Tangent and normal to a circle Questions in English

(D) The equation of a tangent to the parabola $y^2=x$ with slope $m$ is given by $y = mx + \frac{1}{4m}$.
For this to be a tangent to the circle $x^2+y^2-6y+4=0$,the perpendicular distance from the center $(0, 3)$ to the line $mx - y + \frac{1}{4m} = 0$ must equal the radius of the circle.
The circle equation can be written as $x^2 + (y-3)^2 = 5$,so the radius is $\sqrt{5}$.
Thus,$\frac{|m(0) - 3 + \frac{1}{4m}|}{\sqrt{m^2 + 1}} = \sqrt{5}$.
Squaring both sides: $\frac{(-3 + \frac{1}{4m})^2}{m^2 + 1} = 5$.
$9 - \frac{3}{2m} + \frac{1}{16m^2} = 5m^2 + 5$.
Multiplying by $16m^2$: $144m^2 - 24m + 1 = 80m^4 + 80m^2$.
$80m^4 - 64m^2 + 24m - 1 = 0$.
Testing $m = 1/2$: $80(1/16) - 64(1/4) + 24(1/2) - 1 = 5 - 16 + 12 - 1 = 0$.
So,$m = 1/2$ is a solution.
The equation of the tangent is $y = \frac{1}{2}x + \frac{1}{4(1/2)} = \frac{1}{2}x + \frac{1}{2}$.
Multiplying by $2$: $2y = x + 1$,which simplifies to $x - 2y + 1 = 0$.

(C) Given the curve equation: $x^2+y^2=a^2$.
Differentiating both sides with respect to $x$:
$2x + 2y \frac{dy}{dx} = 0$
$\frac{dy}{dx} = -\frac{x}{y}$.
Since the tangent is parallel to the $x$-axis,the slope of the tangent must be zero:
$\frac{dy}{dx} = 0$
$-\frac{x}{y} = 0 \implies x = 0$.
Substituting $x = 0$ into the curve equation:
$0^2 + y^2 = a^2 \implies y^2 = a^2$.
Since $y \geq 0$,we get $y = a$.
Thus,the point on the curve is $(0, a)$.

(A) The given circle is $x^2+y^2-2x=0$.
Comparing this with the general equation $x^2+y^2+2gx+2fy+c=0$,we get $g=-1$,$f=0$,and $c=0$.
The center of the circle is $(-g, -f) = (1, 0)$.
Any normal to a circle always passes through its center.
The normal is parallel to the line $x+2y-3=0$.
The slope of the line $x+2y-3=0$ is $m = -\frac{1}{2}$.
Since the normal is parallel to this line,the slope of the normal is also $m = -\frac{1}{2}$.
The equation of the line passing through $(1, 0)$ with slope $m = -\frac{1}{2}$ is given by $y - y_1 = m(x - x_1)$.
Substituting the values: $y - 0 = -\frac{1}{2}(x - 1)$.
$2y = -x + 1$,which simplifies to $x + 2y - 1 = 0$.

(B) Let the vertices of the square be $P, Q, R, S$. The sides are $x=-5$ and $y=4$. The center of the square is $C(3,-4)$.
The distance from the center $C(3,-4)$ to the line $x=-5$ is $|3 - (-5)| = 8$. Since the side length is $2 \times 8 = 16$,the other sides are $x=11$ and $y=-12$.
The vertices are $P(-5, 4)$,$S(11, 4)$,$R(11, -12)$,and $Q(-5, -12)$.
The vertices lying on $x=-5$ are $P(-5, 4)$ and $Q(-5, -12)$.
The circumcircle of the square has center $C(3,-4)$ and radius $r = \sqrt{(3 - (-5))^2 + (-4 - 4)^2} = \sqrt{8^2 + (-8)^2} = \sqrt{64 + 64} = \sqrt{128} = 8\sqrt{2}$.
The equation of the circle is $(x-3)^2 + (y+4)^2 = 128$.
The tangent at $P(-5, 4)$ is $(x-3)(-5-3) + (y+4)(4+4) = 128$ $\Rightarrow -8(x-3) + 8(y+4) = 128$ $\Rightarrow -x+3+y+4 = 16$ $\Rightarrow y-x = 9$.
The tangent at $Q(-5, -12)$ is $(x-3)(-5-3) + (y+4)(-12+4) = 128$ $\Rightarrow -8(x-3) - 8(y+4) = 128$ $\Rightarrow -x+3-y-4 = 16$ $\Rightarrow -x-y = 17$ $\Rightarrow x+y = -17$.
Solving $y-x=9$ and $x+y=-17$: Adding them gives $2y = -8 \Rightarrow y = -4$. Substituting $y=-4$ into $y-x=9$ gives $-4-x=9 \Rightarrow x = -13$.
The point of intersection is $(-13, -4)$.

(C) The center of the circle is $(-g, -f)$.
Since the line $x-2y-6=0$ is a normal,it passes through the center:
$-g - 2(-f) - 6 = 0$ $\Rightarrow -g + 2f = 6$ $\Rightarrow g = 2f - 6$.
Given that the line $y=2$ touches the circle,the distance from the center $(-g, -f)$ to the line $y=2$ is equal to the radius $r$.
$r = |-f - 2| = \sqrt{g^2 + f^2 - (-8)} = \sqrt{g^2 + f^2 + 8}$.
Squaring both sides: $(f+2)^2 = g^2 + f^2 + 8$.
$f^2 + 4f + 4 = g^2 + f^2 + 8 \Rightarrow 4f - 4 = g^2$.
Substitute $g = 2f - 6$ into the equation:
$4f - 4 = (2f - 6)^2 = 4f^2 - 24f + 36$.
$4f^2 - 28f + 40 = 0 \Rightarrow f^2 - 7f + 10 = 0$.
$(f-2)(f-5) = 0 \Rightarrow f = 2$ or $f = 5$.
If $f = 2$,then $g = 2(2) - 6 = -2$. The radius $r = |-2 - 2| = 4$.
If $f = 5$,then $g = 2(5) - 6 = 4$. The radius $r = |-5 - 2| = 7$.

(C) The equation of the circle is $x^2+y^2=1$. The radius $r=1$ and the center is $(0,0)$.
Let the slope of the tangent be $m$. The equation of a line passing through $P(\sqrt{2}, \sqrt{2})$ with slope $m$ is $y - \sqrt{2} = m(x - \sqrt{2})$,which simplifies to $mx - y + \sqrt{2}(1-m) = 0$.
Since this line is a tangent to the circle $x^2+y^2=1$,the perpendicular distance from the center $(0,0)$ to the line must be equal to the radius $r=1$.
Using the formula $d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2+b^2}}$,we get $1 = \frac{|m(0) - 1(0) + \sqrt{2}(1-m)|}{\sqrt{m^2+(-1)^2}}$.
This implies $1 = \frac{|\sqrt{2}(1-m)|}{\sqrt{m^2+1}}$.
Squaring both sides,we get $m^2+1 = 2(1-m)^2$.
$m^2+1 = 2(1 - 2m + m^2) = 2 - 4m + 2m^2$.
Rearranging the terms,we get $m^2 - 4m + 1 = 0$.
Using the quadratic formula $m = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$,we get $m = \frac{4 \pm \sqrt{16-4}}{2} = \frac{4 \pm \sqrt{12}}{2} = 2 \pm \sqrt{3}$.

(B) The equation of the circle is $x^2+y^2-6x+2y+c=0$. The center of the circle is $C = (3, -1)$.
Since $x-2y=0$ is a tangent at point $P$,the radius $CP$ is perpendicular to the tangent line.
The slope of the tangent $x-2y=0$ is $m_1 = 1/2$. Thus,the slope of the normal $CP$ is $m_2 = -2$.
The equation of the normal line passing through $C(3, -1)$ with slope $-2$ is $y - (-1) = -2(x - 3)$,which simplifies to $y+1 = -2x+6$,or $2x+y-5=0$.
Point $P$ is the intersection of the tangent $x-2y=0$ and the normal $2x+y-5=0$.
Solving these: $x=2y$,so $2(2y)+y-5=0 \implies 5y=5 \implies y=1$. Then $x=2(1)=2$. So $P = (2, 1)$.
The distance between $P(2, 1)$ and the point $(6, 3)$ is given by the distance formula:
$d = \sqrt{(6-2)^2 + (3-1)^2} = \sqrt{4^2 + 2^2} = \sqrt{16+4} = \sqrt{20} = 2\sqrt{5}$.

(D) Let the equation of the tangent be $y = mx + c$.
For the circle $x^2 + y^2 = 16$,the radius is $r = 4$ and the center is $(0, 0)$.
The condition for the line $mx - y + c = 0$ to be a tangent is $\frac{|c|}{\sqrt{m^2 + 1}} = 4$,so $c^2 = 16(m^2 + 1)$.
For the circle $(x - 9)^2 + y^2 = 16$,the center is $(9, 0)$ and the radius is $r = 4$.
The condition for the line $mx - y + c = 0$ to be a tangent is $\frac{|9m + c|}{\sqrt{m^2 + 1}} = 4$,so $(9m + c)^2 = 16(m^2 + 1)$.
Equating the two expressions for $16(m^2 + 1)$,we get $c^2 = (9m + c)^2$.
This implies $c = -(9m + c)$ or $c = 9m + c$.
Case $1$: $c = 9m + c \implies 9m = 0 \implies m = 0$.
Case $2$: $c = -9m - c \implies 2c = -9m \implies c = -\frac{9m}{2}$.
Substituting $c = -\frac{9m}{2}$ into $c^2 = 16(m^2 + 1)$:
$\frac{81m^2}{4} = 16m^2 + 16 \implies 81m^2 = 64m^2 + 64 \implies 17m^2 = 64 \implies m^2 = \frac{64}{17}$.
Thus,$m = \pm \frac{8}{\sqrt{17}}$.
Comparing with the options,the correct slope is $\frac{8}{\sqrt{17}}$.

(D) The equation of a line passing through $(-1, -2)$ with slope $m$ is $y + 2 = m(x + 1)$,which simplifies to $mx - y + (m - 2) = 0$.
Since this line is tangent to the circle $(x-3)^2 + (y-4)^2 = 4$ with center $(3, 4)$ and radius $r = 2$,the perpendicular distance from the center to the line must equal the radius:
$\frac{|m(3) - 4 + m - 2|}{\sqrt{m^2 + (-1)^2}} = 2$
$\frac{|4m - 6|}{\sqrt{m^2 + 1}} = 2$
$|2m - 3| = \sqrt{m^2 + 1}$
Squaring both sides: $(2m - 3)^2 = m^2 + 1$
$4m^2 - 12m + 9 = m^2 + 1$
$3m^2 - 12m + 8 = 0$
Let $m_1, m_2$ be the roots of this quadratic equation.
Then $m_1 + m_2 = \frac{12}{3} = 4$ and $m_1 m_2 = \frac{8}{3}$.
We know $|m_1 - m_2| = \sqrt{(m_1 + m_2)^2 - 4m_1 m_2} = \sqrt{4^2 - 4(\frac{8}{3})} = \sqrt{16 - \frac{32}{3}} = \sqrt{\frac{48 - 32}{3}} = \sqrt{\frac{16}{3}} = \frac{4}{\sqrt{3}}$.
Therefore,$\sqrt{3}|m_1 - m_2| = \sqrt{3} \times \frac{4}{\sqrt{3}} = 4$.

(B) The center of the circle $(h, k)$ is the intersection point of the two normals.
Solving the system of equations:
$2x - 3y = -5$ $(1)$
$4x - 5y = -7$ $(2)$
Multiply $(1)$ by $2$: $4x - 6y = -10$ $(3)$
Subtract $(3)$ from $(2)$: $(4x - 5y) - (4x - 6y) = -7 - (-10) \implies y = 3$.
Substitute $y = 3$ into $(1)$: $2x - 3(3) = -5 \implies 2x - 9 = -5 \implies 2x = 4 \implies x = 2$.
So,the center of the circle is $(2, 3)$.
The point $(2, 5)$ lies on the circle.
The radius $r$ is the distance between the center $(2, 3)$ and the point $(2, 5)$.
$r = \sqrt{(2 - 2)^2 + (5 - 3)^2} = \sqrt{0^2 + 2^2} = \sqrt{4} = 2$.

(A) The equation of the circle is $x^2+y^2+6x-6y+2=0$.
Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=3, f=-3, c=2$.
The center is $C(-3, 3)$ and the radius $r = \sqrt{g^2+f^2-c} = \sqrt{9+9-2} = \sqrt{16} = 4$.
Let $\theta$ be the angle between the tangents. Given $\theta = 2 \operatorname{Tan}^{-1}\left(\frac{4}{3}\right)$,so $\frac{\theta}{2} = \operatorname{Tan}^{-1}\left(\frac{4}{3}\right)$,which implies $\tan\left(\frac{\theta}{2}\right) = \frac{4}{3}$.
In the right-angled triangle formed by the center,the point $P$,and the point of tangency,we have $\sin\left(\frac{\theta}{2}\right) = \frac{r}{CP}$.
Since $\tan\left(\frac{\theta}{2}\right) = \frac{4}{3}$,we have $\sin\left(\frac{\theta}{2}\right) = \frac{4}{5}$.
Thus,$\frac{4}{CP} = \frac{4}{5}$,which gives $CP = 5$.
The distance $CP^2 = (k - (-3))^2 + (6k - 3)^2 = 5^2 = 25$.
$(k+3)^2 + (6k-3)^2 = 25$.
$k^2 + 6k + 9 + 36k^2 - 36k + 9 = 25$.
$37k^2 - 30k + 18 = 25$.
$37k^2 - 30k - 7 = 0$.
$(37k + 7)(k - 1) = 0$.
Since $k$ must be an integer,$k = 1$.

(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 $g = -2, f = 3, c = 4$.
Centre $C = (-g, -f) = (2, -3)$ and radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{4 + 9 - 4} = 3$.
Since the line $4x - 3y + p = 0$ is tangent to the circle,the perpendicular distance from the centre $(2, -3)$ to the line must be equal to the radius $r = 3$.
$\frac{|4(2) - 3(-3) + p|}{\sqrt{4^2 + (-3)^2}} = 3$ $\Rightarrow \frac{|8 + 9 + p|}{5} = 3$ $\Rightarrow |17 + p| = 15$.
This gives $17 + p = 15 \Rightarrow p = -2$ or $17 + p = -15 \Rightarrow p = -32$.
Given $p + 3 > 0$,we have $p > -3$,so $p = -2$.
The line is $4x - 3y - 2 = 0$.
The point of contact $(h, k)$ lies on the line $4h - 3k - 2 = 0$.
The normal at $(h, k)$ passes through the centre $(2, -3)$. The slope of the tangent is $4/3$,so the slope of the normal is $-3/4$.
$\frac{k - (-3)}{h - 2} = -\frac{3}{4}$ $\Rightarrow 4k + 12 = -3h + 6$ $\Rightarrow 3h + 4k = -6$.
Solving $4h - 3k = 2$ and $3h + 4k = -6$:
Multiply first by $4$ and second by $3$: $16h - 12k = 8$ and $9h + 12k = -18$.
Adding them: $25h = -10 \Rightarrow h = -2/5$.
Substituting $h$: $4(-2/5) - 3k = 2$ $\Rightarrow -8/5 - 2 = 3k$ $\Rightarrow 3k = -18/5$ $\Rightarrow k = -6/5$.
Thus,$h - 2k = -2/5 - 2(-6/5) = -2/5 + 12/5 = 10/5 = 2$.

(A) The equation of the circle is $x^2+y^2-2x-2y-1=0$.
Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=-1, f=-1, c=-1$.
Centre $C = (-g, -f) = (1, 1)$ and radius $r = \sqrt{g^2+f^2-c} = \sqrt{1+1+1} = \sqrt{3}$.
The parametric coordinates of any point on the circle are given by $x = 1 + \sqrt{3}\cos\theta$ and $y = 1 + \sqrt{3}\sin\theta$.
For $P(\frac{\pi}{4})$,$x_1 = 1 + \sqrt{3}\cos(\frac{\pi}{4}) = 1 + \frac{\sqrt{6}}{2}$ and $y_1 = 1 + \sqrt{3}\sin(\frac{\pi}{4}) = 1 + \frac{\sqrt{6}}{2}$.
For $Q(\frac{\pi}{3})$,$x_2 = 1 + \sqrt{3}\cos(\frac{\pi}{3}) = 1 + \frac{\sqrt{3}}{2}$ and $y_2 = 1 + \sqrt{3}\sin(\frac{\pi}{3}) = 1 + \frac{3}{2}$.
The slope of chord $PQ$ is $m = \frac{y_2-y_1}{x_2-x_1} = \frac{(1 + \frac{3}{2}) - (1 + \frac{\sqrt{6}}{2})}{(1 + \frac{\sqrt{3}}{2}) - (1 + \frac{\sqrt{6}}{2})} = \frac{3-\sqrt{6}}{\sqrt{3}-\sqrt{6}}$.
Multiplying numerator and denominator by $(\sqrt{3}+\sqrt{6})$,we get $m = \frac{3\sqrt{3}+3\sqrt{6}-\sqrt{18}-6}{3-6} = \frac{3\sqrt{3}+3\sqrt{6}-3\sqrt{2}-6}{-3} = -\sqrt{3}-\sqrt{6}+\sqrt{2}+2 = 2+\sqrt{2}-\sqrt{3}-\sqrt{6}$.
Since the tangent is parallel to the chord,its slope is equal to the slope of the chord,which is $2+\sqrt{2}-\sqrt{3}-\sqrt{6}$.

(A) The equation of the circle is $x^2+y^2+6x+6y-2=0$. The center $O$ is $(-3, -3)$ and the radius $r = \sqrt{(-3)^2+(-3)^2-(-2)} = \sqrt{9+9+2} = \sqrt{20}$.
Since point $Q$ lies on the $Y$-axis,its $x$-coordinate is $0$. Substituting $x=0$ into the line equation $5x-2y+6=0$,we get $-2y+6=0$,so $y=3$. Thus,$Q$ is $(0, 3)$.
The length of the tangent $PQ$ from point $Q(0, 3)$ to the circle is given by $\sqrt{S_1}$,where $S_1 = x_1^2+y_1^2+6x_1+6y_1-2$.
$PQ = \sqrt{0^2+3^2+6(0)+6(3)-2} = \sqrt{0+9+0+18-2} = \sqrt{25} = 5$.

(B) The equation of the circle is $x^2+y^2-10x=0$. The center of the circle is $(5,0)$ and the radius is $5$.
The equation of the tangent at point $P(9,3)$ is given by $xx_1 + yy_1 - 5(x+x_1) = 0$.
Substituting $(x_1, y_1) = (9,3)$,we get $9x + 3y - 5(x+9) = 0$,which simplifies to $4x + 3y - 45 = 0$.
The tangent intersects the $X$-axis $(y=0)$ at point $A$. Setting $y=0$ in the tangent equation: $4x = 45 \implies x = \frac{45}{4}$. So,$A = (\frac{45}{4}, 0)$.
The normal at $(9,3)$ passes through the center $(5,0)$ and the point $(9,3)$. Its slope is $m = \frac{3-0}{9-5} = \frac{3}{4}$.
The equation of the normal is $y - 0 = \frac{3}{4}(x - 5)$,which simplifies to $3x - 4y - 15 = 0$.
The normal intersects the $X$-axis $(y=0)$ at point $B$. Setting $y=0$ in the normal equation: $3x = 15 \implies x = 5$. So,$B = (5, 0)$.
The triangle is formed by vertices $P(9,3)$,$A(\frac{45}{4}, 0)$,and $B(5,0)$.
The base of the triangle along the $X$-axis is $AB = |\frac{45}{4} - 5| = |\frac{45-20}{4}| = \frac{25}{4}$.
The height of the triangle is the $y$-coordinate of $P$,which is $3$.
Area $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \frac{25}{4} \times 3 = \frac{75}{8}$ sq. units.

(B) The line is $x+2y=1$. The $X$-intercept is $A(1,0)$ and the $Y$-intercept is $B(0, 1/2)$.
The circle passes through $(0,0), (1,0), (0, 1/2)$. The equation of the circle is $x^2+y^2-x-\frac{1}{2}y=0$.
The tangent at the origin $(0,0)$ is found by setting the linear terms to zero: $-x-\frac{1}{2}y=0$,which simplifies to $2x+y=0$.
The perpendicular distance from $A(1,0)$ to $2x+y=0$ is $d_1 = \frac{|2(1)+0|}{\sqrt{2^2+1^2}} = \frac{2}{\sqrt{5}}$.
The perpendicular distance from $B(0, 1/2)$ to $2x+y=0$ is $d_2 = \frac{|2(0)+1/2|}{\sqrt{2^2+1^2}} = \frac{1/2}{\sqrt{5}} = \frac{1}{2\sqrt{5}}$.
The sum of distances is $d_1+d_2 = \frac{2}{\sqrt{5}} + \frac{1}{2\sqrt{5}} = \frac{5}{2\sqrt{5}} = \frac{\sqrt{5}}{2}$.
The radius of the circle $x^2+y^2-x-\frac{1}{2}y=0$ is $r = \sqrt{g^2+f^2-c} = \sqrt{(-1/2)^2+(-1/4)^2-0} = \sqrt{1/4+1/16} = \sqrt{5/16} = \frac{\sqrt{5}}{4}$.
Thus,the sum of distances is $\frac{\sqrt{5}}{2} = 2 \times \frac{\sqrt{5}}{4} = 2r$,which is the diameter of the circle $S$.

(B) The given equation of the circle is $x^2 + y^2 = 16$,which is of the form $x^2 + y^2 = a^2$ with $a^2 = 16$.
We know that the line $y = mx + C$ is a tangent to the circle $x^2 + y^2 = a^2$ if and only if $C^2 = a^2(1 + m^2)$.
Substituting $a^2 = 16$ into the condition,we get $C^2 = 16(1 + m^2)$.
Dividing by $16$,we have $\frac{C^2}{16} = 1 + m^2$.
Rearranging for $m^2$,we get $m^2 = \frac{C^2}{16} - 1 = \frac{C^2 - 16}{16}$.
Taking the square root on both sides,we get $m = \pm \frac{1}{4} \sqrt{C^2 - 16}$.
Thus,the correct option is $B$.

(C) The equation of the pair of tangents from a point $(x_1, y_1)$ to a circle $S=0$ is given by $T^2=SS_1$.
Here,the point is the origin $(0,0)$ and the circle is $S: x^2+y^2+2gx+2fy+g^2=0$.
$S_1$ at $(0,0)$ is $0^2+0^2+2g(0)+2f(0)+g^2 = g^2$.
The tangent $T$ at $(0,0)$ is $x(0)+y(0)+g(x+0)+f(y+0)+g^2 = gx+fy+g^2$.
Substituting into $T^2=SS_1$:
$(gx+fy+g^2)^2 = (x^2+y^2+2gx+2fy+g^2)(g^2)$
$g^2x^2+f^2y^2+g^4+2gfxy+2g^3x+2g^2fy = g^2x^2+g^2y^2+2g^3x+2g^2fy+g^4$
Canceling common terms $g^2x^2, g^4, 2g^3x, 2g^2fy$ from both sides:
$f^2y^2+2gfxy = g^2y^2$
$y^2(g^2-f^2)-2gfxy = 0$
$y[(g^2-f^2)y-2gfx] = 0$
Thus,the equations are $y=0$ and $(g^2-f^2)y-2gfx=0$.

(A) The line $ax+by-1=0$ is a tangent to the circle $x^2+y^2-r^2=0$ with center $(0,0)$ and radius $r$.
The perpendicular distance from the center $(0,0)$ to the line $ax+by-1=0$ must be equal to the radius $r$.
So,$\frac{|a(0)+b(0)-1|}{\sqrt{a^2+b^2}} = r$.
$\frac{1}{\sqrt{a^2+b^2}} = r$.
Squaring both sides,we get $\frac{1}{a^2+b^2} = r^2$,which implies $a^2+b^2 = \frac{1}{r^2}$.
If $r=1$,then $a^2+b^2 = 1$,which means the point $(a,b)$ lies on the circle $x^2+y^2=1$,i.e.,$S_1=0$.

(B) The given equation of the circle is $x^2+y^2-6x-2y+1=0$.
Since the point $(a, 4)$ lies on the circle,we substitute $y=4$ into the circle equation:
$x^2+4^2-6x-2(4)+1=0$
$x^2-6x+16-8+1=0$
$x^2-6x+9=0$
$(x-3)^2=0 \Rightarrow x=3$.
Thus,the point of tangency is $(3, 4)$,so $a=3$.
The equation of the tangent at $(x_1, y_1)$ to the circle $x^2+y^2+2gx+2fy+c'=0$ is $xx_1+yy_1+g(x+x_1)+f(y+y_1)+c'=0$.
Here,$g=-3, f=-1, c'=1$ and $(x_1, y_1)=(3, 4)$.
Substituting these values:
$x(3)+y(4)-3(x+3)-1(y+4)+1=0$
$3x+4y-3x-9-y-4+1=0$
$3y-12=0 \Rightarrow y-4=0$.
Comparing this with $y+c=0$,we get $c=-4$.
Therefore,$ac = 3 \times (-4) = -12$.

(D) The equation of the circle is $x^2+y^2-2x-4y-20=0$.
Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=-1, f=-2, c=-20$.
The centre $A$ is $(-g, -f) = (1, 2)$ and the radius $r = \sqrt{g^2+f^2-c} = \sqrt{1+4+20} = \sqrt{25} = 5$.
The equation of the tangent at $(x_1, y_1)$ is $xx_1 + yy_1 - (x+x_1) - 2(y+y_1) - 20 = 0$.
For point $B(1, 7)$,the tangent is $x(1) + y(7) - (x+1) - 2(y+7) - 20 = 0$,which simplifies to $x + 7y - x - 1 - 2y - 14 - 20 = 0$,so $5y = 35$,or $y = 7$.
For point $D(4, -2)$,the tangent is $x(4) + y(-2) - (x+4) - 2(y-2) - 20 = 0$,which simplifies to $4x - 2y - x - 4 - 2y + 4 - 20 = 0$,so $3x - 4y = 20$.
Substituting $y = 7$ into $3x - 4y = 20$,we get $3x - 4(7) = 20$,so $3x = 48$,which gives $x = 16$.
Thus,the point of intersection $C$ is $(16, 7)$.
The quadrilateral $ABCD$ consists of two congruent right-angled triangles $\triangle ABC$ and $\triangle ADC$,where the right angle is at $B$ and $D$ respectively.
The area of $\triangle ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AB \times BC$.
$AB = r = 5$. $BC = \sqrt{(16-1)^2 + (7-7)^2} = 15$.
Area of $\triangle ABC = \frac{1}{2} \times 5 \times 15 = 37.5$.
Since $\triangle ADC \cong \triangle ABC$,the total area of quadrilateral $ABCD = 2 \times 37.5 = 75$.

(B) Given equation of the circle is $S \equiv x^2+y^2-13=0$.
On differentiating with respect to $x$,we get $2x + 2y \frac{dy}{dx} = 0$,which implies $\frac{dy}{dx} = -\frac{x}{y}$.
The slope of the tangent $m$ at $(2,3)$ is $m = \left. \frac{dy}{dx} \right|_{(2,3)} = -\frac{2}{3}$.
Now,the point $\left(m, -\frac{1}{m}\right)$ becomes $\left(-\frac{2}{3}, \frac{3}{2}\right)$.
To check the position of this point with respect to the circle,we substitute it into the expression $S(x, y) = x^2+y^2-13$:
$S\left(-\frac{2}{3}, \frac{3}{2}\right) = \left(-\frac{2}{3}\right)^2 + \left(\frac{3}{2}\right)^2 - 13 = \frac{4}{9} + \frac{9}{4} - 13 = \frac{16 + 81 - 468}{36} = \frac{97 - 468}{36} = -\frac{371}{36}$.
Since $S\left(-\frac{2}{3}, \frac{3}{2}\right) < 0$,the point lies inside the circle.

(D) The given equation of the circle is $x^2+y^2-6x+4y=12$.
Completing the square,we get $(x-3)^2+(y+2)^2 = 12+9+4 = 25 = 5^2$.
The center is $(3, -2)$ and the radius $r=5$.
$A$ 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 $4x+3y+k=0$ must be equal to the radius $r=5$.
Using the distance formula,$\frac{|4(3)+3(-2)+k|}{\sqrt{4^2+3^2}} = 5$.
$\frac{|12-6+k|}{5} = 5
\Rightarrow |6+k| = 25$.
This gives $6+k = 25$ or $6+k = -25$.
So,$k = 19$ or $k = -31$.
The equations of the tangents are $4x+3y+19=0$ and $4x+3y-31=0$.

(D) The given equation of the circle is $x^2+y^2-6x+8y-144=0$.
Comparing this with the general equation $x^2+y^2+2gx+2fy+c=0$,we get $2g = -6 \Rightarrow g = -3$ and $2f = 8 \Rightarrow f = 4$.
The centre of the circle is $(-g, -f) = (3, -4)$.
We know that the normal to a circle at any point always passes through its centre.
Let the point where the normal meets the circle again be $(x_1, y_1)$.
Since the centre $(3, -4)$ is the midpoint of the chord connecting $(8, 8)$ and $(x_1, y_1)$,we have:
$\frac{x_1+8}{2} = 3$ $\Rightarrow x_1+8 = 6$ $\Rightarrow x_1 = -2$
$\frac{y_1+8}{2} = -4$ $\Rightarrow y_1+8 = -8$ $\Rightarrow y_1 = -16$
Thus,the required point is $(-2, -16)$.

(C) Let $(a, b)$ be the point where the line $4x - 3y + 7 = 0$ touches the circle $x^2 + y^2 - 6x + 4y - 12 = 0$.
The equation of the tangent at $(a, b)$ to the circle is given by $xa + yb - 3(x + a) + 2(y + b) - 12 = 0$.
Rearranging the terms,we get $(a - 3)x + (b + 2)y - (3a - 2b + 12) = 0$.
Comparing this with the given line $4x - 3y + 7 = 0$,we have the ratio of coefficients:
$\frac{a - 3}{4} = \frac{b + 2}{-3} = \frac{-(3a - 2b + 12)}{7} = k$.
From this,$a = 4k + 3$ and $b = -3k - 2$.
Substituting these into the third ratio: $-(3(4k + 3) - 2(-3k - 2) + 12) = 7k$.
$-(12k + 9 + 6k + 4 + 12) = 7k$ $\Rightarrow -(18k + 25) = 7k$ $\Rightarrow -25k = 25$ $\Rightarrow k = -1$.
Substituting $k = -1$ back into the expressions for $a$ and $b$:
$a = 4(-1) + 3 = -1$ and $b = -3(-1) - 2 = 1$.
Thus,the point of contact is $(-1, 1)$.

(C) The equation of the circle is $x^2+y^2=4$. The point $P(1, \sqrt{3})$ lies on the circle.
The equation of the tangent at $P(x_1, y_1)$ is $xx_1 + yy_1 = r^2$,which is $x(1) + y(\sqrt{3}) = 4$,or $x + \sqrt{3}y = 4$.
The tangent meets the $X$-axis at $A(4, 0)$.
The normal at $P(1, \sqrt{3})$ passes through the origin $O(0, 0)$ and has the equation $y = \sqrt{3}x$.
The triangle is formed by the points $O(0, 0)$,$P(1, \sqrt{3})$,and $A(4, 0)$.
The area of $\triangle OPA = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times OA \times y_P = \frac{1}{2} \times 4 \times \sqrt{3} = 2 \sqrt{3}$ sq units.

(C) The given equation of the circle is $x^2+y^2=4$.
On differentiating with respect to $x$,we get $2x + 2y \frac{dy}{dx} = 0$,which implies $\frac{dy}{dx} = -\frac{x}{y}$.
At the point $(1, \sqrt{3})$,the slope of the tangent is $m_t = -\frac{1}{\sqrt{3}}$.
The equation of the tangent at $(1, \sqrt{3})$ is $y - \sqrt{3} = -\frac{1}{\sqrt{3}}(x - 1)$,which simplifies to $x + \sqrt{3}y = 4$.
This tangent intersects the $x$-axis at point $B(4, 0)$.
The slope of the normal at $(1, \sqrt{3})$ is $m_n = -\frac{1}{m_t} = \sqrt{3}$.
The equation of the normal at $(1, \sqrt{3})$ is $y - \sqrt{3} = \sqrt{3}(x - 1)$,which simplifies to $y = \sqrt{3}x$.
This normal passes through the origin $O(0, 0)$.
The triangle formed by the positive $x$-axis,the tangent,and the normal is $\triangle OAB$,where $O(0, 0)$,$A(1, \sqrt{3})$,and $B(4, 0)$.
The area of $\triangle OAB$ is given by $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times OB \times AD$,where $AD$ is the $y$-coordinate of point $A$.
Area $\Delta = \frac{1}{2} \times 4 \times \sqrt{3} = 2\sqrt{3}$.

(D) The line $x+3y=0$ is tangent to the circle at $(0,0)$. The radius of the circle is $r=1$.
The centre $(h,k)$ of the circle lies on the normal to the tangent at $(0,0)$.
The slope of the tangent $x+3y=0$ is $m_t = -\frac{1}{3}$.
The slope of the normal is $m_n = -\frac{1}{m_t} = 3$.
The equation of the normal passing through $(0,0)$ is $y-0 = 3(x-0)$,which simplifies to $y=3x$.
Thus,the centre is of the form $(x, 3x)$.
The distance from the centre $(x, 3x)$ to the point $(0,0)$ must be equal to the radius $r=1$.
$\sqrt{(x-0)^2 + (3x-0)^2} = 1$
$\sqrt{x^2 + 9x^2} = 1$
$\sqrt{10x^2} = 1$
$|x|\sqrt{10} = 1 \Rightarrow x = \pm \frac{1}{\sqrt{10}}$.
If $x = \frac{1}{\sqrt{10}}$,then $y = 3x = \frac{3}{\sqrt{10}}$. The centre is $\left(\frac{1}{\sqrt{10}}, \frac{3}{\sqrt{10}}\right)$.
If $x = -\frac{1}{\sqrt{10}}$,then $y = 3x = -\frac{3}{\sqrt{10}}$. The centre is $\left(-\frac{1}{\sqrt{10}}, -\frac{3}{\sqrt{10}}\right)$.
Comparing with the given options,the correct centre is $\left(\frac{1}{\sqrt{10}}, \frac{3}{\sqrt{10}}\right)$.

(D) The given line is $y=2x+c$,which can be written as $2x-y+c=0$.
Comparing this with $y=mx+c_1$,we have $m=2$.
The equation of the circle is $x^2+y^2=5$,so the radius $r=\sqrt{5}$.
The condition for a line $y=mx+c_1$ to be a tangent to the circle $x^2+y^2=r^2$ is $c_1^2 = r^2(1+m^2)$.
Substituting the values,we get $c^2 = 5(1+2^2)$.
$c^2 = 5(1+4) = 5(5) = 25$.
Therefore,$c = \pm 5$.
Among the given options,the value of $c$ is $5$.

(C) 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 $\sqrt{x_1^2+y_1^2+2gx_1+2fy_1+c}$.
Given the circle equation $x^2+y^2-2x+4y-11=0$ and the point $(1,3)$.
Substituting the values into the formula:
Length $= \sqrt{1^2+3^2-2(1)+4(3)-11}$
$= \sqrt{1+9-2+12-11}$
$= \sqrt{22-13}$
$= \sqrt{9}$
$= 3$

(A) The line $y-3x=0$ is a tangent to the circle. The radius $r$ is the perpendicular distance from the centre $(1,1)$ to the line $3x-y=0$.
$r = \frac{|3(1) - 1(1)|}{\sqrt{3^2 + (-1)^2}} = \frac{|3-1|}{\sqrt{10}} = \frac{2}{\sqrt{10}}$.
Let the other tangent passing through the origin $(0,0)$ be $y=mx$,or $mx-y=0$.
The perpendicular distance from the centre $(1,1)$ to this line must also equal the radius $r$.
$r = \frac{|m(1) - 1(1)|}{\sqrt{m^2 + (-1)^2}} = \frac{|m-1|}{\sqrt{m^2+1}}$.
Equating the two expressions for $r$:
$\frac{|m-1|}{\sqrt{m^2+1}} = \frac{2}{\sqrt{10}}$.
Squaring both sides:
$\frac{(m-1)^2}{m^2+1} = \frac{4}{10} = \frac{2}{5}$.
$5(m^2 - 2m + 1) = 2(m^2 + 1)$.
$5m^2 - 10m + 5 = 2m^2 + 2$.
$3m^2 - 10m + 3 = 0$.
$3m^2 - 9m - m + 3 = 0$.
$3m(m-3) - 1(m-3) = 0$.
$(3m-1)(m-3) = 0$.
So,$m=3$ or $m=\frac{1}{3}$.
Since $m=3$ corresponds to the given tangent $y=3x$,the other tangent is $y=\frac{1}{3}x$,which is $3y=x$.

(B) The given equation of the circle is $x^2+y^2+6x+4y-3=0$.
Comparing this with the general form $x^2+y^2+2gx+2fy+c=0$,we get $g=3$ and $f=2$.
The center of the circle is $(-g, -f) = (-3, -2)$.
The normal to a circle at any point always passes through the center of the circle.
We need to find the equation of the line passing through the center $(-3, -2)$ and the given point $(1, -2)$.
Since the $y$-coordinates of both points are equal to $-2$,the line is a horizontal line given by $y = -2$.
Thus,the equation of the normal is $y+2=0$.

(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 $C(-1, 1)$ and the radius $r=1$.
Since the tangents $x+y+k=0$ and $x+ay+b=0$ are perpendicular,their slopes $m_1 = -1$ and $m_2 = -1/a$ satisfy $m_1 m_2 = -1$. Thus,$(-1)(-1/a) = -1$,which gives $a = -1$.
The distance from the center $(-1, 1)$ to the tangent $x+y+k=0$ is equal to the radius $r=1$:
$\frac{|-1+1+k|}{\sqrt{1^2+1^2}} = 1 \Rightarrow |k| = \sqrt{2}$. Since $k > 1$,we have $k = \sqrt{2}$.
The distance from the center $(-1, 1)$ to the tangent $x-y+b=0$ is also $r=1$:
$\frac{|-1-1+b|}{\sqrt{1^2+(-1)^2}} = 1 \Rightarrow |b-2| = \sqrt{2}$.
This gives $b-2 = \sqrt{2}$ or $b-2 = -\sqrt{2}$. Since $b > 1$,we take $b = 2+\sqrt{2}$.
Finally,$b-k = (2+\sqrt{2}) - \sqrt{2} = 2$.

(A) The equation of the pair of tangents from a point $(x_1, y_1)$ to a circle $S=0$ is given by $SS_1=T^2$.
Here,$S = x^2+y^2-4=0$ and the point is $(4,0)$.
$S_1 = 4^2+0^2-4 = 16-4 = 12$.
$T = x(4)+y(0)-4 = 4x-4$.
Substituting these into $SS_1=T^2$:
$12(x^2+y^2-4) = (4x-4)^2$.
$12(x^2+y^2-4) = 16(x-1)^2$.
Dividing by $4$:
$3(x^2+y^2-4) = 4(x^2-2x+1)$.
$3x^2+3y^2-12 = 4x^2-8x+4$.
$x^2-3y^2-8x+16 = 0$.
$(x^2-8x+16) - 3y^2 = 0$.
$(x-4)^2 - 3y^2 = 0$.
$(x-4)^2 = 3y^2$.
Taking the square root on both sides:
$\sqrt{3}y = \pm(x-4)$.

(D) The equation of the circle is $x^2+y^2-2x+4y+3=0$.
Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=-1, f=2, c=3$.
The radius $r = \sqrt{g^2+f^2-c} = \sqrt{(-1)^2+2^2-3} = \sqrt{1+4-3} = \sqrt{2}$.
The length of the tangent $L$ from point $P(6,-5)$ to the circle is given by $\sqrt{S_1}$.
$L = \sqrt{6^2+(-5)^2-2(6)+4(-5)+3} = \sqrt{36+25-12-20+3} = \sqrt{32} = 4\sqrt{2}$.
In the right-angled triangle $\triangle OAP$,where $O$ is the center and $A$ is the point of contact,$\tan(\frac{\theta}{2}) = \frac{r}{L} = \frac{\sqrt{2}}{4\sqrt{2}} = \frac{1}{4}$.
We know that $\tan \theta = \frac{2\tan(\frac{\theta}{2})}{1-\tan^2(\frac{\theta}{2})} = \frac{2(\frac{1}{4})}{1-(\frac{1}{4})^2} = \frac{1/2}{1-1/16} = \frac{1/2}{15/16} = \frac{1}{2} \times \frac{16}{15} = \frac{8}{15}$.
Therefore,$\cot \theta = \frac{1}{\tan \theta} = \frac{15}{8}$.

(A) Given circle equation: $x^2+y^2-6x+4y+12=0$.
Completing the square,we get $(x-3)^2+(y+2)^2=1$.
This represents a circle with center $(3, -2)$ and radius $r=1$.
Let the slope of the tangent passing through $(1, -3)$ be $m$. The equation of the tangent is $y+3=m(x-1)$,which simplifies to $mx-y-m-3=0$.
The perpendicular distance from the center $(3, -2)$ to the tangent must equal the radius $r=1$:
$\left|\frac{m(3)-(-2)-m-3}{\sqrt{m^2+(-1)^2}}\right|=1$.
$\left|\frac{2m-1}{\sqrt{m^2+1}}\right|=1$.
Squaring both sides: $(2m-1)^2 = m^2+1$.
$4m^2-4m+1 = m^2+1$.
$3m^2-4m = 0$.
$m(3m-4) = 0$.
Thus,the slopes are $m_1=0$ and $m_2=\frac{4}{3}$.
Finally,$9(m_1^2+m_2^2) = 9(0^2 + (\frac{4}{3})^2) = 9(\frac{16}{9}) = 16$.

(A) The centers of $C_1$ and $C_2$ are $O(0,0)$ and $A(1,0)$ with radius $r=1$.
$C_3$ is a unit circle passing through $O(0,0)$ and $A(1,0)$. Let its center be $(h, k)$.
Since it passes through $(0,0)$,$h^2 + k^2 = 1^2 = 1$.
Since it passes through $(1,0)$,$(h-1)^2 + k^2 = 1^2 = 1$.
Subtracting the equations: $h^2 - (h-1)^2 = 0 \implies h^2 - (h^2 - 2h + 1) = 0 \implies 2h - 1 = 0 \implies h = 1/2$.
Then $(1/2)^2 + k^2 = 1 \implies k^2 = 3/4 \implies k = \sqrt{3}/2$ (since center is above $X$-axis).
So $C_3$ has center $(1/2, \sqrt{3}/2)$ and radius $1$.
$C_1$ has center $(0,0)$ and radius $1$. The line $L: ax + by + c = 0$ is tangent to $C_1$ if $|c|/\sqrt{a^2+b^2} = 1 \implies c^2 = a^2 + b^2$.
It is tangent to $C_3$ if $|a/2 + b\sqrt{3}/2 + c|/\sqrt{a^2+b^2} = 1 \implies |a + b\sqrt{3} + 2c| = 2\sqrt{a^2+b^2}$.
Substituting $c^2 = a^2+b^2$,we test the options. For option $A$: $\sqrt{3}x - y + 2 = 0$,$a=\sqrt{3}, b=-1, c=2$. $a^2+b^2 = 3+1=4=c^2$. Tangent to $C_1$.
For $C_3$: $|\sqrt{3}(1/2) - (\sqrt{3}/2) + 2| = |2| = 2$. $\sqrt{a^2+b^2} = 2$. $2/2 = 1$. Tangent to $C_3$.
Checking intersection with $C_2$ (center $(1,0)$,radius $1$): Distance from $(1,0)$ to $\sqrt{3}x - y + 2 = 0$ is $|\sqrt{3}(1) - 0 + 2|/\sqrt{3+1} = |\sqrt{3}+2|/2 \approx 3.73/2 > 1$. It does not intersect $C_2$.

(C) The equation of the circle is $x^2+y^2-2x+4y+3=0$.
Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=-1, f=2, c=3$.
The center of the circle is $O(-g, -f) = (1, -2)$ and the radius $r = \sqrt{g^2+f^2-c} = \sqrt{(-1)^2+(2)^2-3} = \sqrt{1+4-3} = \sqrt{2}$.
Let $P$ be the point $(6, -5)$. The length of the tangent $PA$ is given by $\sqrt{S_1}$,where $S_1 = x_1^2+y_1^2-2x_1+4y_1+3$.
$PA = \sqrt{(6)^2+(-5)^2-2(6)+4(-5)+3} = \sqrt{36+25-12-20+3} = \sqrt{32} = 4\sqrt{2}$.
In the right-angled triangle $\triangle OAP$,where $\angle OAP = 90^\circ$,we have $\tan(\theta/2) = \frac{OA}{AP} = \frac{r}{PA} = \frac{\sqrt{2}}{4\sqrt{2}} = \frac{1}{4}$.
Using the formula $\tan \theta = \frac{2\tan(\theta/2)}{1-\tan^2(\theta/2)}$,we get:
$\tan \theta = \frac{2(1/4)}{1-(1/4)^2} = \frac{1/2}{1-1/16} = \frac{1/2}{15/16} = \frac{1}{2} \times \frac{16}{15} = \frac{8}{15}$.

(C) Given,from point $P(1,1)$,tangents $PA$ and $PB$ are drawn to the circle $x^2+y^2+gx+gy-2=0$.
The length of the tangent $PA = \sqrt{S_1} = \sqrt{1^2+1^2+g(1)+g(1)-2} = \sqrt{1+1+g+g-2} = \sqrt{2g}$.
The centre $C$ of the circle is $(-\frac{g}{2}, -\frac{g}{2})$ and the radius $r = AC = \sqrt{(-\frac{g}{2})^2 + (-\frac{g}{2})^2 - (-2)} = \sqrt{\frac{g^2}{4} + \frac{g^2}{4} + 2} = \sqrt{\frac{g^2}{2} + 2} = \sqrt{\frac{g^2+4}{2}} = \frac{\sqrt{g^2+4}}{\sqrt{2}}$.
Since $PA$ is a tangent,$\angle PAC = 90^\circ$. The area of $\triangle PAC = \frac{1}{2} \times PA \times AC = \frac{1}{2} \times \sqrt{2g} \times \frac{\sqrt{g^2+4}}{\sqrt{2}} = \frac{1}{2} \sqrt{g(g^2+4)} = \frac{1}{2} \sqrt{g^3+4g}$.
The quadrilateral $PACB$ consists of two congruent triangles $\triangle PAC$ and $\triangle PBC$.
Therefore,the area of quadrilateral $PACB = 2 \times \text{Area}(\triangle PAC) = 2 \times \frac{1}{2} \sqrt{g^3+4g} = \sqrt{g^3+4g}$.

(C) The given lines are $3x + 4y - 14 = 0$ and $6x + 8y + 7 = 0$.
To make the coefficients of $x$ and $y$ the same,divide the second equation by $2$:
$3x + 4y + \frac{7}{2} = 0$.
Since these lines are parallel tangents to the circle,the diameter of the circle is equal to the distance between these two parallel lines.
The distance $d$ between two parallel lines $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$ is given by $d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}$.
Here,$a = 3$,$b = 4$,$c_1 = -14$,and $c_2 = \frac{7}{2}$.
$d = \frac{|-14 - \frac{7}{2}|}{\sqrt{3^2 + 4^2}} = \frac{|-\frac{28}{2} - \frac{7}{2}|}{\sqrt{9 + 16}} = \frac{|-\frac{35}{2}|}{5} = \frac{35}{2 \times 5} = \frac{7}{2}$.
The radius $r$ of the circle is half of the distance between the parallel tangents:
$r = \frac{d}{2} = \frac{7/2}{2} = \frac{7}{4}$.

(B) The condition 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 is $2g_1g_2 + 2f_1f_2 = c_1 + c_2$.
Applying this to the given circles:
$1$) $2g(2) + 2f(2) = c + 7 \Rightarrow 4g + 4f = c + 7$
$2$) $2g(-2) + 2f(2) = c + 7 \Rightarrow -4g + 4f = c + 7$
$3$) $2g(-2) + 2f(-2) = c + 7 \Rightarrow -4g - 4f = c + 7$
Subtracting $(2)$ from $(1)$: $8g = 0 \Rightarrow g = 0$.
Adding $(2)$ and $(3)$: $-8f = 2(c + 7) \Rightarrow -4f = c + 7$. Since $4g + 4f = c + 7$ and $g=0$,we have $4f = c + 7$. Thus,$c + 7 = -(c + 7) \Rightarrow c = -7$.
Substituting $c = -7$ into $4f = c + 7$,we get $4f = 0 \Rightarrow f = 0$.
The circle equation is $S: x^2 + y^2 - 7 = 0$.
The equation of the tangent at $(x_1, y_1)$ is $xx_1 + yy_1 + c = 0$.
At $(\sqrt{3}, 2)$,the tangent is $\sqrt{3}x + 2y - 7 = 0$.

(B) For the circle $x^2+y^2-4x-8y+16=0$,the center $C_1 = (2, 4)$ and radius $r_1 = \sqrt{2^2+4^2-16} = 2$.
For the circle $x^2+y^2-6x-16y+64=0$,the center $C_2 = (3, 8)$ and radius $r_2 = \sqrt{3^2+8^2-64} = 3$.
Let the common tangent be $mx-y+c=0$.
The perpendicular distance from the center to the tangent equals the radius:
$\left|\frac{2m-4+c}{\sqrt{1+m^2}}\right| = 2 \Rightarrow c = 2\sqrt{1+m^2}-2m+4$
$\left|\frac{3m-8+c}{\sqrt{1+m^2}}\right| = 3 \Rightarrow c = 3\sqrt{1+m^2}-3m+8$
Equating the two expressions for $c$:
$2\sqrt{1+m^2}-2m+4 = 3\sqrt{1+m^2}-3m+8$
$\sqrt{1+m^2} = m-4$
Squaring both sides:
$1+m^2 = m^2-8m+16$
$8m = 15 \Rightarrow m = \frac{15}{8}$.

(A) The coordinate of $P$ is $\left(5 \cos \frac{2 \pi}{3}, 5 \sin \frac{2 \pi}{3}\right) = \left(-\frac{5}{2}, \frac{5 \sqrt{3}}{2}\right)$.
The slope of the tangent at $\theta = \frac{2 \pi}{3}$ is $m = -\cot \left(\frac{2 \pi}{3}\right) = \frac{1}{\sqrt{3}}$.
The equation of the tangent line is $y - \frac{5 \sqrt{3}}{2} = \frac{1}{\sqrt{3}}\left(x + \frac{5}{2}\right)$,which simplifies to $x - \sqrt{3}y + 10 = 0$.
To find the intersection point $Q$ with the line $y = 6$,we substitute $y = 6$ into the tangent equation: $x - 6\sqrt{3} + 10 = 0 \Rightarrow x = 6\sqrt{3} - 10$. Thus,$Q = (6\sqrt{3} - 10, 6)$.
To find the intersection point $R$ with the line $x = -6$,we substitute $x = -6$ into the tangent equation: $-6 - \sqrt{3}y + 10 = 0$ $\Rightarrow \sqrt{3}y = 4$ $\Rightarrow y = \frac{4}{\sqrt{3}}$. Thus,$R = (-6, \frac{4}{\sqrt{3}})$.
The third vertex of the triangle is $S = (-6, 6)$.
The area of $\triangle RSQ$ is given by $\frac{1}{2} |x_R(y_S - y_Q) + x_S(y_Q - y_R) + x_Q(y_R - y_S)|$.
Area $= \frac{1}{2} |(-6)(6 - 6) + (-6)(6 - \frac{4}{\sqrt{3}}) + (6\sqrt{3} - 10)(\frac{4}{\sqrt{3}} - 6)|$.
Area $= \frac{1}{2} |0 - 36 + \frac{24}{\sqrt{3}} + (24 - 36\sqrt{3} - \frac{40}{\sqrt{3}} + 60)| = \frac{1}{2} |48 - 36\sqrt{3} - \frac{16}{\sqrt{3}}| = |24 - 18\sqrt{3} - \frac{8}{\sqrt{3}}| = |\frac{24\sqrt{3} - 54 - 8}{\sqrt{3}}| = \frac{62 - 24\sqrt{3}}{\sqrt{3}}$.