Tangent and normal to a circle Questions in English

(B) The circle is $(x-2)^{2}+(y-3)^{2}=25$. The center is $(2,3)$ and the radius is $5$.
At point $(5,7)$,the slope of the radius is $m_r = \frac{7-3}{5-2} = \frac{4}{3}$.
The slope of the tangent is $m_t = -\frac{1}{m_r} = -\frac{3}{4}$.
The equation of the tangent is $y-7 = -\frac{3}{4}(x-5) \implies 4y-28 = -3x+15 \implies 3x+4y-43=0$.
The $x-$intercept of the tangent is found by setting $y=0$: $3x = 43 \implies x = \frac{43}{3}$.
The slope of the normal is $m_n = \frac{4}{3}$.
The equation of the normal is $y-7 = \frac{4}{3}(x-5) \implies 3y-21 = 4x-20 \implies 4x-3y+1=0$.
The $x-$intercept of the normal is found by setting $y=0$: $4x = -1 \implies x = -\frac{1}{4}$.
The triangle is formed by the points $(-\frac{1}{4}, 0)$,$(\frac{43}{3}, 0)$,and $(5, 7)$.
The base of the triangle on the $x-$axis is $b = \frac{43}{3} - (-\frac{1}{4}) = \frac{172+3}{12} = \frac{175}{12}$.
The height of the triangle is the $y-$coordinate of the point $(5,7)$,which is $h = 7$.
Area $A = \frac{1}{2} \times b \times h = \frac{1}{2} \times \frac{175}{12} \times 7 = \frac{1225}{24}$.
Therefore,$24A = 1225$.

(A) Let the point of contact on the parabola $y^{2}=4x$ be $A(t^{2}, 2t)$.
The equation of the tangent to the parabola at $A(t^{2}, 2t)$ is $ty = x + t^{2}$,which can be written as $x - ty + t^{2} = 0$.
Since this line is also tangent to the circle $(x-3)^{2} + y^{2} = 9$ with center $(3, 0)$ and radius $r=3$,the perpendicular distance from the center $(3, 0)$ to the line must be equal to the radius.
$\left| \frac{3 - t(0) + t^{2}}{\sqrt{1^{2} + (-t)^{2}}} \right| = 3$
$\left| 3 + t^{2} \right| = 3\sqrt{1 + t^{2}}$
Squaring both sides: $(3 + t^{2})^{2} = 9(1 + t^{2})$
$9 + t^{4} + 6t^{2} = 9 + 9t^{2}$
$t^{4} - 3t^{2} = 0$
$t^{2}(t^{2} - 3) = 0$
Since the points lie in the first quadrant,$t > 0$,so $t^{2} = 3$,which gives $t = \sqrt{3}$.
Thus,the point of contact on the parabola is $A(t^{2}, 2t) = (3, 2\sqrt{3})$. So,$a=3$ and $b=2\sqrt{3}$.
The equation of the tangent line is $x - \sqrt{3}y + 3 = 0$.
The point of contact on the circle $B(c, d)$ is the foot of the perpendicular from the center $(3, 0)$ to the tangent line $x - \sqrt{3}y + 3 = 0$.
Using the formula for the foot of the perpendicular $(c, d)$ from $(x_{1}, y_{1})$ to $Ax + By + C = 0$:
$\frac{c-3}{1} = \frac{d-0}{-\sqrt{3}} = -\frac{1(3) - \sqrt{3}(0) + 3}{1^{2} + (-\sqrt{3})^{2}} = -\frac{6}{4} = -\frac{3}{2}$
$c - 3 = -\frac{3}{2} \Rightarrow c = 3 - \frac{3}{2} = \frac{3}{2}$
$d = -\sqrt{3} \times (-\frac{3}{2}) = \frac{3\sqrt{3}}{2}$
We need to find $2(a+c) = 2(3 + \frac{3}{2}) = 2(\frac{9}{2}) = 9$.

(D) The equation of the tangent to the parabola $y^{2} = 4Ax$ at point $(x_{1}, y_{1})$ is $yy_{1} = 2A(x + x_{1})$.
Here,$4A = 8$,so $A = 2$.
At point $(2, -4)$,the tangent equation is $y(-4) = 4(x + 2)$.
$-4y = 4x + 8 \Rightarrow x + y + 2 = 0$.
This line $x + y + 2 = 0$ is also tangent to the circle $x^{2} + y^{2} = a$.
The perpendicular distance from the center $(0, 0)$ to the line $x + y + 2 = 0$ must equal the radius $\sqrt{a}$.
Distance $d = \frac{|0 + 0 + 2|}{\sqrt{1^{2} + 1^{2}}} = \frac{2}{\sqrt{2}} = \sqrt{2}$.
Since $d = \sqrt{a}$,we have $\sqrt{a} = \sqrt{2}$,which implies $a = 2$.

(C) Given the lines $L_{1}: 4x + 3y + 2 = 0$ and $L_{2}: 3x - 4y - 11 = 0$. The intersection point $Q$ of these two lines is the point of tangency.
Solving the system of equations:
$4x + 3y = -2$ (multiply by $4$): $16x + 12y = -8$
$3x - 4y = 11$ (multiply by $3$): $9x - 12y = 33$
Adding these gives $25x = 25$,so $x = 1$.
Substituting $x = 1$ into $4(1) + 3y = -2$,we get $3y = -6$,so $y = -2$. Thus,$Q = (1, -2)$.
The line $L_{1}$ is the normal to the circle at $Q$ because it passes through the center $P$. The slope of $L_{2}$ is $m = \frac{3}{4}$. The slope of the normal $L_{1}$ is $-\frac{4}{3}$.
The center $P$ lies on $L_{1}$ at a distance of $5$ units from $Q$ along the normal. The unit vector along the normal $L_{1}$ (direction $(3, -4)$) is $(\frac{3}{5}, -\frac{4}{5})$.
Since the circle lies below the $x$-axis,the center $P = Q + 5(\frac{3}{5}, -\frac{4}{5}) = (1 + 3, -2 - 4) = (4, -6)$.
Now,calculate the distance of $P(4, -6)$ from the line $5x - 12y + 51 = 0$:
Distance $= \left| \frac{5(4) - 12(-6) + 51}{\sqrt{5^2 + (-12)^2}} \right| = \left| \frac{20 + 72 + 51}{13} \right| = \frac{143}{13} = 11$.

(C) The equation of the circle is $x^{2}+y^{2}-2x-4y=0$. The center is $C(1, 2)$ and the radius $R = \sqrt{1^{2}+2^{2}} = \sqrt{5}$.
The tangent at $O(0,0)$ is $x(0)+y(0)-(x+0)-2(y+0)=0$,which simplifies to $x+2y=0$.
The tangent at $P(1+\sqrt{5}, 2)$ is $x(1+\sqrt{5})+y(2)-(x+1+\sqrt{5})-2(y+2)=0$,which simplifies to $x\sqrt{5} - 5 - \sqrt{5} = 0$,so $x = 1+\sqrt{5}$.
Substituting $x = 1+\sqrt{5}$ into $x+2y=0$,we get $1+\sqrt{5}+2y=0$,so $y = -\frac{1+\sqrt{5}}{2}$.
Thus,$Q = (1+\sqrt{5}, -\frac{1+\sqrt{5}}{2})$.
The length of the tangent $L = OQ = \sqrt{(1+\sqrt{5}-0)^{2} + (-\frac{1+\sqrt{5}}{2}-0)^{2}} = \sqrt{(1+\sqrt{5})^{2} + \frac{(1+\sqrt{5})^{2}}{4}} = \sqrt{\frac{5(1+\sqrt{5})^{2}}{4}} = \frac{\sqrt{5}(1+\sqrt{5})}{2} = \frac{\sqrt{5}+5}{2}$.
The area of $\triangle OPQ$ is given by $\frac{RL^{3}}{R^{2}+L^{2}}$.
Substituting $R=\sqrt{5}$ and $L=\frac{5+\sqrt{5}}{2}$,the area simplifies to $\frac{5+3\sqrt{5}}{2}$.

(B) The normals to the circle are given by the equations:
$y+2x=\sqrt{11}+7\sqrt{7}$ $(i)$
$2y+x=2\sqrt{11}+6\sqrt{7}$ $(ii)$
Solving these equations for the intersection point $(h, k)$,which is the center of the circle:
From $(i)$,$y = \sqrt{11}+7\sqrt{7}-2x$.
Substituting into $(ii)$: $2(\sqrt{11}+7\sqrt{7}-2x)+x = 2\sqrt{11}+6\sqrt{7}$
$2\sqrt{11}+14\sqrt{7}-4x+x = 2\sqrt{11}+6\sqrt{7}$
$-3x = -8\sqrt{7} \implies x = h = \frac{8\sqrt{7}}{3}$
Substituting $h$ back into $(i)$: $y = k = \sqrt{11}+7\sqrt{7}-2(\frac{8\sqrt{7}}{3}) = \sqrt{11} + \frac{21\sqrt{7}-16\sqrt{7}}{3} = \sqrt{11} + \frac{5\sqrt{7}}{3}$.
Thus,the center is $(h, k) = (\frac{8\sqrt{7}}{3}, \sqrt{11}+\frac{5\sqrt{7}}{3})$.
The radius $r$ is the perpendicular distance from the center $(h, k)$ to the tangent line $\sqrt{11}y-3x-(\frac{5\sqrt{77}}{3}+11) = 0$.
$r = \frac{|\sqrt{11}(\sqrt{11}+\frac{5\sqrt{7}}{3}) - 3(\frac{8\sqrt{7}}{3}) - (\frac{5\sqrt{77}}{3}+11)|}{\sqrt{(\sqrt{11})^{2}+(-3)^{2}}}$
$r = \frac{|11 + \frac{5\sqrt{77}}{3} - 8\sqrt{7} - \frac{5\sqrt{77}}{3} - 11|}{\sqrt{11+9}} = \frac{|-8\sqrt{7}|}{\sqrt{20}} = \frac{8\sqrt{7}}{2\sqrt{5}} = 4\sqrt{\frac{7}{5}}$.
Then $r^{2} = 16 \times \frac{7}{5} = \frac{112}{5}$.
Now,calculate $(5h-8k)^{2}+5r^{2}$:
$5h = 5(\frac{8\sqrt{7}}{3}) = \frac{40\sqrt{7}}{3}$
$8k = 8(\sqrt{11}+\frac{5\sqrt{7}}{3}) = 8\sqrt{11} + \frac{40\sqrt{7}}{3}$
$5h-8k = -8\sqrt{11} \implies (5h-8k)^{2} = 64 \times 11 = 704$.
$5r^{2} = 5 \times \frac{112}{5} = 112$.
$(5h-8k)^{2}+5r^{2} = 704 + 112 = 816$.

(A) The circle $C_{1}$ has its center at $(2, 0)$ and radius $2$. Its equation is $(x-2)^2 + y^2 = 4$,which simplifies to $x^2 + y^2 - 4x = 0$.
Given the line $y = 2x$,we substitute this into the circle equation: $x^2 + (2x)^2 - 4x = 0$,so $5x^2 - 4x = 0$. Thus,$x = 0$ (at $O$) or $x = 4/5$ (at $A$).
For $x = 4/5$,$y = 2(4/5) = 8/5$. So,$A = (4/5, 8/5)$.
The slope of $OA$ is $m = 2$. Let $\theta$ be the angle $OA$ makes with the $x$-axis,so $\tan \theta = 2$.
The tangent to $C_{2}$ at $A$ is perpendicular to the radius $OA$. Let the tangent line be $L$. Since $OA$ has slope $2$,the tangent $L$ has slope $-1/2$.
The equation of the tangent at $A(4/5, 8/5)$ is $y - 8/5 = -1/2(x - 4/5)$,which simplifies to $y - 8/5 = -1/2x + 2/5$,or $x + 2y = 4$.
The $x$-intercept $P$ is found by setting $y=0$,giving $x=4$,so $P = (4, 0)$.
The $y$-intercept $Q$ is found by setting $x=0$,giving $2y=4$,so $y=2$,$Q = (0, 2)$.
The distance $AP = \sqrt{(4 - 4/5)^2 + (0 - 8/5)^2} = \sqrt{(16/5)^2 + (-8/5)^2} = \sqrt{256/25 + 64/25} = \sqrt{320/25} = \frac{8\sqrt{5}}{5}$.
The distance $QA = \sqrt{(0 - 4/5)^2 + (2 - 8/5)^2} = \sqrt{(-4/5)^2 + (2/5)^2} = \sqrt{16/25 + 4/25} = \sqrt{20/25} = \frac{2\sqrt{5}}{5}$.
The ratio $QA : AP = \frac{2\sqrt{5}}{5} : \frac{8\sqrt{5}}{5} = 2 : 8 = 1 : 4$.

(C) The radius of the circle $r = \sqrt{\frac{169}{4}} = \frac{13}{2}$.
Let $C$ be the center of the circle and $M$ be the midpoint of the chord $AB$. Since $AB = 12$,$AM = 6$.
In the right-angled triangle $\triangle AMC$,$AC = \frac{13}{2}$ and $AM = 6$.
Using the Pythagorean theorem,$CM = \sqrt{AC^2 - AM^2} = \sqrt{(\frac{13}{2})^2 - 6^2} = \sqrt{\frac{169}{4} - 36} = \sqrt{\frac{169-144}{4}} = \sqrt{\frac{25}{4}} = \frac{5}{2}$.
Let $\angle ACM = \theta$. Then $\sin \theta = \frac{AM}{AC} = \frac{6}{13/2} = \frac{12}{13}$ and $\cos \theta = \frac{CM}{AC} = \frac{5/2}{13/2} = \frac{5}{13}$.
In $\triangle PAC$,$\angle PAC = 90^\circ$ because $PA$ is a tangent. Thus,$\triangle PAC$ is a right-angled triangle.
In $\triangle PAC$,$AM$ is the altitude to the hypotenuse $PC$. By property of right triangles,$AM^2 = PM \cdot MC$.
$6^2 = PM \cdot \frac{5}{2} \implies 36 = PM \cdot \frac{5}{2} \implies PM = \frac{72}{5}$.
The distance of point $P$ from chord $AB$ is $PM$.
Therefore,$5(PM) = 5 \cdot \frac{72}{5} = 72$.

(C) Let $O$ be the center of the circle. $OR$ is the angle bisector of $\angle PRQ$. Let $M$ be the intersection of $PQ$ and $OR$. Since $RP=RQ$,$\triangle RPQ$ is an isosceles triangle,and $RM \perp PQ$ and $PM = MQ = \frac{1}{2} PQ = 3$.
In right-angled $\triangle RPM$,by Pythagoras theorem:
$RM^2 = PR^2 - PM^2 = 5^2 - 3^2 = 25 - 9 = 16 \Rightarrow RM = 4$.
Let $\angle PRM = \theta$. Then $\tan \theta = \frac{PM}{RM} = \frac{3}{4}$.
In right-angled $\triangle OPR$ (where $\angle OPR = 90^\circ$ because $PR$ is a tangent),$\angle ORP = \theta$. Thus,$\tan \theta = \frac{OP}{PR} = \frac{r}{5}$.
Equating the two expressions for $\tan \theta$:
$\frac{r}{5} = \frac{3}{4} \Rightarrow r = \frac{15}{4}$.

(A) Since $PR$ is the diameter,$\angle PQR = 90^\circ$. Thus,the product of the slopes of $PQ$ and $QR$ is $-1$.
$m_{PQ} = \frac{10-2}{9-(-3)} = \frac{8}{12} = \frac{2}{3}$.
$m_{QR} = \frac{4-10}{\alpha-9} = \frac{-6}{\alpha-9}$.
Since $m_{PQ} \cdot m_{QR} = -1$,we have $\frac{2}{3} \cdot \left(\frac{-6}{\alpha-9}\right) = -1$ $\Rightarrow \frac{-4}{\alpha-9} = -1$ $\Rightarrow \alpha-9 = 4$ $\Rightarrow \alpha = 13$.
So,$R = (13, 4)$. The center $O$ of the circle is the midpoint of $PR$,$O = \left(\frac{-3+13}{2}, \frac{2+4}{2}\right) = (5, 3)$.
The tangent at $Q(9, 10)$ is perpendicular to the radius $OQ$. $m_{OQ} = \frac{10-3}{9-5} = \frac{7}{4}$.
Slope of tangent $QS = -\frac{4}{7}$.
Equation of $QS$: $y-10 = -\frac{4}{7}(x-9)$ $\Rightarrow 7y - 70 = -4x + 36$ $\Rightarrow 4x + 7y = 106 \quad (1)$.
The tangent at $R(13, 4)$ is perpendicular to the radius $OR$. $m_{OR} = \frac{4-3}{13-5} = \frac{1}{8}$.
Slope of tangent $RS = -8$.
Equation of $RS$: $y-4 = -8(x-13)$ $\Rightarrow y-4 = -8x + 104$ $\Rightarrow 8x + y = 108 \quad (2)$.
Solving $(1)$ and $(2)$: From $(2)$,$y = 108 - 8x$. Substitute into $(1)$:
$4x + 7(108 - 8x) = 106$ $\Rightarrow 4x + 756 - 56x = 106$ $\Rightarrow 52x = 650$ $\Rightarrow x = \frac{650}{52} = 12.5 = \frac{25}{2}$.
$y = 108 - 8(\frac{25}{2}) = 108 - 100 = 8$.
$S = (12.5, 8)$. Since $S$ lies on $2x - ky = 1$:
$2(12.5) - k(8) = 1$ $\Rightarrow 25 - 8k = 1$ $\Rightarrow 8k = 24$ $\Rightarrow k = 3$.

(D) The equation of the circle is $x^2 + y^2 - 3x + 10y - 15 = 0$.
The equation of the tangent at $A(x_1, y_1)$ is $xx_1 + yy_1 - \frac{3}{2}(x + x_1) + 5(y + y_1) - 15 = 0$.
For $A(4, -11)$: $4x - 11y - \frac{3}{2}(x + 4) + 5(y - 11) - 15 = 0$ $\Rightarrow 8x - 22y - 3x - 12 + 10y - 110 - 30 = 0$ $\Rightarrow 5x - 12y - 152 = 0$.
For $B(8, -5)$: $8x - 5y - \frac{3}{2}(x + 8) + 5(y - 5) - 15 = 0$ $\Rightarrow 16x - 10y - 3x - 24 + 10y - 50 - 30 = 0$ $\Rightarrow 13x = 104$ $\Rightarrow x = 8$.
Substituting $x = 8$ into the first tangent equation: $5(8) - 12y - 152 = 0$ $\Rightarrow 40 - 152 = 12y$ $\Rightarrow 12y = -112$ $\Rightarrow y = -\frac{28}{3}$.
So,$C = (8, -\frac{28}{3})$.
The line $AB$ passes through $(4, -11)$ and $(8, -5)$. The slope $m = \frac{-5 - (-11)}{8 - 4} = \frac{6}{4} = \frac{3}{2}$.
The equation of line $AB$ is $y + 5 = \frac{3}{2}(x - 8)$ $\Rightarrow 2y + 10 = 3x - 24$ $\Rightarrow 3x - 2y - 34 = 0$.
The radius $r$ is the perpendicular distance from $C(8, -\frac{28}{3})$ to the line $3x - 2y - 34 = 0$:
$r = \frac{|3(8) - 2(-\frac{28}{3}) - 34|}{\sqrt{3^2 + (-2)^2}} = \frac{|24 + \frac{56}{3} - 34|}{\sqrt{13}} = \frac{|\frac{56}{3} - 10|}{\sqrt{13}} = \frac{|\frac{26}{3}|}{\sqrt{13}} = \frac{26}{3 \sqrt{13}} = \frac{2 \sqrt{13}}{3}$.

(A) The curves are $S_1: y^2=2x$ and $S_2: x^2+y^2=4x$.
The point $P(2,2)$ lies on both curves.
Tangent $T_1$ to $S_1$ at $P(2,2)$ is given by $y(2) = x+2$,which simplifies to $x-2y+2=0$.
Tangent $T_2$ to $S_2$ at $P(2,2)$ is given by $x(2)+y(2) = 2(x+2)$,which simplifies to $2x+2y = 2x+4$,or $y=2$.
The third line is $L_3: x+y+2=0$.
To find the vertices of the triangle:
Intersection of $T_1$ and $T_2$: $x-2(2)+2=0 \Rightarrow x=2$. So $P(2,2)$.
Intersection of $T_1$ and $L_3$: $x-2y+2=0$ and $x+y+2=0$. Subtracting gives $3y=0 \Rightarrow y=0, x=-2$. So $Q(-2,0)$.
Intersection of $T_2$ and $L_3$: $y=2$ and $x+y+2=0$ $\Rightarrow x+2+2=0$ $\Rightarrow x=-4$. So $R(-4,2)$.
The vertices are $P(2,2)$,$Q(-2,0)$,and $R(-4,2)$.
Side lengths:
$PQ = \sqrt{(2 - (-2))^2 + (2-0)^2} = \sqrt{4^2 + 2^2} = \sqrt{16+4} = \sqrt{20}$.
$QR = \sqrt{(-2 - (-4))^2 + (0-2)^2} = \sqrt{2^2 + (-2)^2} = \sqrt{4+4} = \sqrt{8}$.
$RP = \sqrt{(2 - (-4))^2 + (2-2)^2} = \sqrt{6^2 + 0^2} = 6$.
Area of $\Delta PQR = \frac{1}{2} |x_P(y_Q - y_R) + x_Q(y_R - y_P) + x_R(y_P - y_Q)| = \frac{1}{2} |2(0-2) + (-2)(2-2) + (-4)(2-0)| = \frac{1}{2} |-4 + 0 - 8| = \frac{1}{2} |-12| = 6$.
Circumradius $r = \frac{abc}{4\Delta} = \frac{\sqrt{20} \cdot \sqrt{8} \cdot 6}{4 \cdot 6} = \frac{\sqrt{160}}{4} = \frac{4\sqrt{10}}{4} = \sqrt{10}$.
Therefore,$r^2 = 10$.

(D) The equation of the circle is $x^2 + y^2 - 2x + y - 5 = 0$.
Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -1$,$f = \frac{1}{2}$,and $c = -5$.
The center of the circle is $C(-g, -f) = (1, -\frac{1}{2})$.
The radius $r$ is $\sqrt{g^2 + f^2 - c} = \sqrt{1 + \frac{1}{4} + 5} = \sqrt{\frac{25}{4}} = \frac{5}{2}$.
Let $L$ be the length of the tangent $PR = QR$. $L = \sqrt{S_1} = \sqrt{(\frac{9}{4})^2 + 2^2 - 2(\frac{9}{4}) + 2 - 5} = \sqrt{\frac{81}{16} + 4 - \frac{9}{2} + 2 - 5} = \sqrt{\frac{81 - 72 + 16}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4}$.
The distance $d$ from the center $C$ to the point $R$ is $d = \sqrt{(\frac{9}{4} - 1)^2 + (2 + \frac{1}{2})^2} = \sqrt{(\frac{5}{4})^2 + (\frac{5}{2})^2} = \sqrt{\frac{25}{16} + \frac{25}{4}} = \sqrt{\frac{125}{16}} = \frac{5\sqrt{5}}{4}$.
The area of $\triangle PQR$ is given by $\frac{r L^3}{r^2 + L^2} = \frac{(\frac{5}{2}) \cdot (\frac{5}{4})^3}{(\frac{5}{2})^2 + (\frac{5}{4})^2} = \frac{\frac{5}{2} \cdot \frac{125}{64}}{\frac{25}{4} + \frac{25}{16}} = \frac{\frac{625}{128}}{\frac{100 + 25}{16}} = \frac{625}{128} \cdot \frac{16}{125} = \frac{5}{8}$.

(B) The equation of the circle is $x^2+y^2-6x+4y+8=0$. The center of the circle is $C(3, -2)$.
Since $OP$ and $OQ$ are tangents from the origin $O(0, 0)$ to the circle,the angle $\angle OPO = 90^\circ$ and $\angle OQO = 90^\circ$.
Thus,$OP$ and $OQ$ subtend a right angle at the origin $O$ and at the center $C(3, -2)$.
The circle with diameter $OC$ is the circumcircle of $\triangle OPQ$.
The equation of the circle with diameter $OC$ where $O(0, 0)$ and $C(3, -2)$ is:
$(x-0)(x-3) + (y-0)(y+2) = 0$
$x^2 - 3x + y^2 + 2y = 0$
$x^2 + y^2 - 3x + 2y = 0$
Given that this circle passes through $(\alpha, \frac{1}{2})$,we substitute these coordinates into the equation:
$\alpha^2 + (\frac{1}{2})^2 - 3\alpha + 2(\frac{1}{2}) = 0$
$\alpha^2 + \frac{1}{4} - 3\alpha + 1 = 0$
$\alpha^2 - 3\alpha + \frac{5}{4} = 0$
$4\alpha^2 - 12\alpha + 5 = 0$
$4\alpha^2 - 10\alpha - 2\alpha + 5 = 0$
$2\alpha(2\alpha - 5) - 1(2\alpha - 5) = 0$
$(2\alpha - 1)(2\alpha - 5) = 0$
Therefore,$\alpha = \frac{1}{2}$ or $\alpha = \frac{5}{2}$.
Comparing with the options,the correct value is $\frac{5}{2}$.

(B) The equation of the ellipse is $15x^2 + 19y^2 = 285$,which can be written as $\frac{x^2}{19} + \frac{y^2}{15} = 1$.
Here,$a^2 = 19$ and $b^2 = 15$.
The equation of a tangent to the ellipse with slope $m$ is $y = mx \pm \sqrt{a^2m^2 + b^2}$,which is $y = mx \pm \sqrt{19m^2 + 15}$.
This line is also a tangent to the circle $x^2 + y^2 = 4^2 = 16$,so its perpendicular distance from the center $(0,0)$ must be equal to the radius $4$.
$\left| \frac{\pm \sqrt{19m^2 + 15}}{\sqrt{m^2 + 1}} \right| = 4$.
Squaring both sides,we get $\frac{19m^2 + 15}{m^2 + 1} = 16$.
$19m^2 + 15 = 16m^2 + 16$,which simplifies to $3m^2 = 1$,so $m = \pm \frac{1}{\sqrt{3}}$.
The angle $\theta$ that the tangent makes with the $x$-axis is given by $\tan \theta = |m| = \frac{1}{\sqrt{3}}$,so $\theta = \frac{\pi}{6}$.
The minor axis of the ellipse is the $y$-axis (since $b^2 < a^2$).
The angle between the tangent and the $y$-axis is $\frac{\pi}{2} - \theta = \frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3}$.

(A) The centre $(\alpha, \beta)$ lies on the normal line $4x + 3y = 1$,so $4\alpha + 3\beta = 1$,which gives $\beta = \frac{1 - 4\alpha}{3}$.
The distance from the centre $(\alpha, \beta)$ to the two tangents $3x + 4y - 24 = 0$ and $3x - 4y - 32 = 0$ must be equal to the radius $r$.
Thus,$r = \left| \frac{3\alpha + 4\beta - 24}{5} \right| = \left| \frac{3\alpha - 4\beta - 32}{5} \right|$.
Substituting $\beta = \frac{1 - 4\alpha}{3}$:
$r = \left| \frac{3\alpha + 4(\frac{1 - 4\alpha}{3}) - 24}{5} \right| = \left| \frac{9\alpha + 4 - 16\alpha - 72}{15} \right| = \left| \frac{-7\alpha - 68}{15} \right|$.
$r = \left| \frac{3\alpha - 4(\frac{1 - 4\alpha}{3}) - 32}{5} \right| = \left| \frac{9\alpha - 4 + 16\alpha - 96}{15} \right| = \left| \frac{25\alpha - 100}{15} \right| = \left| \frac{5(\alpha - 4)}{3} \right|$.
Equating the two expressions for $r$:
$\left| \frac{-7\alpha - 68}{15} \right| = \left| \frac{25\alpha - 100}{15} \right|$ $\Rightarrow |-7\alpha - 68| = |25\alpha - 100|$.
Case $1$: $-7\alpha - 68 = 25\alpha - 100$ $\Rightarrow 32 = 32\alpha$ $\Rightarrow \alpha = 1$.
Then $\beta = \frac{1 - 4(1)}{3} = -1$. Radius $r = \left| \frac{25(1) - 100}{15} \right| = \left| \frac{-75}{15} \right| = 5$. Since $r < 8$,this is valid.
Case $2$: $-7\alpha - 68 = -(25\alpha - 100)$ $\Rightarrow -7\alpha - 68 = -25\alpha + 100$ $\Rightarrow 18\alpha = 168$ $\Rightarrow \alpha = \frac{28}{3}$.
Then $\beta = \frac{1 - 4(28/3)}{3} = \frac{3 - 112}{9} = -\frac{109}{9}$. Radius $r = \left| \frac{25(28/3) - 100}{15} \right| = \left| \frac{700/3 - 300/3}{15} \right| = \frac{400}{45} = \frac{80}{9} \approx 8.88$. Since $r > 8$,this is rejected.
Thus,$\alpha = 1, \beta = -1, r = 5$.
$\alpha - \beta + r = 1 - (-1) + 5 = 7$.

(D) The equation of the circle is $(x-\alpha)^2+(y-\beta)^2=50$,so the center is $C(\alpha, \beta)$ and the radius $r = \sqrt{50} = 5 \sqrt{2}$.
Since the circle touches the line $x+y=0$ at point $P$,the perpendicular distance from the center $C(\alpha, \beta)$ to the line $x+y=0$ must be equal to the radius $r$.
Distance $d = \frac{|\alpha+\beta|}{\sqrt{1^2+1^2}} = \frac{|\alpha+\beta|}{\sqrt{2}}$.
Setting $d = r$,we get $\frac{|\alpha+\beta|}{\sqrt{2}} = 5 \sqrt{2}$.
$|\alpha+\beta| = 5 \sqrt{2} \times \sqrt{2} = 10$.
Since $\alpha, \beta > 0$,we have $\alpha+\beta = 10$.
Therefore,$(\alpha+\beta)^2 = 10^2 = 100$.

(A) The center of the circle is the intersection of the two diameters $2x - 3y = 5$ and $3x - 4y = 7$.
Multiplying the first equation by $4$ and the second by $3$,we get $8x - 12y = 20$ and $9x - 12y = 21$.
Subtracting the first from the second gives $x = 1$. Substituting $x = 1$ into $2(1) - 3y = 5$,we get $-3y = 3$,so $y = -1$. Thus,the center $C$ is $(1, -1)$.
The line $AB$ passes through $A\left(-\frac{22}{7}, -4\right)$ and $B\left(-\frac{1}{7}, 3\right)$. The slope $m$ of $AB$ is $\frac{3 - (-4)}{-1/7 - (-22/7)} = \frac{7}{21/7} = \frac{7}{3} = \frac{7}{3}$.
The equation of line $AB$ is $y - 3 = \frac{7}{3}(x + \frac{1}{7})$,which simplifies to $3y - 9 = 7x + 1$,or $7x - 3y + 10 = 0$.
Since the line intersects the circle at only one point $P$,it is a tangent to the circle at $P$. Thus,$CP$ is perpendicular to $AB$.
The slope of $CP$ is $-\frac{1}{7/3} = -\frac{3}{7}$. The equation of line $CP$ passing through $C(1, -1)$ is $y + 1 = -\frac{3}{7}(x - 1)$,which simplifies to $7y + 7 = -3x + 3$,or $3x + 7y + 4 = 0$.
Solving the system $7x - 3y = -10$ and $3x + 7y = -4$:
Multiply the first by $7$ and the second by $3$: $49x - 21y = -70$ and $9x + 21y = -12$.
Adding these gives $58x = -82$,so $x = \alpha = -\frac{41}{29}$.
Substituting $x = -\frac{41}{29}$ into $3x + 7y = -4$: $3(-\frac{41}{29}) + 7y = -4 \implies -\frac{123}{29} + 7y = -4 \implies 7y = -4 + \frac{123}{29} = \frac{-116 + 123}{29} = \frac{7}{29} \implies y = \beta = \frac{1}{29}$.
Finally,$17\beta - \alpha = 17(\frac{1}{29}) - (-\frac{41}{29}) = \frac{17 + 41}{29} = \frac{58}{29} = 2$.

(B) The given circle is $x^2+y^2-6x-4y-11=0$. Comparing this with the general equation $x^2+y^2+2gx+2fy+c=0$,we get the center $C(-g, -f) = (3, 2)$.
Since $PA$ and $PB$ are tangents from point $P(1,8)$ to the circle,the radii $CA$ and $CB$ are perpendicular to the tangents at points $A$ and $B$ respectively. Thus,$\angle PAC = 90^\circ$ and $\angle PBC = 90^\circ$.
This implies that the points $A$ and $B$ lie on a circle with $PC$ as the diameter. The triangle $PAB$ is inscribed in this circle,so the circumcircle of $\triangle PAB$ is the circle with diameter $PC$.
The equation of a circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is $(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0$.
Using $P(1,8)$ and $C(3,2)$ as diameter endpoints:
$(x-1)(x-3) + (y-8)(y-2) = 0$
$x^2 - 4x + 3 + y^2 - 10y + 16 = 0$
$x^2 + y^2 - 4x - 10y + 19 = 0$.
Thus,the correct option is $B$.

(B) The parabola is $y^2 = 4x$,so $a = 1$. The end points of the latus rectum are $(1, 2)$ and $(1, -2)$.
The equation of the normal to the parabola $y^2 = 4ax$ at $(at^2, 2at)$ is $y = -tx + 2at + at^3$. For $y^2 = 4x$,$a = 1$,so the normal is $y = -tx + 2t + t^3$.
At $(1, 2)$,$t = 1$,so the normal is $y = -x + 2 + 1$,which simplifies to $x + y = 3$.
At $(1, -2)$,$t = -1$,so the normal is $y = -(-1)x + 2(-1) + (-1)^3$,which simplifies to $y = x - 3$,or $x - y = 3$.
The normals are $x + y - 3 = 0$ and $x - y - 3 = 0$. These lines are tangents to the circle $(x - 3)^2 + (y + 2)^2 = r^2$ with center $(3, -2)$.
The perpendicular distance from the center $(3, -2)$ to the line $x + y - 3 = 0$ is $r = \frac{|3 + (-2) - 3|}{\sqrt{1^2 + 1^2}} = \frac{|-2|}{\sqrt{2}} = \sqrt{2}$.
Thus,$r^2 = 2$.

(C) The center of the circle is $P(0, \alpha)$ and the radius is $r$. The line $2x - y = 0$ is tangent to the circle at $A_1$.
The perpendicular distance from the center $(0, \alpha)$ to the line $2x - y = 0$ is equal to the radius $r$:
$\frac{|2(0) - \alpha|}{\sqrt{2^2 + (-1)^2}} = r \implies \frac{|-\alpha|}{\sqrt{5}} = r \implies \alpha = r\sqrt{5}$.
Given $\alpha + r = 5 + \sqrt{5}$,substitute $\alpha = r\sqrt{5}$:
$r\sqrt{5} + r = 5 + \sqrt{5} \implies r(\sqrt{5} + 1) = \sqrt{5}(\sqrt{5} + 1) \implies r = \sqrt{5}$.
Then $\alpha = \sqrt{5} \cdot \sqrt{5} = 5$. So,the center is $P(0, 5)$.
The point $A_1$ is the foot of the perpendicular from $P(0, 5)$ to the line $2x - y = 0$:
$\frac{x - 0}{2} = \frac{y - 5}{-1} = -\frac{2(0) - 5}{2^2 + (-1)^2} = \frac{5}{5} = 1$.
$x = 2(1) = 2$ and $y - 5 = -1 \implies y = 4$. Thus,$A_1 = (2, 4)$.
Since $A_1 B_1$ is a diameter,the center $P(0, 5)$ is the midpoint of $A_1 B_1$. Let $B_1 = (x_1, y_1)$:
$\frac{x_1 + 2}{2} = 0 \implies x_1 = -2$.
$\frac{y_1 + 4}{2} = 5 \implies y_1 + 4 = 10 \implies y_1 = 6$.
Thus,$B_1 = (-2, 6)$.
Matching the results:
$(P) \alpha = 5 \rightarrow (4)$
$(Q) r = \sqrt{5} \rightarrow (2)$
$(R) A_1 = (2, 4) \rightarrow (5)$
$(S) B_1 = (-2, 6) \rightarrow (3)$
Therefore,$(P) \rightarrow (4), (Q) \rightarrow (2), (R) \rightarrow (5), (S) \rightarrow (3)$.

(B) The given curve is $x^2 = y - 6$,which can be written as $y = x^2 + 6$.
To find the slope of the tangent at $(1, 7)$,we differentiate $y$ with respect to $x$: $\frac{dy}{dx} = 2x$.
At the point $(1, 7)$,the slope $m = 2(1) = 2$.
The equation of the tangent line at $(1, 7)$ is $y - 7 = 2(x - 1)$,which simplifies to $2x - y + 5 = 0$.
This line touches the circle $x^2 + y^2 + 16x + 12y + C = 0$. The center of the circle is $(-8, -6)$ and the radius $r = \sqrt{(-8)^2 + (-6)^2 - C} = \sqrt{64 + 36 - C} = \sqrt{100 - C}$.
The perpendicular distance from the center $(-8, -6)$ to the line $2x - y + 5 = 0$ must be equal to the radius $r$:
$r = \frac{|2(-8) - (-6) + 5|}{\sqrt{2^2 + (-1)^2}} = \frac{|-16 + 6 + 5|}{\sqrt{5}} = \frac{|-5|}{\sqrt{5}} = \sqrt{5}$.
Squaring both sides,we get $r^2 = 5$,so $100 - C = 5$,which implies $C = 95$.

(D) Given curve is $xy = 100$. Differentiating with respect to $x$,we get $y + x \frac{dy}{dx} = 0$,so $\frac{dy}{dx} = -\frac{y}{x}$.
At point $(5, 20)$,the slope of the tangent $m_1 = -\frac{20}{5} = -4$.
The equation of the tangent is $y - 20 = -4(x - 5) \implies y - 20 = -4x + 20 \implies 4x + y - 40 = 0$.
The slope of the normal $m_2 = -\frac{1}{m_1} = \frac{1}{4}$.
The equation of the normal is $y - 20 = \frac{1}{4}(x - 5) \implies 4y - 80 = x - 5 \implies x - 4y + 75 = 0$.
The combined equation is $(4x + y - 40)(x - 4y + 75) = 0$.
Expanding this: $4x^2 - 16xy + 300x + xy - 4y^2 + 75y - 40x + 160y - 3000 = 0$.
Simplifying: $4x^2 - 15xy - 4y^2 + 260x + 235y - 3000 = 0$.
Comparing this with the given options,none of them match the result.

(D) Given the curve equation: $ax^2 = 2y^2 - b$.
Differentiating both sides with respect to $x$:
$2ax = 4y \frac{dy}{dx} \implies \frac{dy}{dx} = \frac{2ax}{4y} = \frac{ax}{2y}$.
At the point $(1, -1)$,the slope of the tangent is:
$m = \frac{a(1)}{2(-1)} = -\frac{a}{2}$.
The given line is $x + y = 0$,which can be written as $y = -x$. The slope of this line is $-1$.
Since the line touches the curve at $(1, -1)$,their slopes must be equal:
$-\frac{a}{2} = -1 \implies a = 2$.
Now,substitute $a = 2$ and the point $(1, -1)$ into the original curve equation:
$2(1)^2 = 2(-1)^2 - b$
$2 = 2 - b \implies b = 0$.
Thus,the values are $a = 2$ and $b = 0$.

(C) Given the parametric equations of the curve are $x=3 \cos \theta$ and $y=3 \sin \theta$.
Squaring and adding these equations,we get $x^2+y^2 = 9(\cos^2 \theta + \sin^2 \theta) = 9$,which represents a circle with radius $r=3$.
The point of tangency at $\theta = \frac{\pi}{4}$ is given by $x_1 = 3 \cos(\frac{\pi}{4}) = \frac{3}{\sqrt{2}}$ and $y_1 = 3 \sin(\frac{\pi}{4}) = \frac{3}{\sqrt{2}}$.
The equation of the tangent to the circle $x^2+y^2=r^2$ at point $(x_1, y_1)$ is $xx_1 + yy_1 = r^2$.
Substituting the values,we get $x(\frac{3}{\sqrt{2}}) + y(\frac{3}{\sqrt{2}}) = 9$.
Multiplying both sides by $\frac{\sqrt{2}}{3}$,we obtain $x+y = 3\sqrt{2}$.

(C) The normals to a circle intersect at its center. Given $(x - 1)(y - 2) = 0$,the center of the circle is $(1, 2)$.
Since the line $3x + 4y = 6$ is a tangent to the circle,the radius $r$ is the perpendicular distance from the center $(1, 2)$ to the line $3x + 4y - 6 = 0$.
$r = \frac{|3(1) + 4(2) - 6|}{\sqrt{3^2 + 4^2}} = \frac{|3 + 8 - 6|}{\sqrt{25}} = \frac{5}{5} = 1$.
The equation of the circle with center $(h, k) = (1, 2)$ and radius $r = 1$ is $(x - h)^2 + (y - k)^2 = r^2$.
$(x - 1)^2 + (y - 2)^2 = 1^2$.
$x^2 - 2x + 1 + y^2 - 4y + 4 = 1$.
$x^2 + y^2 - 2x - 4y + 4 = 0$.

(B) The equation of the circle is $x^2+y^2=4$. The point $P(\sqrt{3}, 1)$ lies on the circle.
The equation of the tangent at $P(x_1, y_1)$ is $xx_1+yy_1=r^2$,which gives $\sqrt{3}x+y=4$.
For the tangent,the $X$-intercept is found by setting $y=0$,giving $x = \frac{4}{\sqrt{3}}$. So the point is $A(\frac{4}{\sqrt{3}}, 0)$.
The normal at any point on a circle passes through the center $(0, 0)$. The line passing through $(0, 0)$ and $(\sqrt{3}, 1)$ is $y = \frac{1}{\sqrt{3}}x$,or $x - \sqrt{3}y = 0$.
The $X$-intercept of the normal is at the origin $O(0, 0)$.
The triangle is formed by the vertices $O(0, 0)$,$A(\frac{4}{\sqrt{3}}, 0)$,and $P(\sqrt{3}, 1)$.
The base of the triangle along the $X$-axis is $OA = \frac{4}{\sqrt{3}}$.
The height of the triangle is the $y$-coordinate of $P$,which is $1$.
Area $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \frac{4}{\sqrt{3}} \times 1 = \frac{2}{\sqrt{3}}$ sq. units.

(A) The given circle is $x^2+y^2=36$,which has center $(0,0)$ and radius $r=6$.
The given line is $5x+y=2$,which can be written as $y=-5x+2$. The slope of this line is $m_1=-5$.
The tangent is perpendicular to this line,so its slope $m$ must satisfy $m \times (-5) = -1$,which gives $m = \frac{1}{5}$.
The equation of a line with slope $m$ is $y = mx + c$,or $mx - y + c = 0$. Here,$\frac{1}{5}x - y + c = 0$,which simplifies to $x - 5y + 5c = 0$.
The distance from the center $(0,0)$ to the tangent line $x - 5y + 5c = 0$ must be equal to the radius $r=6$.
Using the distance formula $d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2+b^2}}$,we have $6 = \frac{|0 - 0 + 5c|}{\sqrt{1^2+(-5)^2}}$.
$6 = \frac{|5c|}{\sqrt{26}} \implies |5c| = 6\sqrt{26} \implies 5c = \pm 6\sqrt{26}$.
Substituting $5c$ back into the equation $x - 5y + 5c = 0$,we get $x - 5y \pm 6\sqrt{26} = 0$.

(A) The equation of the circle is $x^2+y^2=36$,so the radius $r=6$ and the center is $(0,0)$.
Any line perpendicular to $5x+y-2=0$ will have the form $x-5y+k=0$.
The distance from the center $(0,0)$ to the tangent line $x-5y+k=0$ must be equal to the radius $r=6$.
Using the distance formula $d = \frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}$,we get $6 = \frac{|1(0)-5(0)+k|}{\sqrt{1^2+(-5)^2}}$.
$6 = \frac{|k|}{\sqrt{26}}$,which implies $|k| = 6\sqrt{26}$.
Thus,$k = \pm 6\sqrt{26}$.
The equations of the tangents are $x-5y \pm 6\sqrt{26}=0$.

(D) The equation of the circle is $x^2+y^2+6x-4y-12=0$.
Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=3, f=-2, c=-12$.
The center $C$ is $(-g, -f) = (-3, 2)$ and the radius $r = \sqrt{g^2+f^2-c} = \sqrt{9+4+12} = \sqrt{25} = 5$.
The distance $d$ from point $P(-4, -5)$ to the center $C(-3, 2)$ is $d = \sqrt{(-3 - (-4))^2 + (2 - (-5))^2} = \sqrt{1^2 + 7^2} = \sqrt{50} = 5\sqrt{2}$.
Let $\theta$ be the angle between the tangent and the line joining the center to the point $P$. Then $\sin \theta = \frac{r}{d} = \frac{5}{5\sqrt{2}} = \frac{1}{\sqrt{2}}$,so $\theta = 45^\circ$ or $\frac{\pi}{4}$ radians.
The angle between the two tangents is $2\alpha$,where $\alpha = 90^\circ - \theta = 45^\circ$.
The area of the quadrilateral formed by the center,the two points of tangency,and point $P$ is $2 \times (\frac{1}{2} \times r \times \sqrt{d^2-r^2}) = r\sqrt{d^2-r^2} = 5 \times \sqrt{50-25} = 5 \times 5 = 25$.
The area of the two circular sectors is $2 \times (\frac{1}{2} r^2 \theta) = r^2 \theta = 25 \times \frac{\pi}{4}$.
The area enclosed between the tangents and the circle is $25 - \frac{25\pi}{4} = 25\left(1 - \frac{\pi}{4}\right) = 25\left(\frac{4-\pi}{4}\right)$ sq. units.

(C) The equation of the circle is $x^2+y^2=2^2$,so the radius $r=2$ and the center $O=(0,0)$.
Point $P$ is $(-4,0)$. The distance $OP = 4$.
In the right-angled triangle $\triangle OAP$ (where $\angle OAP = 90^\circ$ because $PA$ is a tangent),
$OA = r = 2$ and $OP = 4$.
Using the Pythagorean theorem,$AP = \sqrt{OP^2 - OA^2} = \sqrt{4^2 - 2^2} = \sqrt{16-4} = \sqrt{12} = 2\sqrt{3}$.
The area of $\triangle OAP = \frac{1}{2} \times OA \times AP = \frac{1}{2} \times 2 \times 2\sqrt{3} = 2\sqrt{3}$ sq. units.
Since the quadrilateral $PAOB$ is composed of two congruent triangles $\triangle OAP$ and $\triangle OBP$,the total area is $2 \times \text{Area}(\triangle OAP) = 2 \times 2\sqrt{3} = 4\sqrt{3}$ sq. units.

(A) The equation of the circle is $(x-7)^2+(y+1)^2=25$.
The center of the circle is $C(7, -1)$ and the radius $r = 5$.
The distance $d$ from the origin $O(0, 0)$ to the center $C(7, -1)$ is $d = \sqrt{7^2 + (-1)^2} = \sqrt{49+1} = \sqrt{50}$.
Let $\theta$ be the angle between the tangents. The angle between the tangents is given by $2 \alpha$,where $\sin(\alpha) = \frac{r}{d}$.
Thus,$\sin(\alpha) = \frac{5}{\sqrt{50}}$.
Therefore,the angle between the tangents is $2 \arcsin\left(\frac{5}{\sqrt{50}}\right)$.

(A) Let the equation of the tangent to the parabola $y^2 = 4x$ be $y = mx + \frac{a}{m}$,where $a = 1$. So,$y = mx + \frac{1}{m}$.
This can be rewritten as $mx - y + \frac{1}{m} = 0$.
This line is also a tangent to the circle $(x-3)^2 + y^2 = 9$,which has center $(3, 0)$ and radius $r = 3$.
The perpendicular distance from the center $(3, 0)$ to the line $mx - y + \frac{1}{m} = 0$ must be equal to the radius $3$.
$\frac{|m(3) - 0 + \frac{1}{m}|}{\sqrt{m^2 + (-1)^2}} = 3$
$|3m + \frac{1}{m}| = 3\sqrt{m^2 + 1}$
Squaring both sides: $(3m + \frac{1}{m})^2 = 9(m^2 + 1)$
$9m^2 + 6 + \frac{1}{m^2} = 9m^2 + 9$
$\frac{1}{m^2} = 3 \implies m^2 = \frac{1}{3} \implies m = \pm \frac{1}{\sqrt{3}}$.
Since the tangent is above the $X$-axis,we consider the positive slope $m = \frac{1}{\sqrt{3}}$.
Substituting $m = \frac{1}{\sqrt{3}}$ into the tangent equation: $y = \frac{1}{\sqrt{3}}x + \frac{1}{1/\sqrt{3}} = \frac{x}{\sqrt{3}} + \sqrt{3}$.
Multiplying by $\sqrt{3}$: $\sqrt{3}y = x + 3$,or $x - \sqrt{3}y + 3 = 0$.

(A) The equation of the tangent to the circle $x^2+y^2=5$ at the point $(x_1, y_1) = (1, -2)$ is given by $x x_1 + y y_1 = r^2$.
Substituting the values,we get $x(1) + y(-2) = 5$,which simplifies to $x - 2y = 5$.
For the second circle $x^2 + y^2 - 8x + 6y + 20 = 0$,the center is $(4, -3)$ and the radius is $\sqrt{4^2 + (-3)^2 - 20} = \sqrt{16 + 9 - 20} = \sqrt{5}$.
The point of contact on the second circle must lie on the line $x - 2y = 5$.
Checking the options:
For $(3, -1)$: $3 - 2(-1) = 3 + 2 = 5$. This point satisfies the tangent equation.
For $(3, 1)$: $3 - 2(1) = 1 \neq 5$.
For $(-3, -1)$: $-3 - 2(-1) = -1 \neq 5$.
For $(-3, 1)$: $-3 - 2(1) = -5 \neq 5$.
Thus,the point of contact is $(3, -1)$.

(B) The equation of the tangent to the circle $x^2+y^2=a^2$ at the point $P(\theta)$ is given by $x \cos \theta + y \sin \theta = a$.
Here,the radius $a = 5$ and the angle $\theta = \frac{\pi}{3}$.
Substituting these values into the equation:
$x \cos \frac{\pi}{3} + y \sin \frac{\pi}{3} = 5$
$x \left( \frac{1}{2} \right) + y \left( \frac{\sqrt{3}}{2} \right) = 5$
Multiplying the entire equation by $2$,we get:
$x + \sqrt{3} y = 10$.

(C) The given circle equation is $x^2+y^2-6x-5y-1=0$. Comparing this with $x^2+y^2+2gx+2fy+c=0$,we get $g=-3$ and $f=-2.5$.
The center of the circle is $(-g, -f) = (3, 2.5)$.
Let the other end of the diameter be $(x_2, y_2)$. Since the center $(3, 2.5)$ is the midpoint of the diameter with endpoints $(-1, 3)$ and $(x_2, y_2)$,we have:
$\frac{-1+x_2}{2} = 3 \Rightarrow x_2 = 7$
$\frac{3+y_2}{2} = 2.5 \Rightarrow y_2 = 2$
So,the other end is $(7, 2)$.
The equation of the tangent to the circle $x^2+y^2+2gx+2fy+c=0$ at point $(x_1, y_1)$ is given by $xx_1 + yy_1 + g(x+x_1) + f(y+y_1) + c = 0$.
Substituting $(x_1, y_1) = (7, 2)$,$g=-3$,$f=-2.5$,and $c=-1$:
$7x + 2y - 3(x+7) - 2.5(y+2) - 1 = 0$
$7x + 2y - 3x - 21 - 2.5y - 5 - 1 = 0$
$4x - 0.5y - 27 = 0$
Multiplying by $2$,we get $8x - y - 54 = 0$.

(A) The equation of the circle is $x^2+y^2=4$,so the radius $r = 2$.
The given line is $x+2y+3=0$,which can be written as $y = -\frac{1}{2}x - \frac{3}{2}$.
The slope of this line is $m = -\frac{1}{2}$.
Since the tangents are parallel to this line,their slope is also $m = -\frac{1}{2}$.
The equation of a tangent to the circle $x^2+y^2=r^2$ with slope $m$ is given by $y = mx \pm r\sqrt{1+m^2}$.
Substituting $m = -\frac{1}{2}$ and $r = 2$:
$y = -\frac{1}{2}x \pm 2\sqrt{1 + (-\frac{1}{2})^2}$
$y = -\frac{1}{2}x \pm 2\sqrt{1 + \frac{1}{4}}$
$y = -\frac{1}{2}x \pm 2\sqrt{\frac{5}{4}}$
$y = -\frac{1}{2}x \pm 2 \times \frac{\sqrt{5}}{2}$
$y = -\frac{1}{2}x \pm \sqrt{5}$
Multiplying by $2$:
$2y = -x \pm 2\sqrt{5}$
$x+2y = \pm 2\sqrt{5}$

(D) The circle $x^2+y^2=(8)^2$ has radius $r=8$ and center $(0,0)$.
The point $P$ on the circle corresponding to the parameter $\theta = \frac{2\pi}{3}$ has coordinates:
$P \equiv (8 \cos \frac{2\pi}{3}, 8 \sin \frac{2\pi}{3}) = (8 \times -\frac{1}{2}, 8 \times \frac{\sqrt{3}}{2}) = (-4, 4\sqrt{3})$.
The equation of the tangent to the circle $x^2+y^2=r^2$ at point $(x_1, y_1)$ is $xx_1 + yy_1 = r^2$.
Substituting $(x_1, y_1) = (-4, 4\sqrt{3})$ and $r^2 = 64$:
$-4x + 4\sqrt{3}y = 64$.
Dividing by $-4$:
$x - \sqrt{3}y = -16 \Rightarrow x - \sqrt{3}y + 16 = 0$.

(A) The given equation of the circle is $x^{2}+y^{2}-6x-4y+9=0$.
Comparing this with the general form $x^{2}+y^{2}+2gx+2fy+c=0$,we get $g=-3, f=-2, c=9$.
The center of the circle is $(-g, -f) = (3, 2)$ and the radius $r = \sqrt{g^{2}+f^{2}-c} = \sqrt{(-3)^{2}+(-2)^{2}-9} = \sqrt{9+4-9} = \sqrt{4} = 2$.
To check if the line $4x+3y-8=0$ is a tangent,we calculate the perpendicular distance $d$ from the center $(3, 2)$ to the line:
$d = \left|\frac{4(3)+3(2)-8}{\sqrt{4^{2}+3^{2}}}\right| = \left|\frac{12+6-8}{\sqrt{16+9}}\right| = \left|\frac{10}{5}\right| = 2$.
Since the perpendicular distance $d$ is equal to the radius $r$ $(d=r=2)$,the line is a tangent to the circle.

(C) The equation of the circle is $x^{2}+y^{2}-6x+4y-12=0$.
Comparing with $x^{2}+y^{2}+2gx+2fy+c=0$,we get $g=-3, f=2, c=-12$.
The center is $(-g, -f) = (3, -2)$ and the radius $r = \sqrt{g^{2}+f^{2}-c} = \sqrt{9+4+12} = \sqrt{25} = 5$.
Any line parallel to $4x+3y+5=0$ is of the form $4x+3y+k=0$.
The distance from the center $(3, -2)$ to the tangent line is equal to the radius:
$\frac{|4(3)+3(-2)+k|}{\sqrt{4^{2}+3^{2}}} = 5$
$\frac{|12-6+k|}{5} = 5$
$|6+k| = 25$
$6+k = 25$ or $6+k = -25$
$k = 19$ or $k = -31$.
Thus,the equations of the tangents are $4x+3y+19=0$ and $4x+3y-31=0$.

(A) The equation of the circle is $x^{2} + y^{2} = r^{2}$.
Any point $P$ on the circle can be represented as $(r \cos \theta, r \sin \theta)$.
The slope of the tangent at $P$ is given by differentiating $x^{2} + y^{2} = r^{2}$ with respect to $x$:
$2x + 2y \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{x}{y}$.
At point $P(r \cos \theta, r \sin \theta)$,the slope of the tangent $m_{t} = -\frac{r \cos \theta}{r \sin \theta} = -\cot \theta$.
The slope of the normal $m_{n}$ is $-\frac{1}{m_{t}} = \frac{1}{\cot \theta} = \tan \theta = \frac{\sin \theta}{\cos \theta}$.
The equation of the normal at $(r \cos \theta, r \sin \theta)$ is:
$y - r \sin \theta = \frac{\sin \theta}{\cos \theta} (x - r \cos \theta)$.
$y \cos \theta - r \sin \theta \cos \theta = x \sin \theta - r \sin \theta \cos \theta$.
$x \sin \theta - y \cos \theta = 0$.

(B) The equation of the pair of tangents from a point $(x_{1}, y_{1})$ to the circle $S: x^{2}+y^{2}-r^{2}=0$ is given by $SS_{1}=T^{2}$.
Here,$S = x^{2}+y^{2}-4$,$S_{1} = 3^{2}+2^{2}-4 = 9$,and $T = 3x+2y-4$.
Substituting these,we get $(x^{2}+y^{2}-4)(9) = (3x+2y-4)^{2}$.
Expanding the equation: $9x^{2}+9y^{2}-36 = 9x^{2}+4y^{2}+16+12xy-24x-16y$.
Simplifying: $5y^{2}-12xy+24x+16y-52 = 0$.
For a pair of lines $ax^{2}+2hxy+by^{2}+2gx+2fy+c=0$,the slopes $m_{1}, m_{2}$ satisfy $bm^{2}+2hm+a=0$ (by dividing by $x^{2}$ and setting $m=y/x$).
Here,$b=5$,$2h=-12$ (so $h=-6$),and $a=24$ (coefficient of $x^{2}$ is $0$,but we treat the quadratic in $y/x$).
Actually,dividing $5y^{2}-12xy+24x+16y-52=0$ by $x^{2}$ gives $5m^{2}-12m+24/x+16m/x-52/x^{2}=0$. This approach is complex.
Using the condition for tangents $y-2 = m(x-3) \Rightarrow mx-y+(2-3m)=0$.
The distance from $(0,0)$ to the line is $r=2$: $\frac{|2-3m|}{\sqrt{m^{2}+1}} = 2$.
Squaring both sides: $(2-3m)^{2} = 4(m^{2}+1)$.
$4-12m+9m^{2} = 4m^{2}+4$.
$5m^{2}-12m = 0$.
$m(5m-12) = 0$.
Thus,the slopes are $m_{1} = 0$ and $m_{2} = \frac{12}{5}$.
Therefore,$|m_{1}-m_{2}| = |0 - \frac{12}{5}| = \frac{12}{5}$.

(A) Given the circle equation $x^{2}+y^{2}=13$.
Since the abscissa of the points is $x=2$,we substitute this into the circle equation:
$2^{2}+y^{2}=13$
$4+y^{2}=13$
$y^{2}=9$
$y=\pm 3$.
So,the points of contact are $(2, 3)$ and $(2, -3)$.
The equation of the tangent to the circle $x^{2}+y^{2}=r^{2}$ at point $(x_{1}, y_{1})$ is given by $xx_{1}+yy_{1}=r^{2}$.
For point $(2, 3)$: $2x+3y=13$.
For point $(2, -3)$: $2x-3y=13$.
Thus,the equations are $2x+3y=13$ and $2x-3y=13$.

(C) The circle is $x^2 + y^2 = 25$,so its center $O$ is $(0, 0)$ and radius $r = 5$.
Length of the tangent segment from $P(1, 7)$ to the circle is given by $\sqrt{x_1^2 + y_1^2 - r^2} = \sqrt{1^2 + 7^2 - 25} = \sqrt{1 + 49 - 25} = \sqrt{25} = 5$.
In the quadrilateral $PQOR$,$PQ$ and $PR$ are tangents,so $PQ = PR = 5$.
Also,$OQ = OR = 5$ (radii of the circle).
Since the tangent is perpendicular to the radius at the point of contact,$\angle OQP = \angle ORP = 90^{\circ}$.
The quadrilateral $PQOR$ consists of two congruent right-angled triangles,$\triangle PQO$ and $\triangle PRO$.
Area of $\triangle PQO = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times PQ \times OQ = \frac{1}{2} \times 5 \times 5 = 12.5 \text{ sq. units}$.
Total area of quadrilateral $PQOR = 2 \times \text{Area of } \triangle PQO = 2 \times 12.5 = 25 \text{ sq. units}$.

(C) The circle is $x^2+y^2=2^2$,so the radius $r = 2$.
The distance $OP = \sqrt{(-4-0)^2 + (0-0)^2} = 4$.
In the right-angled triangle $\triangle PBO$,the length of the tangent $PB = \sqrt{OP^2 - OB^2} = \sqrt{4^2 - 2^2} = \sqrt{16-4} = \sqrt{12} = 2\sqrt{3}$.
The area of the quadrilateral $PAOB$ is the sum of the areas of $\triangle PBO$ and $\triangle PAO$.
Since $\triangle PBO \cong \triangle PAO$,the area of $PAOB = 2 \times \text{Area}(\triangle PBO)$.
Area of $\triangle PBO = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times PB \times OB = \frac{1}{2} \times 2\sqrt{3} \times 2 = 2\sqrt{3}$.
Therefore,the area of the quadrilateral $PAOB = 2 \times 2\sqrt{3} = 4\sqrt{3}$ sq. units.

(D) The equation of the circle is $x^2+y^2=25$,so the center is $O(0,0)$ and the radius $r=5$.
Let $P$ be the point $(1,7)$. The distance $OP = \sqrt{1^2+7^2} = \sqrt{1+49} = \sqrt{50} = 5\sqrt{2}$.
Let $T$ be the point of tangency. In the right-angled triangle $\triangle OPT$,$\sin(\angle OPT) = \frac{OT}{OP} = \frac{5}{5\sqrt{2}} = \frac{1}{\sqrt{2}}$.
Therefore,$\angle OPT = 45^{\circ}$.
The angle between the two tangents is $2 \times \angle OPT = 2 \times 45^{\circ} = 90^{\circ}$ or $\frac{\pi}{2}$.

(C) The equation of a tangent to the parabola $y^{2}=4x$ is given by $y=mx+\frac{a}{m}$,where $a=1$. Thus,$y=mx+\frac{1}{m}$.
This line is also tangent to the circle $(x-3)^{2}+y^{2}=9$,which has center $(3,0)$ and radius $r=3$.
The perpendicular distance from the center $(3,0)$ to the line $mx-y+\frac{1}{m}=0$ must equal the radius $3$:
$\frac{|m(3)-0+\frac{1}{m}|}{\sqrt{m^{2}+(-1)^{2}}}=3$
$|3m+\frac{1}{m}|=3\sqrt{m^{2}+1}$
Squaring both sides:
$(3m+\frac{1}{m})^{2}=9(m^{2}+1)$
$9m^{2}+6+\frac{1}{m^{2}}=9m^{2}+9$
$\frac{1}{m^{2}}=3$ $\Rightarrow m^{2}=\frac{1}{3}$ $\Rightarrow m=\pm\frac{1}{\sqrt{3}}$.
Since the tangent touches the parabola above the $x$-axis,we choose $m=\frac{1}{\sqrt{3}}$.
Substituting $m$ into the tangent equation:
$y=\frac{1}{\sqrt{3}}x+\sqrt{3}$
Multiplying by $\sqrt{3}$ gives $\sqrt{3}y=x+3$.

(B) The equation of the circle is $2x^{2} + 2y^{2} - 3x + 4y = 0$. Dividing by $2$,we get the standard form: $x^{2} + y^{2} - \frac{3}{2}x + 2y = 0$.
Here,$AB$ represents the length of the tangent from point $B(2, 1)$ to the circle.
The length of the tangent from a point $(x_{1}, y_{1})$ to a circle $x^{2} + y^{2} + 2gx + 2fy + c = 0$ is given by $\sqrt{x_{1}^{2} + y_{1}^{2} + 2gx_{1} + 2fy_{1} + c}$.
Substituting the coordinates of $B(2, 1)$ into the circle equation:
$AB = \sqrt{(2)^{2} + (1)^{2} - \frac{3}{2}(2) + 2(1)}$
$AB = \sqrt{4 + 1 - 3 + 2}$
$AB = \sqrt{4} = 2 \text{ units}$.

(C) Given the circle equation: $x^2 + y^2 = 16x$
Rearranging the terms: $x^2 - 16x + y^2 = 0$
Completing the square: $(x - 8)^2 + y^2 = 64$
Thus,the center is $(8, 0)$ and the radius $r = 8$.
Since the line $3x + 4y - k = 0$ touches the circle,the perpendicular distance from the center $(8, 0)$ to the line must equal the radius $r = 8$.
Using the distance formula: $d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$
$8 = \frac{|3(8) + 4(0) - k|}{\sqrt{3^2 + 4^2}}$
$8 = \frac{|24 - k|}{\sqrt{25}}$
$8 = \frac{|24 - k|}{5}$
$|24 - k| = 40$
Case $1$: $24 - k = 40 \Rightarrow k = -16$
Case $2$: $24 - k = -40 \Rightarrow k = 64$
Therefore,the values of $k$ are $-16$ and $64$.

(A) Any line passing through the point $(-5, -4)$ with slope $m$ is given by $y+4 = m(x+5)$,which simplifies to $mx - y + (5m - 4) = 0$.
For the circle $x^{2}+y^{2}+4x+6y+8=0$,the center is $(-2, -3)$ and the radius $r = \sqrt{(-2)^{2} + (-3)^{2} - 8} = \sqrt{4+9-8} = \sqrt{5}$.
Since the line is a tangent,the perpendicular distance from the center $(-2, -3)$ to the line must equal the radius:
$\frac{|m(-2) - (-3) + (5m - 4)|}{\sqrt{m^{2} + (-1)^{2}}} = \sqrt{5}$
$\frac{|3m - 1|}{\sqrt{m^{2} + 1}} = \sqrt{5}$
Squaring both sides: $(3m - 1)^{2} = 5(m^{2} + 1)$
$9m^{2} - 6m + 1 = 5m^{2} + 5$
$4m^{2} - 6m - 4 = 0$
$2m^{2} - 3m - 2 = 0$
$(2m + 1)(m - 2) = 0$
So,$m = 2$ or $m = -\frac{1}{2}$.
For $m = 2$: $2x - y + (5(2) - 4) = 0 \Rightarrow 2x - y + 6 = 0$.
For $m = -\frac{1}{2}$: $-\frac{1}{2}x - y + (5(-\frac{1}{2}) - 4) = 0$ $\Rightarrow -x - 2y - 5 - 8 = 0$ $\Rightarrow x + 2y + 13 = 0$.