Chord of contact of tangent and common chord Questions in English

(D) Given circles are $S_1: x^2+y^2-2x-2y-7=0$ and $S_2: x^2+y^2+4x+2y+k=0$.
Since they cut orthogonally,$2g_1g_2 + 2f_1f_2 = c_1 + c_2$.
Here,$g_1=-1, f_1=-1, c_1=-7$ and $g_2=2, f_2=1, c_2=k$.
$2(-1)(2) + 2(-1)(1) = -7 + k$ $\Rightarrow -4 - 2 = -7 + k$ $\Rightarrow k = 1$.
The equation of the common chord is $S_1 - S_2 = 0$.
$(x^2+y^2-2x-2y-7) - (x^2+y^2+4x+2y+1) = 0$ $\Rightarrow -6x - 4y - 8 = 0$ $\Rightarrow 3x + 2y + 4 = 0$.
For circle $S_1$,center $C = (1, 1)$ and radius $r = \sqrt{1^2+1^2-(-7)} = \sqrt{9} = 3$.
The length of the perpendicular from $C(1, 1)$ to the chord $3x+2y+4=0$ is $d = \frac{|3(1)+2(1)+4|}{\sqrt{3^2+2^2}} = \frac{9}{\sqrt{13}}$.
The length of the common chord is $2\sqrt{r^2-d^2} = 2\sqrt{3^2 - (\frac{9}{\sqrt{13}})^2} = 2\sqrt{9 - \frac{81}{13}} = 2\sqrt{\frac{117-81}{13}} = 2\sqrt{\frac{36}{13}} = \frac{12}{\sqrt{13}}$ units.

(D) The equations of the circles are $x^2 + y^2 = b^2$ and $(x - a)^2 + y^2 = b^2$.
Since the line $y = mx - b \sqrt{1 + m^2}$ is tangent to the first circle,it is a standard form for a tangent to $x^2 + y^2 = b^2$.
For the second circle $(x - a)^2 + y^2 = b^2$,the tangent line equation is $y = m(x - a) \pm b \sqrt{1 + m^2}$.
Since the line is common,we equate the two forms:
$mx - b \sqrt{1 + m^2} = mx - ma \pm b \sqrt{1 + m^2}$.
Taking the positive sign for the common tangent,we get:
$ma = 2b \sqrt{1 + m^2}$.
Squaring both sides:
$m^2 a^2 = 4b^2 (1 + m^2) = 4b^2 + 4b^2 m^2$.
$m^2 (a^2 - 4b^2) = 4b^2$.
$m^2 = \frac{4b^2}{a^2 - 4b^2}$.
Taking the positive square root:
$m = \frac{2b}{\sqrt{a^2 - 4b^2}}$.

(D) For circle $C_1: x^2+y^2-2x-2y-2=0$,center $O_1(1, 1)$ and radius $r_1 = \sqrt{1^2+1^2-(-2)} = 2$.
For circle $C_2: x^2+y^2+4x+6y+12=0$,center $O_2(-2, -3)$ and radius $r_2 = \sqrt{(-2)^2+(-3)^2-12} = 1$.
The direct common tangents intersect at the external center of similitude $S$,which divides $O_1O_2$ externally in the ratio $r_1:r_2 = 2:1$.
$S = \left(\frac{2(-2)-1(1)}{2-1}, \frac{2(-3)-1(1)}{2-1}\right) = (-5, -7)$.
Any line through $S(-5, -7)$ is $y+7 = m(x+5) \implies mx-y+5m-7=0$.
The distance from $O_1(1, 1)$ to this line is equal to $r_1=2$:
$\frac{|m(1)-1+5m-7|}{\sqrt{m^2+1}} = 2 \implies |6m-8| = 2\sqrt{m^2+1} \implies |3m-4| = \sqrt{m^2+1}$.
Squaring both sides: $9m^2-24m+16 = m^2+1 \implies 8m^2-24m+15=0$.
The combined equation of the tangents is $(y+7-m(x+5))(y+7-m'(x+5)) = 0$,where $m, m'$ are roots of $8m^2-24m+15=0$.
Let $Y = y+7$ and $X = x+5$. Then $m^2X^2 - mX(Y) + Y^2 = 0$ is not the form; rather,the equation is $(Y-mX)(Y-m'X) = Y^2 - (m+m')XY + mm'X^2 = 0$.
Here $m+m' = 3$ and $mm' = 15/8$.
$Y^2 - 3XY + \frac{15}{8}X^2 = 0 \implies 15X^2 - 24XY + 8Y^2 = 0$.
Substituting $X=x+5, Y=y+7$: $15(x+5)^2 - 24(x+5)(y+7) + 8(y+7)^2 = 0$.
$15(x^2+10x+25) - 24(xy+7x+5y+35) + 8(y^2+14y+49) = 0$.
$15x^2+150x+375 - 24xy-168x-120y-840 + 8y^2+112y+392 = 0$.
$15x^2-24xy+8y^2-18x-8y-73=0$.

(C) The equation of the common chord of two circles $S_1=0$ and $S_2=0$ is given by $S_1-S_2=0$.
Given circles are $S_1: x^2+y^2+5kx+2y+k=0$ and $S_2: x^2+y^2+kx+\frac{3}{2}y-\frac{1}{2}=0$.
The common chord is $(x^2+y^2+5kx+2y+k) - (x^2+y^2+kx+\frac{3}{2}y-\frac{1}{2}) = 0$.
Simplifying this,we get $4kx + \frac{1}{2}y + k + \frac{1}{2} = 0$,which is $8kx + y + 2k + 1 = 0$.
We are given that the line $4x+5y-k=0$ is the common chord.
Comparing the coefficients of the two equations:
$\frac{8k}{4} = \frac{1}{5} = \frac{2k+1}{-k}$.
From $\frac{8k}{4} = \frac{1}{5}$,we get $2k = \frac{1}{5}$,so $k = \frac{1}{10}$.
From $\frac{1}{5} = \frac{2k+1}{-k}$,we get $-k = 10k+5$,so $11k = -5$,which means $k = -\frac{5}{11}$.
Since the values of $k$ are not consistent,there is no value of $k$ for which the given line is the common chord.

(A) The equations of the circles are $S_1: x^2+y^2+x-3y-10=0$ and $S_2: x^2+y^2+2x-y-20=0$.
The common chord is given by $S_1 - S_2 = 0$,which is $(x^2+y^2+x-3y-10) - (x^2+y^2+2x-y-20) = 0$.
This simplifies to $-x-2y+10=0$,or $x+2y-10=0$.
The family of circles passing through the intersection of $S_1$ and $S_2$ is $S_1 + \lambda(S_1 - S_2) = 0$.
$(x^2+y^2+x-3y-10) + \lambda(x+2y-10) = 0$.
$x^2+y^2+(1+\lambda)x+(-3+2\lambda)y+(-10-10\lambda) = 0$.
The center of this circle is $(-\frac{1+\lambda}{2}, \frac{3-2\lambda}{2})$.
Since the common chord $x+2y-10=0$ is the diameter,the center must lie on it:
$-\frac{1+\lambda}{2} + 2(\frac{3-2\lambda}{2}) - 10 = 0$.
$-1-\lambda + 6-4\lambda - 20 = 0 \implies -5\lambda - 15 = 0 \implies \lambda = -3$.
Substituting $\lambda = -3$ into the circle equation:
$x^2+y^2+(1-3)x+(-3-6)y+(-10+30) = 0$.
$x^2+y^2-2x-9y+20=0$.
Comparing with $x^2+y^2+\alpha x+\beta y+\gamma=0$,we get $\alpha=-2, \beta=-9, \gamma=20$.
Thus,$\alpha+2\beta+\gamma = -2 + 2(-9) + 20 = -2 - 18 + 20 = 0$.

(B) The equation of circle $C_1$ is $x^2+y^2=16$.
The equation of circle $C_2$ is $(x-\alpha)^2+(y-\beta)^2=25$,which expands to $x^2-2\alpha x+\alpha^2+y^2-2\beta y+\beta^2=25$.
The equation of the common chord is given by $C_2-C_1=0$,which is $-2\alpha x-2\beta y+\alpha^2+\beta^2=9$.
The slope of the common chord is $m = -\frac{-2\alpha}{-2\beta} = -\frac{\alpha}{\beta} = \frac{3}{4}$.
Let $\alpha = -3\lambda$ and $\beta = 4\lambda$.
For the common chord to have maximum length,it must be the diameter of the smaller circle $C_1$,meaning the chord must pass through the centre of $C_1$,which is $(0,0)$.
Substituting $(0,0)$ into the chord equation: $-2\alpha(0)-2\beta(0)+\alpha^2+\beta^2=9$,so $\alpha^2+\beta^2=9$.
Substituting $\alpha$ and $\beta$ in terms of $\lambda$: $(-3\lambda)^2+(4\lambda)^2=9$ $\Rightarrow 25\lambda^2=9$ $\Rightarrow \lambda = \pm \frac{3}{5}$.
If $\lambda = \frac{3}{5}$,then $\alpha = -\frac{9}{5}$ and $\beta = \frac{12}{5}$,so $\alpha+\beta = \frac{3}{5}$.
If $\lambda = -\frac{3}{5}$,then $\alpha = \frac{9}{5}$ and $\beta = -\frac{12}{5}$,so $\alpha+\beta = -\frac{3}{5}$.

(D) Let $S \equiv x^2+y^2+3x+5y+4=0$ and $S' \equiv x^2+y^2+5x+3y+4=0$.
The equation of the common chord is $S-S'=0$.
$\Rightarrow (x^2+y^2+3x+5y+4) - (x^2+y^2+5x+3y+4) = 0$
$\Rightarrow -2x+2y=0$
$\Rightarrow x-y=0$.
The center of $S=0$ is $C\left(-\frac{3}{2}, -\frac{5}{2}\right)$.
The radius $r$ is $\sqrt{\left(\frac{3}{2}\right)^2 + \left(\frac{5}{2}\right)^2 - 4} = \sqrt{\frac{9}{4} + \frac{25}{4} - 4} = \sqrt{\frac{34}{4} - \frac{16}{4}} = \sqrt{\frac{18}{4}} = \sqrt{\frac{9}{2}} = \frac{3}{\sqrt{2}}$.
Let $p$ be the length of the perpendicular from $C$ to the line $x-y=0$:
$p = \frac{|-\frac{3}{2} - (-\frac{5}{2})|}{\sqrt{1^2+(-1)^2}} = \frac{|1|}{\sqrt{2}} = \frac{1}{\sqrt{2}}$.
The length of the common chord $AB = 2\sqrt{r^2-p^2}$.
$AB = 2\sqrt{\frac{9}{2} - \frac{1}{2}} = 2\sqrt{\frac{8}{2}} = 2\sqrt{4} = 2 \times 2 = 4$.

(D) The equation of the radical axis of the two circles is obtained by subtracting the two equations: $(x^2+y^2+ax+by+c) - (x^2+y^2+bx+ay+c) = 0$.
This simplifies to $(a-b)x + (b-a)y = 0$,which is $(a-b)(x-y) = 0$.
Assuming $a \neq b$,the radical axis is the line $x-y=0$,or $y=x$.
Substituting $y=x$ into the first circle equation: $x^2+x^2+ax+bx+c=0 \Rightarrow 2x^2+(a+b)x+c=0$.
Let the roots be $x_1$ and $x_2$. Then $x_1+x_2 = -\frac{a+b}{2}$ and $x_1x_2 = \frac{c}{2}$.
The points of intersection are $(x_1, x_1)$ and $(x_2, x_2)$.
The length of the common chord is the distance between these points: $\sqrt{(x_2-x_1)^2 + (x_2-x_1)^2} = \sqrt{2(x_2-x_1)^2} = \sqrt{2} |x_2-x_1|$.
Using $|x_2-x_1| = \sqrt{(x_1+x_2)^2 - 4x_1x_2} = \sqrt{\frac{(a+b)^2}{4} - 2c} = \frac{\sqrt{(a+b)^2-8c}}{2}$.
Thus,the length is $\sqrt{2} \cdot \frac{\sqrt{(a+b)^2-8c}}{2} = \frac{\sqrt{(a+b)^2-8c}}{\sqrt{2}}$.
Therefore,Statement-$I$ is false.
Statement-$II$ is a standard theorem in geometry of circles,which is true.
Hence,Statement-$I$ is false and Statement-$II$ is true.

(B) Given circles are $S_1: x^2+y^2-2x+4y-4=0$ and $S_2: x^2+y^2+4x-6y-3=0$.
The equation of the common chord is given by $S_1 - S_2 = 0$.
$(x^2+y^2-2x+4y-4) - (x^2+y^2+4x-6y-3) = 0$.
$-6x + 10y - 1 = 0$,which simplifies to $6x - 10y + 1 = 0$.
The perpendicular distance from the point $(x_1, y_1) = (1, 2)$ to the line $Ax + By + C = 0$ is given by $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$.
Substituting the values: $d = \frac{|6(1) - 10(2) + 1|}{\sqrt{6^2 + (-10)^2}}$.
$d = \frac{|6 - 20 + 1|}{\sqrt{36 + 100}} = \frac{|-13|}{\sqrt{136}} = \frac{13}{\sqrt{136}}$ units.

(D) The equation of the common chord of two circles $S_1 = 0$ and $S_2 = 0$ is given by $S_1 - S_2 = 0$. \\ Given circles are: \\ $S_1: x^2+y^2+2x+3y+1=0$ \\ $S_2: x^2+y^2-5x-6y+4=0$ \\ Subtracting $S_2$ from $S_1$: \\ $(x^2+y^2+2x+3y+1) - (x^2+y^2-5x-6y+4) = 0$ \\ $(2x - (-5x)) + (3y - (-6y)) + (1 - 4) = 0$ \\ $7x + 9y - 3 = 0$ \\ Thus,the equation of the common chord is $7x+9y-3=0$.

(C) The equation of a circle passing through the intersection of circles $S_1: x^2+y^2-8x=0$ and $S_2: x^2+y^2-9=0$ is given by $S_1 + \lambda S_2 = 0$,where $\lambda \neq -1$.
$(x^2+y^2-8x) + \lambda(x^2+y^2-9) = 0$
$(1+\lambda)x^2 + (1+\lambda)y^2 - 8x - 9\lambda = 0$
$x^2 + y^2 - \frac{8}{1+\lambda}x - \frac{9\lambda}{1+\lambda} = 0$.
The center of this circle is $C\left(\frac{4}{1+\lambda}, 0\right)$.
The equation of the common chord is $S_1 - S_2 = 0$,which is $(x^2+y^2-8x) - (x^2+y^2-9) = 0$,i.e.,$8x = 9$.
Since the common chord is the diameter,the center $C$ must lie on the line $8x = 9$.
$8\left(\frac{4}{1+\lambda}\right) = 9$
$32 = 9(1+\lambda) = 9 + 9\lambda$
$9\lambda = 23 \Rightarrow \lambda = \frac{23}{9}$.
Substituting $\lambda = \frac{23}{9}$ into the equation $(1+\lambda)x^2 + (1+\lambda)y^2 - 8x - 9\lambda = 0$:
$(1 + \frac{23}{9})x^2 + (1 + \frac{23}{9})y^2 - 8x - 9(\frac{23}{9}) = 0$
$\frac{32}{9}x^2 + \frac{32}{9}y^2 - 8x - 23 = 0$
Multiplying by $9$,we get $32x^2 + 32y^2 - 72x - 207 = 0$.
Thus,option $C$ is correct.

(D) The equations of the circles are $S_1: x^2+y^2+3x+5y+4=0$ and $S_2: x^2+y^2+5x+3y+4=0$.
The equation of the common chord is given by $S_1 - S_2 = 0$.
$(x^2+y^2+3x+5y+4) - (x^2+y^2+5x+3y+4) = 0$.
$-2x + 2y = 0$,which simplifies to $y = x$.
Substituting $y = x$ into $S_1$,we get $x^2 + x^2 + 3x + 5x + 4 = 0$,so $2x^2 + 8x + 4 = 0$,or $x^2 + 4x + 2 = 0$.
The roots are $x = \frac{-4 \pm \sqrt{16-8}}{2} = -2 \pm \sqrt{2}$.
Since $y = x$,the intersection points are $P(-2+\sqrt{2}, -2+\sqrt{2})$ and $Q(-2-\sqrt{2}, -2-\sqrt{2})$.
The length of the chord $PQ$ is $\sqrt{(-2-\sqrt{2} - (-2+\sqrt{2}))^2 + (-2-\sqrt{2} - (-2+\sqrt{2}))^2} = \sqrt{(-2\sqrt{2})^2 + (-2\sqrt{2})^2} = \sqrt{8 + 8} = \sqrt{16} = 4$.

(A) Given circles are $S_1 \equiv x^2+y^2+2x+2y+1=0$ and $S_2 \equiv x^2+y^2+4x+6y+4=0$.
The equation of the common chord is $S_1 - S_2 = 0$:
$(x^2+y^2+2x+2y+1) - (x^2+y^2+4x+6y+4) = 0$
$-2x - 4y - 3 = 0 \Rightarrow 2x + 4y + 3 = 0$.
The equation of a circle passing through the intersection of $S_1$ and $S_2$ is $S_1 + \lambda(S_1 - S_2) = 0$ (or $S_1 + \lambda(L) = 0$ where $L$ is the common chord).
$x^2+y^2+2x+2y+1 + \lambda(2x+4y+3) = 0$
$x^2+y^2+2x(1+\lambda) + 2y(1+2\lambda) + (1+3\lambda) = 0$.
The center of this circle is $(- (1+\lambda), -(1+2\lambda))$.
Since the common chord $2x+4y+3=0$ is a diameter,the center must lie on it:
$2(-(1+\lambda)) + 4(-(1+2\lambda)) + 3 = 0$
$-2 - 2\lambda - 4 - 8\lambda + 3 = 0$
$-10\lambda - 3 = 0 \Rightarrow \lambda = -\frac{3}{10}$.
Substituting $\lambda = -\frac{3}{10}$ into the circle equation:
$x^2+y^2+2x(1-\frac{3}{10}) + 2y(1-\frac{6}{10}) + (1-\frac{9}{10}) = 0$
$x^2+y^2+\frac{14x}{10} + \frac{8y}{10} + \frac{1}{10} = 0$
$10x^2+10y^2+14x+8y+1=0$.

(A) Let the two circles be $S_1 \equiv x^2+y^2-6x-7=0$ and $S_2 \equiv x^2+y^2-10x+16=0$.
The common chord is given by $S_1 - S_2 = 0$.
$(x^2+y^2-6x-7) - (x^2+y^2-10x+16) = 0$
$4x - 23 = 0 \Rightarrow x = \frac{23}{4}$.
The points of intersection are found by substituting $x = \frac{23}{4}$ into $S_1$:
$(\frac{23}{4})^2 + y^2 - 6(\frac{23}{4}) - 7 = 0$
$\frac{529}{16} + y^2 - \frac{138}{4} - 7 = 0$
$y^2 = \frac{138}{4} + 7 - \frac{529}{16} = \frac{552 + 112 - 529}{16} = \frac{135}{16}$.
So,$y = \pm \frac{3\sqrt{15}}{4}$.
The diameter endpoints are $(\frac{23}{4}, \frac{3\sqrt{15}}{4})$ and $(\frac{23}{4}, -\frac{3\sqrt{15}}{4})$.
The equation of the circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is $(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0$.
$(x-\frac{23}{4})(x-\frac{23}{4}) + (y-\frac{3\sqrt{15}}{4})(y+\frac{3\sqrt{15}}{4}) = 0$
$(x-\frac{23}{4})^2 + y^2 - \frac{135}{16} = 0$
$x^2 - \frac{23}{2}x + \frac{529}{16} + y^2 - \frac{135}{16} = 0$
$x^2 + y^2 - \frac{23}{2}x + \frac{394}{16} = 0$
$x^2 + y^2 - \frac{23}{2}x + \frac{197}{8} = 0$
Multiplying by $8$: $8x^2 + 8y^2 - 92x + 197 = 0$.

(B) The equations of the circles are $S_1: x^2+y^2+2x+2y+1=0$ and $S_2: x^2+y^2+4x+3y+2=0$.
The equation of the common chord is given by $S_1 - S_2 = 0$.
$(x^2+y^2+2x+2y+1) - (x^2+y^2+4x+3y+2) = 0$.
$-2x-y-1=0$,which simplifies to $2x+y+1=0$.
The center of circle $S_1$ is $(-1, -1)$ and its radius $r_1 = \sqrt{(-1)^2+(-1)^2-1} = \sqrt{1+1-1} = 1$.
The perpendicular distance $d$ from the center $(-1, -1)$ to the line $2x+y+1=0$ is $d = \frac{|2(-1)+(-1)+1|}{\sqrt{2^2+1^2}} = \frac{|-2|}{\sqrt{5}} = \frac{2}{\sqrt{5}}$.
The length of the common chord is $2\sqrt{r_1^2-d^2} = 2\sqrt{1^2-(\frac{2}{\sqrt{5}})^2} = 2\sqrt{1-\frac{4}{5}} = 2\sqrt{\frac{1}{5}} = \frac{2}{\sqrt{5}}$.
The common chord is the diameter of the required circle,so its diameter $D = \frac{2}{\sqrt{5}}$.
The radius of the required circle is $R = \frac{D}{2} = \frac{1}{\sqrt{5}}$.

(A) Given,$S \equiv x^2+y^2-2x-4y+1=0$.
For the $y$-axis,set $x=0$,which gives $y^2-4y+1=0$.
Solving for $y$,we get $y = \frac{4 \pm \sqrt{16-4}}{2} = 2 \pm \sqrt{3}$.
Since $OA > OB$,we have $A(0, 2+\sqrt{3})$ and $B(0, 2-\sqrt{3})$.
The radical axis of $S=0$ and $S^{\prime}=0$ is given by $S-S^{\prime}=0$.
$(x^2+y^2-2x-4y+1) - (x^2+y^2-4x-2y+4) = 0 \Rightarrow 2x-2y-3=0$.
For the $y$-axis,set $x=0$,which gives $-2y-3=0 \Rightarrow y = -\frac{3}{2}$.
So,$C$ is $(0, -\frac{3}{2})$.
Let the ratio in which $C$ divides $AB$ be $k:1$.
Using the section formula for the $y$-coordinate: $-\frac{3}{2} = \frac{k(2-\sqrt{3}) + 1(2+\sqrt{3})}{k+1}$.
$-3(k+1) = 2(k(2-\sqrt{3}) + 2+\sqrt{3})$.
$-3k-3 = 4k - 2k\sqrt{3} + 4 + 2\sqrt{3}$.
$-3k-4k + 2k\sqrt{3} = 4+3+2\sqrt{3}$.
$k(2\sqrt{3}-7) = 7+2\sqrt{3}$.
$k = \frac{7+2\sqrt{3}}{2\sqrt{3}-7} = \frac{7+2\sqrt{3}}{-(7-2\sqrt{3})}$.
Thus,the ratio $k:1$ is $(7+2\sqrt{3}) : (-7+2\sqrt{3})$.

(A) The equation of the first circle is $x^2+y^2+2gx+2fy+c=0$.
Dividing the second circle $2x^2+2y^2+3x+8y+2c=0$ by $2$,we get $x^2+y^2+\frac{3}{2}x+4y+c=0$.
The radical axis is obtained by subtracting the two equations: $(2g-\frac{3}{2})x + (2f-4)y = 0$,which simplifies to $(4g-3)x + 2(2f-4)y = 0$,or $(4g-3)x + 4(f-2)y = 0$.
This line touches the circle $x^2+y^2+2x+2y+1=0$,which can be written as $(x+1)^2+(y+1)^2=1$.
The center of this circle is $(-1, -1)$ and the radius is $1$.
The perpendicular distance from the center $(-1, -1)$ to the line $(4g-3)x + 4(f-2)y = 0$ must be equal to the radius $1$:
$\frac{|(4g-3)(-1) + 4(f-2)(-1)|}{\sqrt{(4g-3)^2 + (4(f-2))^2}} = 1$.
$|-(4g-3) - 4(f-2)| = \sqrt{(4g-3)^2 + 16(f-2)^2}$.
Squaring both sides: $(4g-3)^2 + 16(f-2)^2 + 8(4g-3)(f-2) = (4g-3)^2 + 16(f-2)^2$.
This simplifies to $8(4g-3)(f-2) = 0$,which implies $(4g-3)(f-2) = 0$.

(B) The given circle is $x^2+y^2+6x+4y-12=0$.
Comparing this with $x^2+y^2+2gx+2fy+c=0$,we get $g=3, f=2, c=-12$.
The radius $r = \sqrt{g^2+f^2-c} = \sqrt{9+4+12} = \sqrt{25} = 5$.
Let the distance between the centers of the two circles be $d$.
The length of the common chord is $L = 2\sqrt{r^2 - (d/2)^2} = 2\sqrt{17}$.
Thus,$\sqrt{25 - d^2/4} = \sqrt{17} \implies 25 - d^2/4 = 17 \implies d^2/4 = 8 \implies d^2 = 32 \implies d = 4\sqrt{2}$.
Let $\theta$ be the angle between the circles. The formula for the angle between two circles of equal radius $r$ is $\cos \theta = 1 - \frac{d^2}{2r^2}$.
Substituting the values,$\cos \theta = 1 - \frac{32}{2(25)} = 1 - \frac{32}{50} = 1 - \frac{16}{25} = \frac{9}{25}$.
Therefore,$\theta = \operatorname{Cos}^{-1}\left(\frac{9}{25}\right)$.
However,the angle between two circles is defined as the angle between their tangents at the point of intersection. For two circles of equal radius,the angle $\theta$ is given by $2 \sin(\theta/2) = L/r$.
$2 \sin(\theta/2) = \frac{2\sqrt{17}}{5} \implies \sin(\theta/2) = \frac{\sqrt{17}}{5}$.
Using $\cos \theta = 1 - 2\sin^2(\theta/2) = 1 - 2(17/25) = 1 - 34/25 = -9/25$.
Since we need the acute angle,we consider the geometry where $\cos \theta = |1 - d^2/(2r^2)| = 7/25$ is not correct,the standard formula yields $\theta = 2 \operatorname{Sin}^{-1}(\frac{d}{2r}) = 2 \operatorname{Sin}^{-1}(\frac{4\sqrt{2}}{10}) = 2 \operatorname{Sin}^{-1}(\frac{2\sqrt{2}}{5})$.
Re-evaluating,the angle $\theta$ between circles is $2 \operatorname{Sin}^{-1}(\frac{L}{2r}) = 2 \operatorname{Sin}^{-1}(\frac{\sqrt{17}}{5})$.

(D) The given equations of the circles are $C_1: x^2+y^2-6x-4y+9=0$ and $C_2: x^2+y^2+2x+2y-7=0$.
Subtracting $C_1$ from $C_2$, we get the equation of the common tangent (radical axis) at the point of contact:
$(x^2+y^2+2x+2y-7) - (x^2+y^2-6x-4y+9) = 0$
$8x+6y-16=0 \implies 4x+3y=8$.
From this, $y = \frac{8-4x}{3}$.
Substituting this into $C_1$: $x^2 + (\frac{8-4x}{3})^2 - 6x - 4(\frac{8-4x}{3}) + 9 = 0$.
Multiplying by $9$: $9x^2 + (64 - 64x + 16x^2) - 54x - 12(8-4x) + 81 = 0$.
$25x^2 - 70x + 49 = 0 \implies (5x-7)^2 = 0$.
Thus, $x = \alpha = \frac{7}{5}$.
Then $y = \beta = \frac{8-4(7/5)}{3} = \frac{40-28}{15} = \frac{12}{15} = \frac{4}{5}$.
We need to find $7\beta$.
$7\beta = 7 \times \frac{4}{5} = \frac{28}{5}$.
Since $\alpha = \frac{7}{5}$, we have $4\alpha = 4 \times \frac{7}{5} = \frac{28}{5}$.
Therefore, $7\beta = 4\alpha$.

(C) Given circles are $S_1: x^2+y^2-2x-4y-4=0$ and $S_2: x^2+y^2+2x+4y-11=0$.
To find the common chord,subtract $S_2$ from $S_1$:
$(x^2+y^2-2x-4y-4) - (x^2+y^2+2x+4y-11) = 0$
$-4x-8y+7 = 0$
$4x+8y-7 = 0$.
Since the radical axis (common chord) exists,the circles intersect at two distinct points lying on the line $4x+8y-7=0$.

(C) Let the given circles be $S_1: x^2+y^2+2x+3y+2=0$ and $S_2: x^2+y^2+2x-3y-4=0$.
The equation of the common chord is given by $S_1 - S_2 = 0$.
$(x^2+y^2+2x+3y+2) - (x^2+y^2+2x-3y-4) = 0$
$6y + 6 = 0 \Rightarrow y = -1$.
Substituting $y = -1$ into $S_1 = 0$:
$x^2 + (-1)^2 + 2x + 3(-1) + 2 = 0$
$x^2 + 1 + 2x - 3 + 2 = 0$
$x^2 + 2x = 0 \Rightarrow x(x+2) = 0$.
So,$x = 0$ or $x = -2$.
The endpoints of the diameter are $(0, -1)$ and $(-2, -1)$.
The equation of the circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is $(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0$.
$(x-0)(x+2) + (y+1)(y+1) = 0$
$x^2 + 2x + y^2 + 2y + 1 = 0$.

(C) The equation of the circle is $x^2+y^2-4x+2y-4=0$. The center is $C(2, -1)$ and the radius is $r = \sqrt{2^2 + (-1)^2 - (-4)} = \sqrt{4+1+4} = 3$.
Let $P = (-3, 1)$. The circle with diameter $PC$ has the equation $(x - (-3))(x - 2) + (y - 1)(y - (-1)) = 0$.
This simplifies to $(x+3)(x-2) + (y-1)(y+1) = 0$,which is $x^2+x-6 + y^2-1 = 0$,or $x^2+y^2+x-7=0$.
The points of contact $A$ and $B$ lie on this circle because $\angle PAC = 90^\circ$ and $\angle PBC = 90^\circ$.
Thus,the circumcircle of $\triangle PAB$ is the circle with diameter $PC$,which is $x^2+y^2+x-7=0$.

(C) Given circles are $S_1 \equiv x^2 + y^2 + 2x + 2y + 1 = 0$ and $S_2 \equiv x^2 + y^2 - 2x - 2y + 1 = 0$.
Centre of $S_1$ is $C_1 = (-1, -1)$ and radius $r_1 = \sqrt{(-1)^2 + (-1)^2 - 1} = 1$.
Centre of $S_2$ is $C_2 = (1, 1)$ and radius $r_2 = \sqrt{(1)^2 + (1)^2 - 1} = 1$.
The distance between centres $C_1 C_2 = \sqrt{(1 - (-1))^2 + (1 - (-1))^2} = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}$.
Since $C_1 C_2 > r_1 + r_2$ $(2\sqrt{2} > 2)$,the circles are separate.
The transverse common tangents intersect at the internal centre of similitude $P$,which divides $C_1 C_2$ in the ratio $r_1 : r_2 = 1 : 1$.
$P = \left( \frac{1(-1) + 1(1)}{1+1}, \frac{1(-1) + 1(1)}{1+1} \right) = (0, 0)$.
The pair of tangents from $(0, 0)$ to $S_1$ is given by $T^2 = S_1 S_{11}$,where $T = x(0) + y(0) + 1(x+0) + 1(y+0) + 1 = x + y + 1$ and $S_{11} = 0^2 + 0^2 + 2(0) + 2(0) + 1 = 1$.
Thus,$(x + y + 1)^2 = (x^2 + y^2 + 2x + 2y + 1)(1)$.
$x^2 + y^2 + 1 + 2xy + 2x + 2y = x^2 + y^2 + 2x + 2y + 1$.
$2xy = 0 \Rightarrow xy = 0$.

(C) Let the coordinates of points $P$ and $Q$ be $(h, k)$ and $(p, q)$ respectively.
The equation of the chord of contact of the point $P(h, k)$ with respect to the circle $x^2+y^2=a^2$ is $xh+yk=a^2$.
Since this chord passes through $Q(p, q)$,we have $ph+qk=a^2$.
The lengths of the tangents from $P$ and $Q$ are given by $l_1 = \sqrt{h^2+k^2-a^2}$ and $l_2 = \sqrt{p^2+q^2-a^2}$.
Squaring these,we get $l_1^2 = h^2+k^2-a^2$ and $l_2^2 = p^2+q^2-a^2$.
The distance $PQ$ is given by $PQ = \sqrt{(h-p)^2+(k-q)^2} = \sqrt{h^2+k^2+p^2+q^2-2(hp+kq)}$.
Substituting $hp+kq=a^2$,we get $PQ = \sqrt{(h^2+k^2)+(p^2+q^2)-2a^2}$.
Substituting $h^2+k^2 = l_1^2+a^2$ and $p^2+q^2 = l_2^2+a^2$,we get $PQ = \sqrt{(l_1^2+a^2)+(l_2^2+a^2)-2a^2} = \sqrt{l_1^2+l_2^2}$.
Thus,option $(C)$ is correct.

(B) The equation of the circle is $x^2+y^2-8x-10y+5=0$.
Completing the square,we get $(x-4)^2+(y-5)^2 = 16+25-5 = 36 = 6^2$.
Thus,the center $C$ is $(4, 5)$ and the radius $r = 6$.
The length of the tangent $PA$ from $P(-2, -3)$ is $\sqrt{S_1} = \sqrt{(-2)^2+(-3)^2-8(-2)-10(-3)+5} = \sqrt{4+9+16+30+5} = \sqrt{64} = 8$.
In the right-angled triangle $\triangle CAP$,$\angle CAP = 90^\circ$.
Let $\angle APC = \frac{\theta}{2}$. Then $\tan(\frac{\theta}{2}) = \frac{CA}{PA} = \frac{6}{8} = \frac{3}{4}$.
Using the formula $\tan \theta = \frac{2 \tan(\frac{\theta}{2})}{1-\tan^2(\frac{\theta}{2})}$,we have:
$\tan \theta = \frac{2(\frac{3}{4})}{1-(\frac{3}{4})^2} = \frac{\frac{3}{2}}{1-\frac{9}{16}} = \frac{\frac{3}{2}}{\frac{7}{16}} = \frac{3}{2} \times \frac{16}{7} = \frac{24}{7}$.

(D) Given circle: $x^2+y^2-2gx-2hy+h^2=0$.
Center $C = (g, h)$,radius $r = \sqrt{g^2+h^2-h^2} = |g|$.
Length of tangent $OP = OQ = \sqrt{S_1} = \sqrt{0^2+0^2-2g(0)-2h(0)+h^2} = |h|$.
Area of $\triangle OPQ = \frac{1}{2} \cdot OP \cdot OQ \cdot \sin(2\theta) = \frac{1}{2} h^2 \sin(2\theta)$.
In $\triangle OPC$,$\angle OPC = 90^{\circ}$,so $\tan \theta = \frac{PC}{OP} = \frac{|g|}{|h|}$.
Then $\sin(2\theta) = \frac{2 \tan \theta}{1+\tan^2 \theta} = \frac{2(|g|/|h|)}{1+(g^2/h^2)} = \frac{2|g||h|}{h^2+g^2}$.
Area of $\triangle OPQ = \frac{1}{2} h^2 \cdot \frac{2|g||h|}{h^2+g^2} = \frac{|g|h^2|h|}{h^2+g^2} = \frac{|g|h^3}{h^2+g^2}$ sq. units.
Assuming $g, h > 0$,the area is $\frac{gh^3}{h^2+g^2}$ sq. units.

(D) Let the radius of a circle touching both coordinate axes be $r$. The equation of such a circle is $(x-r)^2 + (y-r)^2 = r^2$.
Since the circle passes through $A=(1,2)$,we have $(1-r)^2 + (2-r)^2 = r^2$.
Expanding this,we get $1 - 2r + r^2 + 4 - 4r + r^2 = r^2$,which simplifies to $r^2 - 6r + 5 = 0$.
Solving for $r$,we get $(r-1)(r-5) = 0$,so $r=1$ or $r=5$.
The two circles are $C_1: (x-1)^2 + (y-1)^2 = 1$ and $C_2: (x-5)^2 + (y-5)^2 = 25$.
The common chord $AB$ is given by $C_1 - C_2 = 0$.
$C_1: x^2 + y^2 - 2x - 2y + 1 = 0$
$C_2: x^2 + y^2 - 10x - 10y + 25 = 0$
Subtracting $C_2$ from $C_1$: $(x^2 - x^2) + (y^2 - y^2) + (-2x + 10x) + (-2y + 10y) + (1 - 25) = 0$,which gives $8x + 8y - 24 = 0$,or $x + y = 3$.
Since $A=(1,2)$ lies on $x+y=3$,the point $B$ is the reflection of $A$ across the line connecting the centers $(1,1)$ and $(5,5)$. The line of centers is $y=x$.
The reflection of $(1,2)$ across $y=x$ is $(2,1)$. Thus $B=(2,1)$.
The length $AB = \sqrt{(2-1)^2 + (1-2)^2} = \sqrt{1^2 + (-1)^2} = \sqrt{2}$.

(C) The equation of the circle is $x^2+y^2=125$.
Given $P(8, \beta)$ is the midpoint of the chord.
The equation of the chord with midpoint $(x_1, y_1)$ is $T=S_1$,which is $xx_1+yy_1=x_1^2+y_1^2$.
Substituting $(8, \beta)$,we get $8x+\beta y = 64+\beta^2$,or $8x+\beta y - (64+\beta^2) = 0$.
The slope of this chord is $m = -\frac{8}{\beta}$.
For $m$ to be an integer,$\beta$ must be a divisor of $8$. Thus,$\beta \in \{ \pm 1, \pm 2, \pm 4, \pm 8 \}$.
Since the point $P(8, \beta)$ must lie inside the circle,$8^2+\beta^2 < 125$,which implies $64+\beta^2 < 125$,so $\beta^2 < 61$.
Checking the values:
If $\beta = \pm 1$,$\beta^2 = 1 < 61$ (Valid,$m = \mp 8$).
If $\beta = \pm 2$,$\beta^2 = 4 < 61$ (Valid,$m = \mp 4$).
If $\beta = \pm 4$,$\beta^2 = 16 < 61$ (Valid,$m = \mp 2$).
If $\beta = \pm 8$,$\beta^2 = 64 > 61$ (Invalid).
Thus,the possible values for $\beta$ are $\pm 1, \pm 2, \pm 4$,which gives a total of $6$ values.

(B) The equation of the common chord of the two circles $C_1: x^2+y^2-2x+2y+1=0$ and $C_2: x^2+y^2-2x-2y-2=0$ is given by $C_1 - C_2 = 0$.
$(x^2+y^2-2x+2y+1) - (x^2+y^2-2x-2y-2) = 0$
$4y + 3 = 0 \Rightarrow y = -\frac{3}{4}$.
Since this common chord is the diameter of circle $S$,the centre of circle $S$ must lie on the line $y = -\frac{3}{4}$.
Among the given options,the points with $y$-coordinate $-\frac{3}{4}$ are $\left(\frac{1}{2}, -\frac{3}{4}\right)$,$\left(1, -\frac{3}{4}\right)$,and $\left(-\frac{1}{2}, -\frac{3}{4}\right)$.
However,the centre of the circle $S$ must also satisfy the condition of being the midpoint of the chord if the chord is a diameter. The common chord is a horizontal line $y = -\frac{3}{4}$. The centre of the circle $S$ is $(1, -\frac{3}{4})$.

(A) Let the circle $S_1 \equiv x^2+y^2-2x-1=0$ have centre $C(1,0)$ and radius $r_1 = \sqrt{2}$.
Let the circle $S_2 \equiv x^2+y^2-2x=0$ have centre $C(1,0)$ and radius $r_2 = 1$.
Since $P$ lies on $S_1$,$PA$ and $PB$ are tangents to $S_2$ from $P$. Thus,$CA \perp PA$ and $CB \perp PB$.
In $\triangle CAB$,$CA = CB = r_2 = 1$. The chord of contact $AB$ is perpendicular to $CP$.
Let $M$ be the midpoint of $AB$. Since $\triangle CAB$ is an isosceles triangle with $CA=CB$,the circumcentre $O$ of $\triangle CAB$ lies on the line $CP$.
Let $P = (1 + \sqrt{2}\cos\theta, \sqrt{2}\sin\theta)$. The line $CP$ has the equation $y = \tan\theta(x-1)$.
The distance $CM$ from $C(1,0)$ to the chord $AB$ is given by $CM = \frac{r_2^2}{CP} = \frac{1^2}{\sqrt{2}} = \frac{1}{\sqrt{2}}$.
The circumcentre $O(h,k)$ lies on $CP$ at a distance $R_{circum}$ from $C$. In $\triangle CAM$,$CA=1$ and $AM = \sqrt{1^2 - (1/\sqrt{2})^2} = 1/\sqrt{2}$.
The circumradius $R_{circum} = \frac{CA^2}{2CM} = \frac{1^2}{2(1/\sqrt{2})} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.
Thus,the distance of $O(h,k)$ from $C(1,0)$ is $1/\sqrt{2}$.
$(h-1)^2 + k^2 = (1/\sqrt{2})^2 = 1/2$.
$h^2 - 2h + 1 + k^2 = 1/2 \Rightarrow h^2 + k^2 - 2h + 1/2 = 0$.
Multiplying by $2$,we get $2h^2 + 2k^2 - 4h + 1 = 0$.
Replacing $(h,k)$ with $(x,y)$,the locus is $2x^2 + 2y^2 - 4x + 1 = 0$.

(A) The equation of the circle is $x^2+y^2=3$. The equation of the chord of contact for a point $(x_1, y_1)$ with respect to the circle $x^2+y^2=r^2$ is given by $xx_1+yy_1=r^2$.
For point $(2,0)$,$L_1: 2x+0y=3 \Rightarrow 2x-3=0$.
For point $(1,-2)$,$L_2: 1x-2y=3 \Rightarrow x-2y-3=0$.
For point $(4,4)$,$L_3: 4x+4y=3 \Rightarrow 4x+4y-3=0$.
To check for concurrency,we solve $L_1$ and $L_2$: $x=\frac{3}{2}$,$y=\frac{1}{2}(\frac{3}{2}-3) = -\frac{3}{4}$.
Substituting $(\frac{3}{2}, -\frac{3}{4})$ into $L_3$: $4(\frac{3}{2})+4(-\frac{3}{4})-3 = 6-3-3 = 0$.
Since the point satisfies $L_3$,the lines are concurrent.

(C) The line $3x-y+k=0$ touches the circle $x^2+y^2+4x-6y+3=0$.
The center of the circle is $(-2, 3)$ and the radius $r = \sqrt{(-2)^2 + 3^2 - 3} = \sqrt{4+9-3} = \sqrt{10}$.
Since the line is tangent to the circle,the perpendicular distance from the center $(-2, 3)$ to the line $3x-y+k=0$ must be equal to the radius $r$.
$\frac{|3(-2) - (3) + k|}{\sqrt{3^2 + (-1)^2}} = \sqrt{10}$
$|k-9| = \sqrt{10} \times \sqrt{10} = 10$
$k-9 = 10$ or $k-9 = -10$
$k = 19$ or $k = -1$.
Given $k_1 < k_2$,we have $k_1 = -1$ and $k_2 = 19$.
The point is $(k_1, k_2) = (-1, 19)$.
The equation of the chord of contact of a point $(x_1, y_1)$ with respect 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,$g=2, f=-3, c=3, x_1=-1, y_1=19$.
$-x + 19y + 2(x-1) - 3(y+19) + 3 = 0$
$-x + 19y + 2x - 2 - 3y - 57 + 3 = 0$
$x + 16y - 56 = 0$.

(C) The equation of the circle is $x^2+y^2+x+3y=0$.
Let the point of intersection of the tangents at $A$ and $B$ be $P(h, k)$.
The equation of the chord of contact $AB$ is given by $T=0$:
$xh + yk + \frac{x+h}{2} + 3\left(\frac{y+k}{2}\right) = 0$
$x\left(h+\frac{1}{2}\right) + y\left(k+\frac{3}{2}\right) + \frac{h+3k}{2} = 0$.
Comparing this with the given line $x+y+1=0$,we have:
$\frac{h+1/2}{1} = \frac{k+3/2}{1} = \frac{(h+3k)/2}{1}$.
From $h+1/2 = k+3/2$,we get $h-k=1$.
From $h+1/2 = (h+3k)/2$,we get $2h+1 = h+3k$,so $h-3k = -1$.
Solving these equations,we get $k=1$ and $h=2$. So $P=(2, 1)$.
The family of circles passing through $A$ and $B$ is $x^2+y^2+x+3y + \lambda(x+y+1) = 0$.
Since this circle passes through $P(2, 1)$,we have $4+1+2+3 + \lambda(2+1+1) = 0$,which gives $10 + 4\lambda = 0$,so $\lambda = -5/2$.
The equation of the circle is $x^2+y^2+x+3y - \frac{5}{2}(x+y+1) = 0$,which simplifies to $x^2+y^2 - \frac{3}{2}x + \frac{1}{2}y - \frac{5}{2} = 0$.
The centre of this circle is $\left(-\frac{-3/2}{2}, -\frac{1/2}{2}\right) = \left(\frac{3}{4}, -\frac{1}{4}\right)$.

(B) The line $L \equiv -mx+y-1=0$ is a chord of the circle $S \equiv x^2+y^2-1=0$ and it touches the circle $S_1 \equiv x^2+y^2-4x+1=0$.
The center of $S_1$ is $(2, 0)$ and its radius is $r = \sqrt{2^2+0^2-1} = \sqrt{3}$.
Since the line touches the circle $S_1$,the perpendicular distance from the center $(2, 0)$ to the line $L$ must equal the radius $\sqrt{3}$.
$\frac{|-m(2) + 0 - 1|}{\sqrt{(-m)^2 + 1^2}} = \sqrt{3}$
$\frac{|-2m-1|}{\sqrt{m^2+1}} = \sqrt{3}$
Squaring both sides: $\frac{(2m+1)^2}{m^2+1} = 3$
$4m^2 + 4m + 1 = 3m^2 + 3$
$m^2 + 4m - 2 = 0$
Solving for $m$: $m = \frac{-4 \pm \sqrt{16 - 4(1)(-2)}}{2} = \frac{-4 \pm \sqrt{24}}{2} = -2 \pm \sqrt{6}$.
Substituting $m$ into $L$: $y - (-2 \pm \sqrt{6})x - 1 = 0$,which simplifies to $y + (2 \mp \sqrt{6})x - 1 = 0$.
The chord of contact of a point $(h, k)$ with respect to $S \equiv x^2+y^2-1=0$ is $hx + ky - 1 = 0$.
Comparing $hx + ky - 1 = 0$ with $(2 \mp \sqrt{6})x + y - 1 = 0$,we get $h = 2 \mp \sqrt{6}$ and $k = 1$.
Thus,the point is $(2 \pm \sqrt{6}, 1)$.

(A) Let $(h, k)$ be the point of intersection of the tangents. The chord of contact of these tangents is the common chord of the two circles.
Subtracting the equations of the circles $x^2+y^2=12$ and $x^2+y^2-5x+3y-2=0$ gives the equation of the common chord:
$(x^2+y^2-5x+3y-2) - (x^2+y^2-12) = 0$
$-5x+3y+10=0$,which simplifies to $5x-3y-10=0$.
The equation of the chord of contact of the circle $x^2+y^2=12$ with respect to the point $(h, k)$ is $hx+ky=12$,or $hx+ky-12=0$.
Since both equations represent the same line,their coefficients must be proportional:
$\frac{h}{5} = \frac{k}{-3} = \frac{-12}{-10} = \frac{6}{5}$.
Equating the ratios for $k$:
$\frac{k}{-3} = \frac{6}{5} \Rightarrow k = -\frac{18}{5}$.
Thus,the ordinate of the point of intersection is $-\frac{18}{5}$.

(A) For circle $S_1: x^2+y^2-6x-8y+9=0$,the center is $C_1(3, 4)$ and radius $r_1 = \sqrt{3^2+4^2-9} = \sqrt{16} = 4$.
For circle $S_2: x^2+y^2+2x-2y+1=0$,the center is $C_2(-1, 1)$ and radius $r_2 = \sqrt{(-1)^2+1^2-1} = \sqrt{1} = 1$.
The distance between centers $C_1C_2 = \sqrt{(3 - (-1))^2 + (4 - 1)^2} = \sqrt{4^2 + 3^2} = 5$.
Since $C_1C_2 = r_1 + r_2 = 4 + 1 = 5$,the circles touch each other externally.
The transverse common tangent at the point of contact is the radical axis of the two circles,given by $S_1 - S_2 = 0$.
$(x^2+y^2-6x-8y+9) - (x^2+y^2+2x-2y+1) = 0$.
$-8x - 6y + 8 = 0$.
Dividing by $-2$,we get $4x + 3y - 4 = 0$.

(D) Given circles are $S_1 \equiv x^2+y^2-2x+18y+78=0$ and $S_2 \equiv x^2+y^2+8x-6y-200=0$.
Centre $C_1 = (1, -9)$ and radius $r_1 = \sqrt{1^2+(-9)^2-78} = \sqrt{1+81-78} = 2$.
Centre $C_2 = (-4, 3)$ and radius $r_2 = \sqrt{(-4)^2+3^2-(-200)} = \sqrt{16+9+200} = 15$.
Distance $C_1C_2 = \sqrt{(-4-1)^2+(3-(-9))^2} = \sqrt{(-5)^2+12^2} = \sqrt{25+144} = 13$.
Since $r_2 - r_1 = 15 - 2 = 13 = C_1C_2$,the circles touch each other internally.
The equation of the common tangent is $S_1 - S_2 = 0$.
$(x^2+y^2-2x+18y+78) - (x^2+y^2+8x-6y-200) = 0$.
$-10x + 24y + 278 = 0$,which simplifies to $-5x + 12y + 139 = 0$.
Checking option $D$: $-5\left(\frac{-2}{5}\right) + 12\left(\frac{-47}{4}\right) + 139 = 2 - 141 + 139 = 0$.
Thus,the point $\left(\frac{-2}{5}, \frac{-47}{4}\right)$ lies on the common tangent.

(B) The equations of the circles are $C_1: x^2+y^2-6x+5=0$ and $C_2: x^2+y^2+4y-5=0$.
Subtracting the two equations gives the equation of the common chord: $(x^2+y^2-6x+5) - (x^2+y^2+4y-5) = 0$,which simplifies to $-6x-4y+10=0$,or $3x+2y-5=0$.
The center of $C_1$ is $(3, 0)$ and its radius $r_1 = \sqrt{3^2+0^2-5} = \sqrt{4} = 2$.
The distance $d$ from the center $(3, 0)$ to the line $3x+2y-5=0$ is $d = \frac{|3(3)+2(0)-5|}{\sqrt{3^2+2^2}} = \frac{|9-5|}{\sqrt{13}} = \frac{4}{\sqrt{13}}$.
The length of the common chord is $2\sqrt{r_1^2-d^2} = 2\sqrt{2^2 - (\frac{4}{\sqrt{13}})^2} = 2\sqrt{4 - \frac{16}{13}} = 2\sqrt{\frac{52-16}{13}} = 2\sqrt{\frac{36}{13}} = \frac{12}{\sqrt{13}}$.

(D) The given circles are $C_1: x^2+y^2-4x-12=0$ and $C_2: x^2+y^2+4x-12=0$.
Their centers are $A(-2, 0)$ and $B(2, 0)$, and both have radius $r = \sqrt{2^2+0^2+12} = 4$.
The common chord is obtained by subtracting the equations: $(x^2+y^2+4x-12) - (x^2+y^2-4x-12) = 0$, which gives $8x = 0$, or $x = 0$ (the $y$-axis).
The intersection points $C$ and $D$ are found by substituting $x=0$ into $x^2+y^2-4x-12=0$, giving $y^2 = 12$, so $y = \pm 2\sqrt{3}$. Thus, $C(0, 2\sqrt{3})$ and $D(0, -2\sqrt{3})$.
The diagonals of the rhombus are $AB$ (length $d_1 = 4$) and $CD$ (length $d_2 = 4\sqrt{3}$).
The area of the rhombus is $\frac{1}{2} \times d_1 \times d_2 = \frac{1}{2} \times 4 \times 4\sqrt{3} = 8\sqrt{3}$ square units.

(A) The common chord of the circles $S_1: x^2+y^2+2x+3y+1=0$ and $S_2: x^2+y^2+4x+3y+2=0$ is given by $S_1 - S_2 = 0$.
$(x^2+y^2+2x+3y+1) - (x^2+y^2+4x+3y+2) = 0$
$-2x - 1 = 0 \Rightarrow x = -\frac{1}{2}$.
Any circle passing through the intersection of $S_1$ and $S_2$ is given by $S_1 + \lambda S_2 = 0$.
$(x^2+y^2+2x+3y+1) + \lambda(x^2+y^2+4x+3y+2) = 0$
$(1+\lambda)x^2 + (1+\lambda)y^2 + (2+4\lambda)x + (3+3\lambda)y + (1+2\lambda) = 0$.
Dividing by $(1+\lambda)$,the center of this circle is $(-\frac{2+4\lambda}{2(1+\lambda)}, -\frac{3+3\lambda}{2(1+\lambda)})$.
Since the common chord $x = -\frac{1}{2}$ is the diameter,the center must lie on this line.
$-\frac{2+4\lambda}{2(1+\lambda)} = -\frac{1}{2}$ $\Rightarrow 2+4\lambda = 1+\lambda$ $\Rightarrow 3\lambda = -1$ $\Rightarrow \lambda = -\frac{1}{3}$.
Substituting $\lambda = -\frac{1}{3}$ into the equation:
$(1-\frac{1}{3})x^2 + (1-\frac{1}{3})y^2 + (2-\frac{4}{3})x + (3-1)y + (1-\frac{2}{3}) = 0$
$\frac{2}{3}x^2 + \frac{2}{3}y^2 + \frac{2}{3}x + 2y + \frac{1}{3} = 0$
Multiplying by $3$,we get $2x^2+2y^2+2x+6y+1=0$.

(B) The equations of the circles are $S_1: x^2+y^2+4y=0$ and $S_2: x^2+y^2-4x-5=0$.
The equation of the common chord is given by $S_1 - S_2 = 0$,which is $(x^2+y^2+4y) - (x^2+y^2-4x-5) = 0$,simplifying to $4x+4y+5=0$.
Let the circle $S=0$ have its centre at $(h, k)$. Since the common chord is the diameter of $S=0$,the centre $(h, k)$ must lie on the common chord $4x+4y+5=0$. Thus,$4h+4k+5=0$.
Also,the centre of $S=0$ must be the midpoint of the common chord if the circle is uniquely determined by the chord as its diameter. The common chord intersects the line joining the centres of $S_1$ and $S_2$. The centre of $S_1$ is $C_1(0, -2)$ and the centre of $S_2$ is $C_2(2, 0)$.
The line joining the centres $C_1$ and $C_2$ has the equation $y - 0 = \frac{-2-0}{0-2}(x-2)$,which simplifies to $y = x-2$ or $x-y-2=0$.
The centre $(h, k)$ of the circle $S=0$ is the intersection of the common chord $4x+4y+5=0$ and the line of centres $x-y-2=0$.
Solving the system:
$4x+4y = -5$
$x-y = 2 \Rightarrow y = x-2$
Substituting $y$ in the first equation: $4x + 4(x-2) = -5$ $\Rightarrow 8x - 8 = -5$ $\Rightarrow 8x = 3$ $\Rightarrow x = \frac{3}{8}$.
Thus,the abscissa of the centre is $\frac{3}{8}$.

(A) Given equations of circles are $S_1: x^2+y^2-4x-8y+4=0$ and $S_2: x^2+y^2-8x-12y+16=0$.
Equation of the common chord is $S_1-S_2=0$.
$(x^2+y^2-4x-8y+4) - (x^2+y^2-8x-12y+16) = 0$
$4x+4y-12=0 \implies x+y-3=0$.
For $S_1: x^2-4x+y^2-8y+4=0 \implies (x-2)^2+(y-4)^2 = 4+16-4 = 16$.
Centre $O(2, 4)$,radius $r_1 = 4$.
The perpendicular distance $d$ from $O(2, 4)$ to the line $x+y-3=0$ is:
$d = \frac{|2+4-3|}{\sqrt{1^2+1^2}} = \frac{3}{\sqrt{2}}$.
Let $L$ be the length of the common chord. Then $L = 2\sqrt{r_1^2-d^2}$.
$L = 2\sqrt{4^2 - (\frac{3}{\sqrt{2}})^2} = 2\sqrt{16 - \frac{9}{2}} = 2\sqrt{\frac{32-9}{2}} = 2\sqrt{\frac{23}{2}} = \sqrt{4 \times \frac{23}{2}} = \sqrt{46}$.
Thus,the correct option is $A$.

(B) The equation of the common chord of two circles $S_1 = 0$ and $S_2 = 0$ is given by $S_1 - S_2 = 0$.
Given circles are $S_1: x^2+y^2+2x+3y+1=0$ and $S_2: x^2+y^2+4x+3y+2=0$.
Subtracting $S_2$ from $S_1$,we get $(x^2+y^2+2x+3y+1) - (x^2+y^2+4x+3y+2) = 0$,which simplifies to $-2x - 1 = 0$,or $x = -\frac{1}{2}$.
Substitute $x = -\frac{1}{2}$ into the first circle equation: $(-\frac{1}{2})^2 + y^2 + 2(-\frac{1}{2}) + 3y + 1 = 0$.
$\frac{1}{4} + y^2 - 1 + 3y + 1 = 0 \Rightarrow y^2 + 3y + \frac{1}{4} = 0$.
Multiplying by $4$,we get $4y^2 + 12y + 1 = 0$.
Using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $y = \frac{-12 \pm \sqrt{144 - 16}}{8} = \frac{-12 \pm \sqrt{128}}{8} = \frac{-12 \pm 8\sqrt{2}}{8} = -\frac{3}{2} \pm \sqrt{2}$.
The endpoints of the common chord are $(-\frac{1}{2}, -\frac{3}{2} + \sqrt{2})$ and $(-\frac{1}{2}, -\frac{3}{2} - \sqrt{2})$.
The length of the common chord is the distance between these points: $\sqrt{(-\frac{1}{2} - (-\frac{1}{2}))^2 + ((-\frac{3}{2} + \sqrt{2}) - (-\frac{3}{2} - \sqrt{2}))^2} = \sqrt{0^2 + (2\sqrt{2})^2} = 2\sqrt{2}$ units.

(A) The first circle is $x^2+y^2-2x-6y+(10-r^2)=0$. Its center $C_1 = (1, 3)$ and radius $r_1 = \sqrt{1^2+3^2-(10-r^2)} = \sqrt{1+9-10+r^2} = |r|$.
The second circle is $x^2+y^2-8x+2y+8=0$. Its center $C_2 = (4, -1)$ and radius $r_2 = \sqrt{4^2+(-1)^2-8} = \sqrt{16+1-8} = 3$.
The distance between the centers $C_1$ and $C_2$ is $d = \sqrt{(4-1)^2+(-1-3)^2} = \sqrt{3^2+(-4)^2} = \sqrt{9+16} = 5$.
For two circles to have a common chord of non-zero length,they must intersect at two distinct points. This occurs if $|r_1 - r_2| < d < r_1 + r_2$.
Substituting the values: $| |r| - 3 | < 5 < |r| + 3$.
Case $1$: $|r| + 3 > 5 \implies |r| > 2$.
Case $2$: $| |r| - 3 | < 5 \implies -5 < |r| - 3 < 5 \implies -2 < |r| < 8$.
Combining these,we get $2 < |r| < 8$.

(B) Given $S \equiv x^2+y^2-4=0$,so $C_1(0,0)$ and $r_1=2$.
Let the center of $S^{\prime}=0$ be $C_2(h, k)$ and radius $r_2=\frac{5 \sqrt{2}}{2}$.
The equation of $S^{\prime}=0$ is $(x-h)^2+(y-k)^2=\left(\frac{5 \sqrt{2}}{2}\right)^2 = \frac{25}{2}$.
Expanding this,we get $x^2+y^2-2xh-2yk+h^2+k^2-\frac{25}{2}=0$.
The equation of the common chord is $S-S^{\prime}=0$,which is $2xh+2yk-(h^2+k^2-\frac{25}{2})-4=0$,or $2xh+2yk = h^2+k^2-\frac{17}{2}$.
The slope of the common chord is $-\frac{2h}{2k} = -\frac{h}{k}$. Given the slope is $\frac{1}{4}$,we have $-\frac{h}{k} = \frac{1}{4} \Rightarrow k = -4h$.
The common chord has maximum length when it passes through the center of the smaller circle $C_1(0,0)$.
Substituting $(0,0)$ into the equation of the chord: $2(0)h+2(0)k = h^2+k^2-\frac{17}{2} \Rightarrow h^2+k^2 = \frac{17}{2}$.
Substituting $k=-4h$ into $h^2+k^2=\frac{17}{2}$,we get $h^2+(-4h)^2 = \frac{17}{2}$ $\Rightarrow 17h^2 = \frac{17}{2}$ $\Rightarrow h^2 = \frac{1}{2}$ $\Rightarrow h = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2}$.
If $h = \frac{\sqrt{2}}{2}$,then $k = -4(\frac{\sqrt{2}}{2}) = -2\sqrt{2}$.
If $h = -\frac{\sqrt{2}}{2}$,then $k = -4(-\frac{\sqrt{2}}{2}) = 2\sqrt{2}$.
Thus,the center $C_2$ is $\left(\frac{\sqrt{2}}{2}, -2\sqrt{2}\right)$ or $\left(-\frac{\sqrt{2}}{2}, 2\sqrt{2}\right)$.

(A) Given $S_1: x^2+y^2=16$. The center is $B(0,0)$ and the radius is $r_1=4$.
Since the common chord is of maximum length,it must be the diameter of $S_1$.
Let $PQ$ be the common chord. The length of the common chord is $2 \times 4 = 8$.
Let the center of $S_2$ be $A(h, k)$ and its radius be $r_2=5$.
The distance from the center $A(h, k)$ to the common chord $PQ$ is $d = \sqrt{r_2^2 - (\text{half-length of chord})^2} = \sqrt{5^2 - 4^2} = \sqrt{25-16} = 3$.
The common chord $PQ$ passes through the origin $(0,0)$ and has a slope $m = \frac{3}{4}$.
The line $AB$ is perpendicular to the common chord $PQ$.
Thus,the slope of $AB$ is $m' = -\frac{1}{m} = -\frac{4}{3}$.
The distance $AB = 3$.
Using the parametric form of the line $AB$ passing through $(0,0)$ with slope $-\frac{4}{3}$:
$h = 0 \pm 3 \cos \theta$ and $k = 0 \pm 3 \sin \theta$,where $\tan \theta = -\frac{4}{3}$.
Since $\tan \theta = -\frac{4}{3}$,we have $\cos \theta = \mp \frac{3}{5}$ and $\sin \theta = \pm \frac{4}{3} \times \frac{3}{5} = \pm \frac{4}{5}$.
Therefore,$(h, k) = \pm 3 \left(-\frac{3}{5}, \frac{4}{5}\right) = \left(\mp \frac{9}{5}, \pm \frac{12}{5}\right)$.
The centers are $\left(-\frac{9}{5}, \frac{12}{5}\right)$ and $\left(\frac{9}{5}, -\frac{12}{5}\right)$.

(D) Given circles are $C_1: (x-a)^2+y^2=a^2$ and $C_2: x^2+(y-a)^2=a^2$.
Expanding these,we get $x^2-2ax+a^2+y^2=a^2 \Rightarrow x^2+y^2-2ax=0$ and $x^2+y^2-2ay=0$.
The equation of the common chord is $C_1 - C_2 = 0$,which gives $-2ax + 2ay = 0 \Rightarrow x-y=0$.
Substituting $y=x$ into $x^2+y^2-2ax=0$,we get $2x^2-2ax=0 \Rightarrow 2x(x-a)=0$.
Thus,the intersection points are $(0,0)$ and $(a,a)$.
The mid-point of the chord is $(\frac{0+a}{2}, \frac{0+a}{2}) = (\frac{a}{2}, \frac{a}{2})$.
The length of the chord is $\sqrt{(a-0)^2+(a-0)^2} = \sqrt{2a^2} = \sqrt{2}a$.
The distance from the center $(a,0)$ to the line $x-y=0$ is $d = \frac{|a-0|}{\sqrt{1^2+(-1)^2}} = \frac{a}{\sqrt{2}}$.
Since $\frac{a}{\sqrt{2}} \neq \sqrt{2}a$,option $D$ is not true.

(D) Given circles are $S_1: x^2+y^2-3x+y-10=0$ and $S_2: x^2+y^2-x+2y-20=0$.
The equation of the common chord is $S_1 - S_2 = 0$,which is $-2x-y+10=0$ or $2x+y-10=0$.
The family of circles passing through the intersection of $S_1$ and $S_2$ is given by $S_1 + \lambda S_2 = 0$:
$x^2(1+\lambda) + y^2(1+\lambda) - x(3+\lambda) + y(1+2\lambda) - 10 - 20\lambda = 0$.
Dividing by $(1+\lambda)$,the center of this circle is $(\frac{3+\lambda}{2(1+\lambda)}, \frac{-(1+2\lambda)}{2(1+\lambda)})$.
Since the common chord is the diameter,the center must lie on $2x+y-10=0$.
Substituting the center coordinates: $2(\frac{3+\lambda}{2(1+\lambda)}) - \frac{1+2\lambda}{2(1+\lambda)} - 10 = 0$.
Multiplying by $2(1+\lambda)$: $2(3+\lambda) - (1+2\lambda) - 20(1+\lambda) = 0$.
$6 + 2\lambda - 1 - 2\lambda - 20 - 20\lambda = 0$ $\Rightarrow -20\lambda - 15 = 0$ $\Rightarrow \lambda = -\frac{3}{4}$.
Substituting $\lambda = -\frac{3}{4}$ into the family equation:
$(1 - \frac{3}{4})x^2 + (1 - \frac{3}{4})y^2 - (3 - \frac{3}{4})x + (1 - \frac{6}{4})y - 10 + 15 = 0$.
$\frac{1}{4}x^2 + \frac{1}{4}y^2 - \frac{9}{4}x - \frac{2}{4}y + 5 = 0$.
Multiplying by $4$,we get $x^2+y^2-9x-2y+20=0$.

(C) The equation of the common chord is obtained by subtracting the two circle equations: $(x^2+y^2-6x-4y+9) - (x^2+y^2-8x-6y+23) = 0$.
This simplifies to $2x + 2y - 14 = 0$,or $x + y - 7 = 0$.
For the circle $x^2+y^2-6x-4y+9=0$,the center is $(3, 2)$ and the radius $r = \sqrt{3^2 + 2^2 - 9} = \sqrt{9+4-9} = 2$.
The perpendicular distance $d$ from the center $(3, 2)$ to the line $x + y - 7 = 0$ is $d = \frac{|3 + 2 - 7|}{\sqrt{1^2 + 1^2}} = \frac{|-2|}{\sqrt{2}} = \sqrt{2}$.
The length of the common chord is $2\sqrt{r^2 - d^2} = 2\sqrt{2^2 - (\sqrt{2})^2} = 2\sqrt{4 - 2} = 2\sqrt{2}$.