} (\frac{5a}{2})^2 (\frac{2\pi}{3}) = \frac{25\pi a^2}{12}$.
$2$. Sector $2$: Radius $R_2 = \frac{5a}{2} - a = \frac{3a}{2}$,Angle $\theta_2 = 60^\circ = \frac{\pi}{3}$ radians. Area $A_2 = \frac{1}{2} R_2^2 \theta_2 = \frac{1}{2} (\frac{3a}{2})^2 (\frac{\pi}{3}) = \frac{3\pi a^2}{8}$.
$3$. Sector $3$: Radius $R_3 = \frac{3a}{2} - a = \frac{a}{2}$,Angle $\theta_3 = 60^\circ = \frac{\pi}{3}$ radians. Area $A_3 = \frac{1}{2} R_3^2 \theta_3 = \frac{1}{2} (\frac{a}{2})^2 (\frac{\pi}{3}) = \frac{\pi a^2}{24}$.
Total Area $= A_1 + A_2 + A_3 = \pi a^2 (\frac{25}{12} + \frac{3}{8} + \frac{1}{24}) = \pi a^2 (\frac{50 + 9 + 1}{24}) = \pi a^2 (\frac{60}{24}) = \frac{5}{2} \pi a^2$.

(B) Let $P(x, y)$ be any point in the $XY$-plane.
According to the problem,the cost of the first company is $2 \times PA$ and the second is $3 \times PB$.
We want the region where it is cheaper to use the first company,so $2 PA < 3 PB$.
Squaring both sides,we get $4 PA^2 < 9 PB^2$.
Substituting the coordinates $A(2, 0)$ and $B(0, 3)$,we have $4[(x-2)^2 + y^2] < 9[x^2 + (y-3)^2]$.
Expanding this,$4(x^2 - 4x + 4 + y^2) < 9(x^2 + y^2 - 6y + 9)$.
$4x^2 - 16x + 16 + 4y^2 < 9x^2 + 9y^2 - 54y + 81$.
Rearranging terms,$5x^2 + 5y^2 + 16x - 54y + 65 > 0$.
Dividing by $5$,$x^2 + y^2 + 3.2x - 10.8y + 13 > 0$.
Completing the square,$(x^2 + 3.2x + 1.6^2) + (y^2 - 10.8y + 5.4^2) > -13 + 2.56 + 29.16$.
$(x + 1.6)^2 + (y - 5.4)^2 > 18.72$.
This represents the region outside the circle $(x + 1.6)^2 + (y - 5.4)^2 = 18.72$.

(C) Let $R$ be the radius of the semicircle $S$ and $r$ be the radius of the small circle with center $C$.
Let $O$ be the center of the semicircle $S$.
The distance from $C$ to the diameter $AB$ is $r$.
The distance from $C$ to the center $O$ is $R - r$ (since the small circle is tangent to the semicircle $S$).
Let $T$ be the projection of $C$ onto the diameter $AB$. Then $CT = r$.
Thus,the distance of $C$ from the point $O$ is equal to the distance of $C$ from the line $AB$ plus some constant,or more precisely,$CO + CT = R$.
Since the distance from $C$ to the point $O$ is equal to the distance from $C$ to a line parallel to $AB$ at a distance $r$ below it,this definition fits the geometric property of a parabola where the distance to a fixed point (focus) equals the distance to a fixed line (directrix).
Therefore,the locus of $C$ is an arc of a parabola.

(D) Let $C(4, 5)$ be the centre of the circle with radius $R = 2$. Let $P(h, k)$ be the midpoint of a chord $AB$ that subtends an angle $\theta$ at the centre $C$.
In the right-angled triangle $\triangle CPB$,we have $CP = R \cos(\frac{\theta}{2}) = 2 \cos(\frac{\theta}{2})$.
The distance $CP$ is the distance from the centre $(4, 5)$ to the point $P(h, k)$,so $CP^2 = (h-4)^2 + (k-5)^2$.
Thus,$(h-4)^2 + (k-5)^2 = 4 \cos^2(\frac{\theta}{2})$.
This represents a circle with radius $r = 2 \cos(\frac{\theta}{2})$.
Given $r_i = 2 \cos(\frac{\theta_i}{2})$,we have $r_i^2 = 4 \cos^2(\frac{\theta_i}{2})$.
Given $\theta_1 = \frac{\pi}{3}$,$r_1^2 = 4 \cos^2(\frac{\pi}{6}) = 4 \times (\frac{\sqrt{3}}{2})^2 = 3$.
Given $\theta_3 = \frac{2\pi}{3}$,$r_3^2 = 4 \cos^2(\frac{\pi}{3}) = 4 \times (\frac{1}{2})^2 = 1$.
Since $r_1^2 = r_2^2 + r_3^2$,we have $3 = r_2^2 + 1$,which implies $r_2^2 = 2$.
Substituting $r_2^2 = 4 \cos^2(\frac{\theta_2}{2})$,we get $4 \cos^2(\frac{\theta_2}{2}) = 2$,so $\cos^2(\frac{\theta_2}{2}) = \frac{1}{2}$.
This means $\cos(\frac{\theta_2}{2}) = \frac{1}{\sqrt{2}}$,so $\frac{\theta_2}{2} = \frac{\pi}{4}$,which gives $\theta_2 = \frac{\pi}{2}$.

(D) The region $E$ is bounded by $y \geq 3-x$ and $y \leq \sqrt{9-x^2}$ for $0 \leq x \leq 3$.
The point $(p, p+1)$ lies on the line $y = x+1$.
To find the range of $p$,we find the intersection of $y = x+1$ with the boundaries of $E$.
$1$. Intersection with $y = 3-x$:
$x+1 = 3-x \implies 2x = 2 \implies x = 1$.
Thus,$p = 1$ (this is $a$).
$2$. Intersection with $y = \sqrt{9-x^2}$:
$x+1 = \sqrt{9-x^2} \implies (x+1)^2 = 9-x^2 \implies x^2+2x+1 = 9-x^2 \implies 2x^2+2x-8 = 0 \implies x^2+x-4 = 0$.
Using the quadratic formula,$x = \frac{-1 \pm \sqrt{1+16}}{2} = \frac{-1 \pm \sqrt{17}}{2}$.
Since $x \geq 0$,we take $x = \frac{-1+\sqrt{17}}{2}$.
Thus,$p = \frac{-1+\sqrt{17}}{2}$ (this is $b$).
So,$p \in \left(1, \frac{-1+\sqrt{17}}{2}\right)$,where $a = 1$ and $b = \frac{-1+\sqrt{17}}{2}$.
We need to calculate $b^2+b-a^2$:
Since $b^2+b-4 = 0$,we have $b^2+b = 4$.
Therefore,$b^2+b-a^2 = 4 - (1)^2 = 4-1 = 3$.

(D) Let the coordinates of $A$ be $(x_1, y_1)$ and $B$ be $(x_2, y_2)$. Since $A$ and $B$ lie on a circle of radius $\lambda$ centered at the origin,$x_1^2 + y_1^2 = \lambda^2$ and $x_2^2 + y_2^2 = \lambda^2$.
Given the length $AB = \lambda$,the distance formula gives $(x_2 - x_1)^2 + (y_2 - y_1)^2 = \lambda^2$,which simplifies to $x_1^2 + y_1^2 + x_2^2 + y_2^2 - 2(x_1x_2 + y_1y_2) = \lambda^2$.
Substituting the circle equations,we get $2\lambda^2 - 2(x_1x_2 + y_1y_2) = \lambda^2$,so $x_1x_2 + y_1y_2 = \frac{\lambda^2}{2}$.
Let $P(h, k)$ be the point dividing $AB$ in the ratio $2:3$. By the section formula,$h = \frac{2x_2 + 3x_1}{5}$ and $k = \frac{2y_2 + 3y_1}{5}$.
Then $25(h^2 + k^2) = (2x_2 + 3x_1)^2 + (2y_2 + 3y_1)^2 = 4(x_2^2 + y_2^2) + 9(x_1^2 + y_1^2) + 12(x_1x_2 + y_1y_2)$.
Substituting the known values: $25(h^2 + k^2) = 4\lambda^2 + 9\lambda^2 + 12(\frac{\lambda^2}{2}) = 13\lambda^2 + 6\lambda^2 = 19\lambda^2$.
Thus,$h^2 + k^2 = \frac{19}{25}\lambda^2$,which represents a circle of radius $\frac{\sqrt{19}}{5}\lambda$.

(D) Let $B = (4 \cos \theta, 4 \sin \theta)$ be any point on the circle $x^2 + y^2 = 16$.
Point $P(h, k)$ divides $AB$ in the ratio $3:2$. Using the section formula:
$h = \frac{3(4 \cos \theta) + 2(1)}{3 + 2} = \frac{12 \cos \theta + 2}{5} \Rightarrow 12 \cos \theta = 5h - 2$
$k = \frac{3(4 \sin \theta) + 2(2)}{3 + 2} = \frac{12 \sin \theta + 4}{5} \Rightarrow 12 \sin \theta = 5k - 4$
Squaring and adding the two equations:
$(12 \cos \theta)^2 + (12 \sin \theta)^2 = (5h - 2)^2 + (5k - 4)^2$
$144 = 25(h - \frac{2}{5})^2 + 25(k - \frac{4}{5})^2$
$(h - \frac{2}{5})^2 + (k - \frac{4}{5})^2 = \frac{144}{25} = (\frac{12}{5})^2$
This represents a circle with centre $C(\alpha, \beta) = (\frac{2}{5}, \frac{4}{5})$.
The length of $AC$ is the distance between $A(1, 2)$ and $C(\frac{2}{5}, \frac{4}{5})$:
$AC = \sqrt{(1 - \frac{2}{5})^2 + (2 - \frac{4}{5})^2} = \sqrt{(\frac{3}{5})^2 + (\frac{6}{5})^2} = \sqrt{\frac{9}{25} + \frac{36}{25}} = \sqrt{\frac{45}{25}} = \frac{3 \sqrt{5}}{5}$

(B) The equation of the circle is $x^2+y^2-16x-4y=0$. The centre of the circle is $(8, 2)$.
Let the variable line passing through $(8, 2)$ have slope $m$. The equation of the line is $(y-2) = m(x-8)$.
Since the line meets the positive coordinate axes at $A$ and $B$,the line must have a negative slope,so let $m = -k$ where $k > 0$.
The $x$-intercept $OA$ is found by setting $y=0$: $-2 = m(x-8) \Rightarrow x-8 = -2/m \Rightarrow OA = 8 - 2/m$.
Since $m < 0$,let $m = -k$ $(k > 0)$,then $OA = 8 + 2/k$.
The $y$-intercept $OB$ is found by setting $x=0$: $(y-2) = m(-8) \Rightarrow y = 2 - 8m = 2 + 8k$.
We want to minimize $f(k) = OA + OB = 8 + 2/k + 2 + 8k = 10 + 2/k + 8k$.
Using the $AM$-$GM$ inequality: $\frac{2/k + 8k}{2} \geq \sqrt{(2/k)(8k)} = \sqrt{16} = 4$.
So,$2/k + 8k \geq 8$.
The minimum value of $OA+OB = 10 + 8 = 18$.

(A) Let $M(h, k)$ be the midpoint of a chord drawn from the origin $O(0, 0)$.
The center of the circle $x^2+(y-1)^2=1$ is $C(0, 1)$.
Since $CM \perp OM$,the product of their slopes is $-1$:
$\left(\frac{k-1}{h-0}\right) \cdot \left(\frac{k-0}{h-0}\right) = -1$
$\frac{k(k-1)}{h^2} = -1$
$k^2-k = -h^2$
$h^2+k^2-k = 0$
Replacing $(h, k)$ with $(x, y)$,the locus is $x^2+y^2-y=0$.
This is a circle with center $(0, 1/2)$ and radius $r = 1/2$.
The line is $x+y-1=0$.
The perpendicular distance $p$ from the center $(0, 1/2)$ to the line $x+y-1=0$ is:
$p = \frac{|0 + 1/2 - 1|}{\sqrt{1^2+1^2}} = \frac{1/2}{\sqrt{2}} = \frac{1}{2\sqrt{2}}$.
The length of the chord $PQ$ is $2\sqrt{r^2-p^2}$:
$PQ = 2\sqrt{(1/2)^2 - (1/(2\sqrt{2}))^2} = 2\sqrt{1/4 - 1/8} = 2\sqrt{1/8} = 2 \cdot \frac{1}{2\sqrt{2}} = \frac{1}{\sqrt{2}}$.

(B) Let the square $ABCD$ have vertices $A(0,0)$,$B(4,0)$,$C(4,4)$,and $D(0,4)$.
The diagonal $AC$ lies on the line $y=x$.
Since $AEFG$ is a square of side $2$ with $E$ on $AB$ and $F$ on $AC$,the coordinates are $A(0,0)$,$E(2,0)$,$G(0,2)$,and $F(2,2)$.
The circle touches the lines $BC$ $(x=4)$ and $CD$ $(y=4)$.
Let the center of the circle be $O(h,k)$. Since it touches $x=4$ and $y=4$ with radius $r$,the center is $O(4-r, 4-r)$.
The circle passes through $F(2,2)$,so the distance $OF=r$.
Using the distance formula: $(4-r-2)^2 + (4-r-2)^2 = r^2$.
$(2-r)^2 + (2-r)^2 = r^2$.
$2(4 - 4r + r^2) = r^2$.
$8 - 8r + 2r^2 = r^2$.
$r^2 - 8r + 8 = 0$.

(C) Let the point be $P(x, y)$.
According to the problem,the ratio of distances from $P(x, y)$ to $(2, 1)$ and $(1, 3)$ is $5:4$.
So,$\frac{\sqrt{(x-2)^2 + (y-1)^2}}{\sqrt{(x-1)^2 + (y-3)^2}} = \frac{5}{4}$.
Squaring both sides,we get $\frac{(x-2)^2 + (y-1)^2}{(x-1)^2 + (y-3)^2} = \frac{25}{16}$.
$16(x^2 - 4x + 4 + y^2 - 2y + 1) = 25(x^2 - 2x + 1 + y^2 - 6y + 9)$.
$16x^2 - 64x + 80 + 16y^2 - 32y + 16 = 25x^2 - 50x + 25 + 25y^2 - 150y + 225$.
Rearranging the terms,we get $9x^2 + 9y^2 + 14x - 118y + 170 = 0$.
Comparing this with $ax^2 + by^2 + cxy + dx + ey + 170 = 0$,we get $a=9, b=9, c=0, d=14, e=-118$.
Now,calculate $a^2 + 2b + 3c + 4d + e = (9)^2 + 2(9) + 3(0) + 4(14) - 118$.
$= 81 + 18 + 0 + 56 - 118 = 155 - 118 = 37$.

(A) The equation of the circle is $x^2+y^2=169$,which is of the form $x^2+y^2=r^2$ with $r=13$.
The director circle of a circle $x^2+y^2=r^2$ is the locus of points from which mutually perpendicular tangents can be drawn to the circle,given by $x^2+y^2=2r^2$.
For the given circle,$r^2=169$,so the director circle is $x^2+y^2=2(169) = 338$.
Thus,$Statement-2$ is True.
Now,check if the point $(17,7)$ lies on the director circle $x^2+y^2=338$:
$17^2 + 7^2 = 289 + 49 = 338$.
Since the point $(17,7)$ satisfies the equation of the director circle,the tangents drawn from this point to the circle are mutually perpendicular.
Thus,$Statement-1$ is True and $Statement-2$ is the correct explanation for $Statement-1$.

(C) Given,$RS$ is the diameter of $x^2+y^2=1$. Let $P = (\cos \theta, \sin \theta)$.
The tangent at $P$ is $x \cos \theta + y \sin \theta = 1$. The tangent at $S(1,0)$ is $x = 1$.
Solving these,$Q = \left(1, \frac{1-\cos \theta}{\sin \theta}\right) = \left(1, \tan \frac{\theta}{2}\right)$.
The line through $Q$ parallel to $RS$ (the $x$-axis) is $y = \tan \frac{\theta}{2}$.
The normal at $P$ is $y = x \tan \theta$.
Let $E = (h, k)$. Then $k = \tan \frac{\theta}{2}$ and $k = h \tan \theta$.
Using $\tan \theta = \frac{2 \tan(\theta/2)}{1 - \tan^2(\theta/2)}$,we get $k = h \frac{2k}{1-k^2}$.
Thus,$1-k^2 = 2h$,or the locus is $2x = 1-y^2$.
Checking the points:
For $x = 1/3$,$1-y^2 = 2/3$ $\Rightarrow y^2 = 1/3$ $\Rightarrow y = \pm 1/\sqrt{3}$.
Thus,points $(1/3, 1/\sqrt{3})$ and $(1/3, -1/\sqrt{3})$ lie on the locus.
For $x = 1/4$,$1-y^2 = 2/4 = 1/2$ $\Rightarrow y^2 = 1/2$ $\Rightarrow y = \pm 1/\sqrt{2} \neq \pm 1/2$.
Therefore,the locus passes through points $A$ and $C$.

(D) Let $M \equiv (h, k)$. Since $MP$ and $MQ$ are tangents from $M$ to the circles $S_1$ and $S_2$ respectively,and $M$ is the point of contact,$MP = MQ$. Also,the angle $\angle PMQ = 90^{\circ}$.
Thus,the slope of $MP \times$ slope of $MQ = -1$.
$\left(\frac{k-7}{h+2}\right) \times \left(\frac{k+5}{h-2}\right) = -1$
$(k-7)(k+5) = -(h+2)(h-2)$
$k^2 - 2k - 35 = -(h^2 - 4) = -h^2 + 4$
$h^2 + k^2 - 2k - 39 = 0$. Thus,$E_1: x^2 + y^2 - 2y - 39 = 0$.
For $E_2$,let a chord of $E_1$ have midpoint $(h, k)$. The equation of the chord is $T = S_1$,i.e.,$xh + yk - (y + k) - 39 = h^2 + k^2 - 2k - 39$.
Since it passes through $R(1, 1)$,$h + k - (1 + k) - 39 = h^2 + k^2 - 2k - 39 \Rightarrow h^2 + k^2 - h - 2k + 1 = 0$.
Checking options: $(A)$ $(-2, 7)$ gives $4 + 49 - 14 - 39 = 0$,but $P$ is the point of tangency,not $M$. $(D)$ $(0, 3/2)$ gives $0 + 9/4 - 3 - 39 \neq 0$,so it does not lie in $E_1$. Thus $(D)$ is true.

(D) Given equations: $x^2+y^2-8x=0$ $(1)$ and $\frac{x^2}{9}-\frac{y^2}{4}=1$ $(2)$.
From $(2)$,$4x^2-9y^2=36 \Rightarrow 9y^2=4x^2-36$.
Substitute $y^2=8x-x^2$ from $(1)$ into $(2)$: $4x^2-9(8x-x^2)=36$.
$4x^2-72x+9x^2=36 \Rightarrow 13x^2-72x-36=0$.
$(13x+6)(x-6)=0$,so $x=6$ or $x=-\frac{6}{13}$.
For $x=6$,$y^2=8(6)-6^2=48-36=12$,so $y=\pm\sqrt{12}$.
Thus,$A=(6, \sqrt{12})$ and $B=(6, -\sqrt{12})$.
Let $P=(\alpha, \beta)$ be a point on $2x-3y+4=0$,so $2\alpha-3\beta+4=0$.
The centroid $(h, k)$ of $\triangle PAB$ is $h=\frac{6+6+\alpha}{3} = \frac{12+\alpha}{3}$ and $k=\frac{\sqrt{12}-\sqrt{12}+\beta}{3} = \frac{\beta}{3}$.
Then $\alpha=3h-12$ and $\beta=3k$.
Substitute into $2\alpha-3\beta+4=0$: $2(3h-12)-3(3k)+4=0$.
$6h-24-9k+4=0 \Rightarrow 6h-9k=20$.
The locus of the centroid is $6x-9y=20$.

(C) Let the coordinates of the other end of the diameter be $(t, 3-t)$ since it lies on the line $x+y=3$.
Let the centre of the circle be $(h, k)$.
The centre is the midpoint of the diameter with endpoints $(1, 1)$ and $(t, 3-t)$.
So,$h = \frac{1+t}{2}$ and $k = \frac{1+3-t}{2} = \frac{4-t}{2}$.
From these equations,we have $t = 2h-1$ and $t = 4-2k$.
Equating the two expressions for $t$:
$2h-1 = 4-2k$
$2h+2k = 5$.
Replacing $(h, k)$ with $(x, y)$,the locus of the centre is $2x+2y=5$.

(B) The given circle is $x^2+y^2=16$,which has radius $r=4$ and center at the origin $(0,0)$.
Let $P(h,k)$ be the point of intersection of the tangents.
The angle between the tangents is $60^{\circ}$,so the angle between the line joining the center to $P$ and the tangent is $30^{\circ}$.
In the right-angled triangle formed by the center,the point of tangency,and $P$,we have $\sin(30^{\circ}) = \frac{r}{OP}$.
$\frac{1}{2} = \frac{4}{\sqrt{h^2+k^2}}$.
$\sqrt{h^2+k^2} = 8$.
Squaring both sides,we get $h^2+k^2=64$.
Replacing $(h,k)$ with $(x,y)$,the locus is $x^2+y^2=64$.

(B) The locus of the point of intersection of perpendicular tangents to a circle is known as its director circle.
For a circle given by the equation $x^{2}+y^{2}=r^{2}$,the equation of the director circle is $x^{2}+y^{2}=2r^{2}$.
Here,the given circle is $x^{2}+y^{2}=16$,so $r^{2}=16$.
Substituting this into the director circle equation,we get $x^{2}+y^{2}=2(16) = 32$.
Thus,the required locus is $x^{2}+y^{2}=32$.

(B) Let the midpoint of the chord $AB$ be $C(x_{1}, y_{1})$. The origin is $O(0, 0)$.
Since the chord $AB$ subtends a right angle at the origin,$\angle AOB = 90^{\circ}$.
In $\Delta OAB$,$OA = OB = r = 2$ (radius of the circle).
Since $C$ is the midpoint of $AB$,$OC \perp AB$.
In $\Delta OCB$,$\angle COB = \frac{1}{2} \angle AOB = 45^{\circ}$.
Using trigonometry in $\Delta OCB$:
$\cos(45^{\circ}) = \frac{OC}{OB}$
$\frac{1}{\sqrt{2}} = \frac{\sqrt{x_{1}^{2} + y_{1}^{2}}}{2}$
Squaring both sides:
$\frac{1}{2} = \frac{x_{1}^{2} + y_{1}^{2}}{4}$
$x_{1}^{2} + y_{1}^{2} = 2$
Replacing $(x_{1}, y_{1})$ with $(x, y)$,the locus is $x^{2} + y^{2} = 2$.

(A) Let the centroid of $\triangle ABC$ be $(x, y)$.
The coordinates of the vertices are $A(\cos \alpha, \sin \alpha)$,$B(\sin \alpha, -\cos \alpha)$,and $C(1, 2)$.
The centroid $(x, y)$ is given by:
$x = \frac{\cos \alpha + \sin \alpha + 1}{3} \implies 3x - 1 = \cos \alpha + \sin \alpha$
$y = \frac{\sin \alpha - \cos \alpha + 2}{3} \implies 3y - 2 = \sin \alpha - \cos \alpha$
Squaring both equations:
$(3x - 1)^2 = (\cos \alpha + \sin \alpha)^2 = \cos^2 \alpha + \sin^2 \alpha + 2 \sin \alpha \cos \alpha = 1 + 2 \sin \alpha \cos \alpha$
$(3y - 2)^2 = (\sin \alpha - \cos \alpha)^2 = \sin^2 \alpha + \cos^2 \alpha - 2 \sin \alpha \cos \alpha = 1 - 2 \sin \alpha \cos \alpha$
Adding the two squared equations:
$(3x - 1)^2 + (3y - 2)^2 = 1 + 2 \sin \alpha \cos \alpha + 1 - 2 \sin \alpha \cos \alpha = 2$
$9x^2 - 6x + 1 + 9y^2 - 12y + 4 = 2$
$9x^2 + 9y^2 - 6x - 12y + 3 = 0$
Dividing by $3$:
$3(x^2 + y^2) - 2x - 4y + 1 = 0$.

(B) Let the coordinates of point $P$ be $(x, y)$.
Given $PA:PB = 2:1$,we have $PA^2 = 4PB^2$.
Using the distance formula,$PA^2 = (x-2)^2 + (y-1)^2$ and $PB^2 = (x-1)^2 + (y-2)^2$.
Substituting these into the equation:
$(x-2)^2 + (y-1)^2 = 4[(x-1)^2 + (y-2)^2]$
$x^2 - 4x + 4 + y^2 - 2y + 1 = 4[x^2 - 2x + 1 + y^2 - 4y + 4]$
$x^2 + y^2 - 4x - 2y + 5 = 4[x^2 + y^2 - 2x - 4y + 5]$
$x^2 + y^2 - 4x - 2y + 5 = 4x^2 + 4y^2 - 8x - 16y + 20$
Rearranging the terms to one side:
$3x^2 + 3y^2 - 4x - 14y + 15 = 0$.

(C) Let $P = (x, y)$. The given condition is $PA + PB = 4$.
The distance between $A(2, 2)$ and $B(2, -2)$ is $d = \sqrt{(2-2)^2 + (2 - (-2))^2} = \sqrt{0^2 + 4^2} = 4$.
Since the sum of the distances $PA + PB$ is equal to the distance between the fixed points $A$ and $B$ (i.e.,$PA + PB = AB = 4$),the point $P$ must lie on the line segment joining $A$ and $B$.
The points $A(2, 2)$ and $B(2, -2)$ both have an $x$-coordinate of $2$,so the line segment connecting them is a vertical line segment.
Therefore,the locus of $P$ is the segment of the vertical line $x = 2$ between $y = -2$ and $y = 2$.

(C) Let $P = (x, y)$. The given condition is $PA = 2PB$.
$\sqrt{(x-4)^2 + y^2} = 2\sqrt{(x+4)^2 + y^2}$.
Squaring both sides: $(x-4)^2 + y^2 = 4((x+4)^2 + y^2)$.
$x^2 - 8x + 16 + y^2 = 4(x^2 + 8x + 16 + y^2)$.
$x^2 - 8x + 16 + y^2 = 4x^2 + 32x + 64 + 4y^2$.
$3x^2 + 3y^2 + 40x + 48 = 0$.
Dividing by $3$: $x^2 + y^2 + \frac{40}{3}x + 16 = 0$.
This is a circle with center $O' = (-\frac{20}{3}, 0)$ and radius $r = \sqrt{(-\frac{20}{3})^2 - 16} = \sqrt{\frac{400}{9} - 16} = \sqrt{\frac{400-144}{9}} = \sqrt{\frac{256}{9}} = \frac{16}{3}$.
The given line is $3y - 3x - 20 = 0$,which can be written as $y - x - \frac{20}{3} = 0$.
Check if the center $(-\frac{20}{3}, 0)$ lies on the line: $0 - (-\frac{20}{3}) - \frac{20}{3} = 0$. Yes,it does.
Since the line passes through the center of the circle,the chord $CD$ is a diameter.
Distance $CD = 2r = 2 \times \frac{16}{3} = \frac{32}{3}$.

(C) Let the point $P$ be $(x, y)$.
Given that $AP + BP = 2$.
$\sqrt{(x-1)^2 + y^2} + \sqrt{x^2 + (y-1)^2} = 2$.
$\sqrt{(x-1)^2 + y^2} = 2 - \sqrt{x^2 + (y-1)^2}$.
Squaring both sides:
$(x-1)^2 + y^2 = 4 + x^2 + (y-1)^2 - 4\sqrt{x^2 + (y-1)^2}$.
$x^2 - 2x + 1 + y^2 = 4 + x^2 + y^2 - 2y + 1 - 4\sqrt{x^2 + (y-1)^2}$.
$-2x + 2y - 4 = -4\sqrt{x^2 + (y-1)^2}$.
Dividing by $-2$:
$x - y + 2 = 2\sqrt{x^2 + (y-1)^2}$.
Squaring both sides again:
$(x - y + 2)^2 = 4(x^2 + y^2 - 2y + 1)$.
$x^2 + y^2 + 4 - 2xy + 4x - 4y = 4x^2 + 4y^2 - 8y + 4$.
$3x^2 + 2xy + 3y^2 - 4x - 4y = 0$.

(A) Let the coordinates of the variable point $P$ be $(x, y)$.
The distance of $P(x, y)$ from $A(2, -2)$ is given by $PA = \sqrt{(x - 2)^2 + (y + 2)^2}$.
The distance of $P(x, y)$ from the $Y$-axis is $|x|$.
According to the problem,$PA = 2|x|$.
Squaring both sides,we get $PA^2 = 4x^2$.
$(x - 2)^2 + (y + 2)^2 = 4x^2$.
$x^2 - 4x + 4 + y^2 + 4y + 4 = 4x^2$.
Rearranging the terms,we get $3x^2 - y^2 + 4x - 4y - 8 = 0$.

(A) Let the centroid of $\triangle APQ$ be $(h, k)$.
Given vertices are $A(a, 0)$,$P(a \cos \theta, a \sin \theta)$,and $Q(b \sin \theta, -b \cos \theta)$.
The centroid $(h, k)$ is given by:
$h = \frac{a + a \cos \theta + b \sin \theta}{3} \implies 3h - a = a \cos \theta + b \sin \theta$
$k = \frac{0 + a \sin \theta - b \cos \theta}{3} \implies 3k = a \sin \theta - b \cos \theta$
Squaring and adding the two equations:
$(3h - a)^2 + (3k)^2 = (a \cos \theta + b \sin \theta)^2 + (a \sin \theta - b \cos \theta)^2$
$(3h - a)^2 + 9k^2 = a^2(\cos^2 \theta + \sin^2 \theta) + b^2(\sin^2 \theta + \cos^2 \theta) + 2ab \cos \theta \sin \theta - 2ab \sin \theta \cos \theta$
$(3h - a)^2 + 9k^2 = a^2 + b^2$
$9(h - \frac{a}{3})^2 + 9k^2 = a^2 + b^2$
$(h - \frac{a}{3})^2 + k^2 = \frac{a^2 + b^2}{9} = \left(\frac{\sqrt{a^2 + b^2}}{3}\right)^2$
Thus,the locus is a circle with centre at $\left(\frac{a}{3}, 0\right)$ and radius $\frac{\sqrt{a^2 + b^2}}{3}$.

(B) $1$. Find the midpoint $M$ of $AB$: $M = (\frac{4+2}{2}, \frac{3+5}{2}) = (3, 4)$.
$2$. The distance of $P(x, y)$ from $M(3, 4)$ is at most $5$ units: $(x-3)^2 + (y-4)^2 \leq 5^2$,which simplifies to $x^2 + y^2 - 6x - 8y \leq 0$.
$3$. The equation of line $AB$ is $y - 3 = \frac{5-3}{2-4}(x - 4)$ $\Rightarrow y - 3 = -1(x - 4)$ $\Rightarrow x + y - 7 = 0$.
$4$. The origin $(0, 0)$ satisfies $0 + 0 - 7 = -7 < 0$. Since $P$ is on the same side as the origin,$P$ must satisfy $x + y - 7 < 0$.
$5$. Combining these,the locus is $x^2 + y^2 - 6x - 8y \leq 0$ and $x + y - 7 < 0$.

(A) Let the third vertex be $P(x, y)$.
Since the triangle is right-angled at $P$,the angle $\angle P = 90^{\circ}$.
This implies that the point $P$ lies on a circle where the hypotenuse joining $(1, 2)$ and $(4, 5)$ is the diameter.
The equation of a circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is $(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0$.
Substituting the given points $(1, 2)$ and $(4, 5)$:
$(x-1)(x-4) + (y-2)(y-5) = 0$
$x^2 - 5x + 4 + y^2 - 7y + 10 = 0$
$x^2 + y^2 - 5x - 7y + 14 = 0$.
Note: The vertex $P$ cannot be $(1, 2)$ or $(4, 5)$ as it would not form a triangle.