System of circles Questions in English

(A) The equation of a family of circles passing through the intersection of a circle $S = 0$ and a line $L = 0$ is given by $S + \lambda L = 0$.
Here,$S = x^2 + y^2 - a^2 = 0$ and $L = x \cos \alpha + y \sin \alpha - p = 0$.
So,the equation of the circle is $(x^2 + y^2 - a^2) + \lambda(x \cos \alpha + y \sin \alpha - p) = 0$.
Since this circle has the line $x \cos \alpha + y \sin \alpha - p = 0$ as its diameter,the center of this circle must lie on the line.
The center of the circle $x^2 + y^2 + (\lambda \cos \alpha)x + (\lambda \sin \alpha)y - (a^2 + \lambda p) = 0$ is $(-\frac{\lambda \cos \alpha}{2}, -\frac{\lambda \sin \alpha}{2})$.
Substituting this center into the line equation:
$(-\frac{\lambda \cos \alpha}{2}) \cos \alpha + (-\frac{\lambda \sin \alpha}{2}) \sin \alpha - p = 0$
$-\frac{\lambda}{2} (\cos^2 \alpha + \sin^2 \alpha) = p$
$-\frac{\lambda}{2} = p \implies \lambda = -2p$.
Substituting $\lambda = -2p$ into the family equation,we get:
$x^2 + y^2 - a^2 - 2p(x \cos \alpha + y \sin \alpha - p) = 0$.

(B) The equation of a circle passing through the intersection of two circles $S_1 = 0$ and $S_2 = 0$ is given by $S_1 + \lambda S_2 = 0$.
Here,$S_1 = x^2 + y^2 - 6x + 8 = 0$ and $S_2 = x^2 + y^2 - 6 = 0$.
The family of circles is $(x^2 + y^2 - 6x + 8) + \lambda(x^2 + y^2 - 6) = 0$.
Since the circle passes through the point $(1, 1)$,we substitute $x = 1$ and $y = 1$ into the equation:
$(1^2 + 1^2 - 6(1) + 8) + \lambda(1^2 + 1^2 - 6) = 0$
$(1 + 1 - 6 + 8) + \lambda(1 + 1 - 6) = 0$
$4 + \lambda(-4) = 0$
$4 - 4\lambda = 0 \implies \lambda = 1$.
Substituting $\lambda = 1$ back into the family equation:
$(x^2 + y^2 - 6x + 8) + 1(x^2 + y^2 - 6) = 0$
$2x^2 + 2y^2 - 6x + 2 = 0$
Dividing by $2$,we get $x^2 + y^2 - 3x + 1 = 0$.

(C) The angle of intersection $\theta$ between two circles is given by $\cos \theta = \frac{r_1^2 + r_2^2 - d^2}{2r_1r_2}$,where $d$ is the distance between the centers.
For the first circle $x^2 + y^2 - x + y - 8 = 0$,the center $C_1 = (\frac{1}{2}, -\frac{1}{2})$ and radius $r_1 = \sqrt{(\frac{1}{2})^2 + (-\frac{1}{2})^2 - (-8)} = \sqrt{\frac{1}{4} + \frac{1}{4} + 8} = \sqrt{\frac{17}{2}}$.
For the second circle $x^2 + y^2 + 2x + 2y - 11 = 0$,the center $C_2 = (-1, -1)$ and radius $r_2 = \sqrt{(-1)^2 + (-1)^2 - (-11)} = \sqrt{1 + 1 + 11} = \sqrt{13}$.
The distance $d$ between centers $C_1$ and $C_2$ is $d = \sqrt{(\frac{1}{2} - (-1))^2 + (-\frac{1}{2} - (-1))^2} = \sqrt{(\frac{3}{2})^2 + (\frac{1}{2})^2} = \sqrt{\frac{9}{4} + \frac{1}{4}} = \sqrt{\frac{10}{4}} = \sqrt{\frac{5}{2}}$.
Now,$\cos \theta = \frac{\frac{17}{2} + 13 - \frac{5}{2}}{2 \times \sqrt{\frac{17}{2}} \times \sqrt{13}} = \frac{6 + 13}{2 \times \sqrt{\frac{221}{2}}} = \frac{19}{2 \times \frac{\sqrt{221}}{\sqrt{2}}} = \frac{19}{\sqrt{2} \times \sqrt{221}} = \frac{19}{\sqrt{442}}$.
Since $\cos \theta = \frac{19}{\sqrt{442}}$,we have $\sin^2 \theta = 1 - \cos^2 \theta = 1 - \frac{361}{442} = \frac{81}{442}$,so $\sin \theta = \frac{9}{\sqrt{442}}$.
Thus,$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{9/\sqrt{442}}{19/\sqrt{442}} = \frac{9}{19}$.
Therefore,$\theta = \tan^{-1}\left(\frac{9}{19}\right)$.

(B) The equation of the radical axis of two circles $S_1 = 0$ and $S_2 = 0$ is given by $S_1 - S_2 = 0$.
First,normalize the equations so that the coefficients of ${x^2}$ and ${y^2}$ are $1$.
For the first circle: ${S_1} \equiv {x^2} + {y^2} - 3x - 4y + 5 = 0$.
For the second circle,divide by $3$: ${S_2} \equiv {x^2} + {y^2} - \frac{7}{3}x + \frac{8}{3}y + \frac{11}{3} = 0$.
Subtracting $S_2$ from $S_1$:
$({x^2} + {y^2} - 3x - 4y + 5) - ({x^2} + {y^2} - \frac{7}{3}x + \frac{8}{3}y + \frac{11}{3}) = 0$
$(-3 + \frac{7}{3})x + (-4 - \frac{8}{3})y + (5 - \frac{11}{3}) = 0$
$-\frac{2}{3}x - \frac{20}{3}y + \frac{4}{3} = 0$
Multiplying by $-3$: $2x + 20y - 4 = 0$,which simplifies to $x + 10y - 2 = 0$.
The equation of the line is $10y = -x + 2$,or $y = -\frac{1}{10}x + \frac{1}{5}$.
The gradient (slope) is $-\frac{1}{10}$.

(D) The required point is the radical center of the three given circles.
First,write the equations in the standard form $S = 0$:
$S_1: x^2 + y^2 - 12x - 16y + 64 = 0$
$S_2: x^2 + y^2 - 12x + 27 = 0$ (dividing by $3$)
$S_3: x^2 + y^2 - 16x + 81 = 0$
To find the radical center,solve the radical axes $S_1 - S_2 = 0$ and $S_2 - S_3 = 0$:
$S_1 - S_2 = (-12x - 16y + 64) - (-12x + 27) = -16y + 37 = 0 \Rightarrow y = \frac{37}{16}$
$S_2 - S_3 = (-12x + 27) - (-16x + 81) = 4x - 54 = 0 \Rightarrow x = \frac{54}{4} = \frac{27}{2}$
The radical center is $(\frac{27}{2}, \frac{37}{16})$.
Since this point is not among the given options,the correct answer is $(d)$.

(D) Let the equations of the circles be:
$S_1: x^2 + y^2 - 4x - 2y + 6 = 0$
$S_2: x^2 + y^2 - 2x - 4y - 1 = 0$
$S_3: x^2 + y^2 - 12x + 2y + 30 = 0$
The radical axis of $S_1$ and $S_2$ is given by $S_1 - S_2 = 0$:
$(x^2 + y^2 - 4x - 2y + 6) - (x^2 + y^2 - 2x - 4y - 1) = 0$
$-2x + 2y + 7 = 0$ or $2x - 2y = 7$ $(i)$
The radical axis of $S_2$ and $S_3$ is given by $S_2 - S_3 = 0$:
$(x^2 + y^2 - 2x - 4y - 1) - (x^2 + y^2 - 12x + 2y + 30) = 0$
$10x - 6y - 31 = 0$ $(ii)$
Solving equations $(i)$ and $(ii)$:
From $(i)$,$2x = 2y + 7 \implies x = y + 3.5$
Substitute into $(ii)$:
$10(y + 3.5) - 6y - 31 = 0$
$10y + 35 - 6y - 31 = 0$
$4y + 4 = 0 \implies y = -1$
$x = -1 + 3.5 = 2.5$
The radical centre is $(2.5, -1)$,which is not among the given options.

(D) Let the required circle be $S: x^2 + y^2 + 2gx + 2fy + c = 0$.
If a circle $S=0$ bisects the circumference of another circle $S'=0$,then their common chord passes through the centre of $S'=0$. The common chord is given by $S - S' = 0$.
$1$. For $S' = x^2 + y^2 - 1 = 0$,the centre is $(0, 0)$.
The common chord is $(x^2 + y^2 + 2gx + 2fy + c) - (x^2 + y^2 - 1) = 0$,which simplifies to $2gx + 2fy + c + 1 = 0$.
Since it passes through $(0, 0)$,we get $c + 1 = 0$,so $c = -1$.
$2$. For $S' = x^2 + y^2 + 2x - 3 = 0$,the centre is $(-1, 0)$.
The common chord is $(x^2 + y^2 + 2gx + 2fy - 1) - (x^2 + y^2 + 2x - 3) = 0$,which simplifies to $2(g-1)x + 2fy + 2 = 0$.
Since it passes through $(-1, 0)$,we get $2(g-1)(-1) + 2f(0) + 2 = 0$,which implies $-2g + 2 + 2 = 0$,so $2g = 4$ or $g = 2$.
$3$. For $S' = x^2 + y^2 + 2y - 3 = 0$,the centre is $(0, -1)$.
The common chord is $(x^2 + y^2 + 4x + 2fy - 1) - (x^2 + y^2 + 2y - 3) = 0$,which simplifies to $4x + 2(f-1)y + 2 = 0$.
Since it passes through $(0, -1)$,we get $4(0) + 2(f-1)(-1) + 2 = 0$,which implies $-2f + 2 + 2 = 0$,so $2f = 4$ or $f = 2$.
The centre of the required circle is $(-g, -f) = (-2, -2)$.

(C) Let the two circles be $S_1: x^2 + y^2 - 2x + 8y - q = 0$ and $S_2: x^2 + y^2 + 4x + 12y + p = 0$.
The common chord of the two circles is given by $S_1 - S_2 = 0$.
$(x^2 + y^2 - 2x + 8y - q) - (x^2 + y^2 + 4x + 12y + p) = 0$
$-6x - 4y - q - p = 0$
$6x + 4y + (p + q) = 0$
Since the circumference of $S_1$ is bisected by $S_2$,the common chord must be the diameter of $S_1$.
The center of $S_1$ is $(1, -4)$.
Since the diameter passes through the center $(1, -4)$,we substitute these coordinates into the equation of the common chord:
$6(1) + 4(-4) + (p + q) = 0$
$6 - 16 + (p + q) = 0$
$-10 + (p + q) = 0$
$p + q = 10$

(A) Let the radii of the two circles be $r$ and $r_1$. Given that the area of one is twice the area of the other,we have $\pi r_1^2 = 2 \pi r^2$,which implies $r_1^2 = 2r^2$ or $r_1 = \sqrt{2} r$.
For two orthogonal circles with radii $r$ and $r_1$ and distance between centers $d$,the condition for orthogonality is $d^2 = r^2 + r_1^2$.
Substituting the values,we get $d^2 = r^2 + (\sqrt{2} r)^2 = r^2 + 2r^2 = 3r^2$.
Therefore,the distance between their centers is $d = \sqrt{3} r$.

(A) The centers of the two circles are $C_1(-1, 1)$ and $C_2(1, 1)$,and both have radii $r_1 = 1$ and $r_2 = 1$.
The distance between the centers is $d = C_1C_2 = \sqrt{(1 - (-1))^2 + (1 - 1)^2} = \sqrt{2^2 + 0^2} = 2$.
Since the sum of the radii $r_1 + r_2 = 1 + 1 = 2$,and $d = r_1 + r_2$,the circles touch each other externally.
The equation of the common tangent is obtained by subtracting the two circle equations:
$(x^2 + y^2 + 2x - 2y + 1) - (x^2 + y^2 - 2x - 2y + 1) = 0$
$4x = 0 \Rightarrow x = 0$.
Substituting $x = 0$ into the first circle equation:
$0^2 + y^2 + 2(0) - 2y + 1 = 0$
$y^2 - 2y + 1 = 0$ $\Rightarrow (y - 1)^2 = 0$ $\Rightarrow y = 1$.
Thus,the point of contact is $(0, 1)$.

(C) For the first circle $x^2 + y^2 = 4$,the center $C_1 = (0, 0)$ and radius $r_1 = \sqrt{4} = 2$.
For the second circle $x^2 + y^2 - 8x - 8y + 7 = 0$,the center $C_2 = (-(-8/2), -(-8/2)) = (4, 4)$ and radius $r_2 = \sqrt{4^2 + 4^2 - 7} = \sqrt{16 + 16 - 7} = \sqrt{25} = 5$.
The distance between the centers $C_1 C_2 = \sqrt{(4-0)^2 + (4-0)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2} \approx 5.66$.
Since $|r_1 - r_2| < C_1 C_2 < r_1 + r_2$ (i.e.,$|2 - 5| < 5.66 < 2 + 5$,which is $3 < 5.66 < 7$),the circles intersect at two points.
When two circles intersect,they have $2$ direct common tangents and $0$ transverse common tangents.
Therefore,the number of direct common tangents is $2$.

(D) The equation of the circle is $S \equiv x^2 + y^2 - 30x = 0$ and the chord is $L \equiv y + 3x = 0$.
The equation of a circle passing through the intersection of $S = 0$ and $L = 0$ is given by $S + \lambda L = 0$.
$x^2 + y^2 - 30x + \lambda(y + 3x) = 0$
$x^2 + y^2 + (3\lambda - 30)x + \lambda y = 0$
For this circle,the chord $y + 3x = 0$ is the diameter. The center of this circle is $\left( -\frac{3\lambda - 30}{2}, -\frac{\lambda}{2} \right)$.
Since the center must lie on the chord $y + 3x = 0$,we substitute the center coordinates into the chord equation:
$-\frac{\lambda}{2} + 3\left( -\frac{3\lambda - 30}{2} \right) = 0$
$-\lambda - 9\lambda + 90 = 0$
$-10\lambda = -90 \implies \lambda = 9$
Substituting $\lambda = 9$ into the circle equation:
$x^2 + y^2 + (3(9) - 30)x + 9y = 0$
$x^2 + y^2 + (27 - 30)x + 9y = 0$
$x^2 + y^2 - 3x + 9y = 0$

(A) The given circle is $S: x^2 + y^2 - 16 = 0$.
The equation of the chord is $L: 3x + y + 5 = 0$.
The equation of the family of circles passing through the intersection of $S$ and $L$ is $S + \lambda L = 0$.
Thus,$x^2 + y^2 - 16 + \lambda(3x + y + 5) = 0$.
$x^2 + y^2 + 3\lambda x + \lambda y + (5\lambda - 16) = 0$.
The center of this circle is $C = (-\frac{3\lambda}{2}, -\frac{\lambda}{2})$.
Since the chord $3x + y + 5 = 0$ is the diameter of this circle,the center $C$ must lie on the line $3x + y + 5 = 0$.
Substituting the center into the line equation: $3(-\frac{3\lambda}{2}) + (-\frac{\lambda}{2}) + 5 = 0$.
$-\frac{9\lambda}{2} - \frac{\lambda}{2} + 5 = 0$.
$-5\lambda + 5 = 0 \implies \lambda = 1$.
Substituting $\lambda = 1$ into the circle equation: $x^2 + y^2 + 3(1)x + (1)y + 5(1) - 16 = 0$.
$x^2 + y^2 + 3x + y - 11 = 0$.

(D) The equation of any circle passing through the intersection of the circle $S = x^2 + y^2 - 4x - 6y - 21 = 0$ and the line $L = 3x + 4y + 5 = 0$ is given by $S + \lambda L = 0$.
$(x^2 + y^2 - 4x - 6y - 21) + \lambda(3x + 4y + 5) = 0$
Since this circle passes through the point $(1, 2)$,we substitute $x = 1$ and $y = 2$ into the equation:
$(1^2 + 2^2 - 4(1) - 6(2) - 21) + \lambda(3(1) + 4(2) + 5) = 0$
$(1 + 4 - 4 - 12 - 21) + \lambda(3 + 8 + 5) = 0$
$-32 + 16\lambda = 0$
$16\lambda = 32 \Rightarrow \lambda = 2$
Substituting $\lambda = 2$ back into the family of circles equation:
$(x^2 + y^2 - 4x - 6y - 21) + 2(3x + 4y + 5) = 0$
$x^2 + y^2 - 4x - 6y - 21 + 6x + 8y + 10 = 0$
$x^2 + y^2 + 2x + 2y - 11 = 0$
Thus,the required equation is $x^2 + y^2 + 2x + 2y - 11 = 0$.

(B) The first circle is $x^2 + y^2 - 16x - 20y + 164 = r^2$. Rewriting in standard form: $(x - 8)^2 + (y - 10)^2 = r^2$. The center is $A(8, 10)$ and the radius is $R_1 = r$.
The second circle is $(x - 4)^2 + (y - 7)^2 = 36$. The center is $B(4, 7)$ and the radius is $R_2 = 6$.
The distance between the centers $A$ and $B$ is $AB = \sqrt{(8 - 4)^2 + (10 - 7)^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = 5$.
For two circles to intersect at two distinct points,the condition is $|R_1 - R_2| < AB < R_1 + R_2$.
Substituting the values: $|r - 6| < 5 < r + 6$.
From $r + 6 > 5$,we get $r > -1$. Since $r$ is a radius,$r > 0$.
From $|r - 6| < 5$,we get $-5 < r - 6 < 5$,which implies $1 < r < 11$.
Combining these,the condition is $1 < r < 11$.

(D) For the circle $x^2 + y^2 - 2x - 2y - 2 = 0$,the centre $C_1 = (1, 1)$ and radius $r_1 = \sqrt{1^2 + 1^2 - (-2)} = \sqrt{4} = 2$.
For the circle $x^2 + y^2 - 6x - 6y + 14 = 0$,the centre $C_2 = (3, 3)$ and radius $r_2 = \sqrt{3^2 + 3^2 - 14} = \sqrt{18 - 14} = \sqrt{4} = 2$.
The distance between the centres $C_1C_2 = \sqrt{(3-1)^2 + (3-1)^2} = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}$.
In the quadrilateral $PC_1QC_2$,the sides are $C_1P = C_1Q = r_1 = 2$ and $C_2P = C_2Q = r_2 = 2$.
Since all sides are equal to $2$,the quadrilateral is a rhombus.
The diagonals are $C_1C_2 = 2\sqrt{2}$ and $PQ$. The length of the common chord $PQ$ can be found using the formula $PQ = \frac{2r_1r_2}{d} \sin(\theta)$,or by noting that in $\triangle PC_1C_2$,$C_1P=2, C_2P=2, C_1C_2=2\sqrt{2}$. Since $2^2 + 2^2 = (2\sqrt{2})^2$,$\triangle PC_1C_2$ is a right-angled triangle at $P$.
The area of $\triangle PC_1C_2 = \frac{1}{2} \times 2 \times 2 = 2$.
The area of the quadrilateral $PC_1QC_2 = 2 \times (\text{Area of } \triangle PC_1C_2) = 2 \times 2 = 4$ sq. units.

(B) For the circle $x^2 + y^2 = 4$,the center is $C_1(0, 0)$ and the radius is $r_1 = 2$.
For the circle $x^2 + y^2 + 6x + 8y - 24 = 0$,the center is $C_2(-3, -4)$ and the radius is $r_2 = \sqrt{(-3)^2 + (-4)^2 - (-24)} = \sqrt{9 + 16 + 24} = \sqrt{49} = 7$.
The distance between the centers is $d = \sqrt{(-3-0)^2 + (-4-0)^2} = \sqrt{9 + 16} = 5$.
Since $d = |r_2 - r_1| = |7 - 2| = 5$,the circles touch each other internally.
The equation of the common tangent at the point of contact is given by $S_1 - S_2 = 0$.
$(x^2 + y^2 - 4) - (x^2 + y^2 + 6x + 8y - 24) = 0$
$-6x - 8y + 20 = 0$
$3x + 4y - 10 = 0$.
Checking the options,the point $(6, -2)$ satisfies the equation: $3(6) + 4(-2) - 10 = 18 - 8 - 10 = 0$.

(B) The given equations of the circles are $S_{1}: x^{2}+y^{2}-6x+8=0$ and $S_{2}: x^{2}+y^{2}-8y+16-k=0$.
For $S_{1}$,the center $C_{1} = (3, 0)$ and radius $r_{1} = \sqrt{3^{2}+0^{2}-8} = \sqrt{1} = 1$.
For $S_{2}$,the center $C_{2} = (0, 4)$ and radius $r_{2} = \sqrt{0^{2}+4^{2}-(16-k)} = \sqrt{k}$.
The distance between the centers is $d = \sqrt{(3-0)^{2}+(0-4)^{2}} = \sqrt{9+16} = 5$.
Since the circles touch each other,the distance between centers must be equal to the sum or difference of the radii: $d = |r_{1} \pm r_{2}|$.
Case $1$: $d = r_{1}+r_{2}$ $\Rightarrow 5 = 1+\sqrt{k}$ $\Rightarrow \sqrt{k} = 4$ $\Rightarrow k = 16$.
Case $2$: $d = |r_{1}-r_{2}| \Rightarrow 5 = |1-\sqrt{k}|$.
This implies $1-\sqrt{k} = 5$ or $1-\sqrt{k} = -5$.
$1-\sqrt{k} = 5 \Rightarrow \sqrt{k} = -4$ (not possible as $k>0$).
$1-\sqrt{k} = -5$ $\Rightarrow \sqrt{k} = 6$ $\Rightarrow k = 36$.
Comparing the values,the largest value of $k$ is $36$.

(A) The parabola is $y^{2} = 4x$. Comparing with $y^{2} = 4ax$,we get $a = 1$.
The length of the latus rectum is $4a = 4(1) = 4$.
The common chord of the two circles $C_{1}$ and $C_{2}$ is the latus rectum of the parabola,so its length is $4$.
Let $D$ be one endpoint of the latus rectum and $B$ be the midpoint of the common chord. Thus,$DB = \frac{4}{2} = 2$.
Let $r$ be the radius of the circles,$r = 2\sqrt{5}$.
In the right-angled triangle formed by the radius,the distance from the center to the chord,and half the chord length,let $x$ be the distance from the center of a circle to the common chord.
$x^{2} + DB^{2} = r^{2}$
$x^{2} + 2^{2} = (2\sqrt{5})^{2}$
$x^{2} + 4 = 20$
$x^{2} = 16 \implies x = 4$.
The distance between the centers $C_{1}$ and $C_{2}$ is $x + x = 4 + 4 = 8$.

(D) Let the equation of the family of circles passing through the intersection of $S_{1} = x^{2}+y^{2}-6x=0$ and $S_{2} = x^{2}+y^{2}-4y=0$ be $S_{1} + \lambda S_{2} = 0$ for $\lambda \neq -1$.
$(x^{2}+y^{2}-6x) + \lambda(x^{2}+y^{2}-4y) = 0$
$(1+\lambda)x^{2} + (1+\lambda)y^{2} - 6x - 4\lambda y = 0$
Dividing by $(1+\lambda)$,we get $x^{2} + y^{2} - \frac{6}{1+\lambda}x - \frac{4\lambda}{1+\lambda}y = 0$.
The centre of this circle is $\left(\frac{3}{1+\lambda}, \frac{2\lambda}{1+\lambda}\right)$.
Since the centre lies on the line $2x - 3y + 12 = 0$,we substitute the coordinates:
$2\left(\frac{3}{1+\lambda}\right) - 3\left(\frac{2\lambda}{1+\lambda}\right) + 12 = 0$
$6 - 6\lambda + 12(1+\lambda) = 0$
$6 - 6\lambda + 12 + 12\lambda = 0$
$6\lambda = -18 \Rightarrow \lambda = -3$.
Substituting $\lambda = -3$ into the equation of the circle:
$(x^{2}+y^{2}-6x) - 3(x^{2}+y^{2}-4y) = 0$
$-2x^{2} - 2y^{2} - 6x + 12y = 0$
$x^{2} + y^{2} + 3x - 6y = 0$.
Checking the point $(-3, 6)$ in the equation:
$(-3)^{2} + (6)^{2} + 3(-3) - 6(6) = 9 + 36 - 9 - 36 = 0$.
Thus,the circle passes through the point $(-3, 6)$.

(D) For the first circle $x^{2}+y^{2}-10x-10y+41=0$:
Center $C_{1} = (5, 5)$,Radius $r_{1} = \sqrt{5^{2}+5^{2}-41} = \sqrt{25+25-41} = \sqrt{9} = 3$.
For the second circle $x^{2}+y^{2}-24x-10y+160=0$:
Center $C_{2} = (12, 5)$,Radius $r_{2} = \sqrt{12^{2}+5^{2}-160} = \sqrt{144+25-160} = \sqrt{9} = 3$.
The distance between the centers $C_{1}$ and $C_{2}$ is $d = \sqrt{(12-5)^{2}+(5-5)^{2}} = \sqrt{7^{2}+0^{2}} = 7$.
The minimum distance between the two circles is given by $d - (r_{1} + r_{2}) = 7 - (3 + 3) = 7 - 6 = 1$.

(B) The given circle is $x^{2}+y^{2}-2x-6y+6=0$.
Comparing this with the general equation $x^{2}+y^{2}+2gx+2fy+c=0$,we get $g=-1$ and $f=-3$.
The center of the circle is $(-g, -f) = (1, 3)$.
The radius of the circle is $R = \sqrt{g^{2}+f^{2}-c} = \sqrt{(-1)^{2}+(-3)^{2}-6} = \sqrt{1+9-6} = \sqrt{4} = 2$.
Since a diameter of this circle is a chord of another circle $'C'$ with center $(2, 1)$,the radius of the circle $'C'$ (let it be $r$) forms a right-angled triangle with the distance between the centers and the radius of the first circle.
The distance $d$ between the centers $(1, 3)$ and $(2, 1)$ is $d = \sqrt{(2-1)^{2}+(1-3)^{2}} = \sqrt{1^{2}+(-2)^{2}} = \sqrt{1+4} = \sqrt{5}$.
In the right-angled triangle formed,the hypotenuse is the radius $r$ of circle $'C'$,and the other two sides are the distance $d$ and the radius $R$ of the first circle.
Thus,$r^{2} = d^{2} + R^{2} = (\sqrt{5})^{2} + (2)^{2} = 5 + 4 = 9$.
Therefore,$r = 3$.

(A) The equation of a circle $C$ touching the line $x-2y=0$ at $(2,1)$ is given by $(x-2)^{2}+(y-1)^{2}+\lambda(x-2y)=0$.
Expanding this,we get $x^{2}+y^{2}+x(\lambda-4)+y(-2-2\lambda)+5=0$.
The given circle $C_{1}$ is $x^{2}+y^{2}+2y-5=0$.
The common chord $PQ$ is the radical axis $C-C_{1}=0$,which is $x(\lambda-4)+y(-2-2\lambda-2)+10=0$,or $x(\lambda-4)-y(2\lambda+4)+10=0$.
Since $PQ$ is a diameter of $C_{1}$,it must pass through the center of $C_{1}$,which is $(0,-1)$.
Substituting $(0,-1)$ into the equation of $PQ$: $0(\lambda-4)-(-1)(2\lambda+4)+10=0$ $\Rightarrow 2\lambda+4+10=0$ $\Rightarrow 2\lambda=-14$ $\Rightarrow \lambda=-7$.
Substituting $\lambda=-7$ into the equation of $C$: $x^{2}+y^{2}-11x+12y+5=0$.
The center of $C$ is $(\frac{11}{2}, -6)$ and the radius $r$ is $\sqrt{(\frac{11}{2})^{2}+(-6)^{2}-5} = \sqrt{\frac{121}{4}+36-5} = \sqrt{\frac{121+124}{4}} = \sqrt{\frac{245}{4}} = \frac{7\sqrt{5}}{2}$.
Thus,the diameter of $C$ is $2r = 7\sqrt{5}$.

(C) The circle $C_1: x^{2}+y^{2}+6x+8y+16=0$ has center $O_1(-3, -4)$ and radius $r_1 = \sqrt{3^2+4^2-16} = 3$.
The circle $C_2: x^{2}+y^{2}+2(3-\sqrt{3})x+2(4-\sqrt{6})y = k+6\sqrt{3}+8\sqrt{6}$ has center $O_2(\sqrt{3}-3, \sqrt{6}-4)$ and radius $r_2 = \sqrt{(\sqrt{3}-3)^2 + (\sqrt{6}-4)^2 + k + 6\sqrt{3} + 8\sqrt{6}} = \sqrt{3-6\sqrt{3}+9 + 6-8\sqrt{6}+16 + k + 6\sqrt{3} + 8\sqrt{6}} = \sqrt{k+34}$.
Since the circles touch internally,the distance between centers $d = |r_2 - r_1|$.
$d^2 = ((\sqrt{3}-3) - (-3))^2 + ((\sqrt{6}-4) - (-4))^2 = (\sqrt{3})^2 + (\sqrt{6})^2 = 3+6 = 9$.
Thus,$d = 3$.
So,$3 = |\sqrt{k+34} - 3|$,which implies $\sqrt{k+34} = 6$ (since $k>0$),so $k+34 = 36$,$k=2$.
The point of contact $P(\alpha, \beta)$ divides the line segment $O_1O_2$ externally in the ratio $r_1 : r_2 = 3 : 6 = 1 : 2$.
Using the section formula for external division:
$\alpha = \frac{r_1 x_2 - r_2 x_1}{r_1 - r_2} = \frac{1(\sqrt{3}-3) - 2(-3)}{1-2} = \frac{\sqrt{3}-3+6}{-1} = -\sqrt{3}-3$.
$\beta = \frac{r_1 y_2 - r_2 y_1}{r_1 - r_2} = \frac{1(\sqrt{6}-4) - 2(-4)}{1-2} = \frac{\sqrt{6}-4+8}{-1} = -\sqrt{6}-4$.
Therefore,$(\alpha+\sqrt{3})^{2} + (\beta+\sqrt{6})^{2} = (-3)^2 + (-4)^2 = 9 + 16 = 25$.

(C) For two circles to intersect at two distinct points,the distance between their centers $C_1C_2$ must satisfy the condition: $|r_1 - r_2| < C_1C_2 < r_1 + r_2$.
For the first circle $(x+1)^2 + (y+2)^2 = r^2$,the center $C_1 = (-1, -2)$ and radius $r_1 = r$.
For the second circle $x^2 + y^2 - 4x - 4y + 4 = 0$,the center $C_2 = (2, 2)$ and radius $r_2 = \sqrt{2^2 + 2^2 - 4} = \sqrt{4+4-4} = 2$.
The distance between the centers $C_1C_2 = \sqrt{(2 - (-1))^2 + (2 - (-2))^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5$.
Applying the condition $|r - 2| < 5 < r + 2$:
$1$) $|r - 2| < 5$ $\Rightarrow -5 < r - 2 < 5$ $\Rightarrow -3 < r < 7$.
$2$) $5 < r + 2 \Rightarrow r > 3$.
Combining these inequalities,we get $3 < r < 7$.

(C) The given circle is $x^2+y^2-10x+4y+13=0$. Its center $M$ is $(5, -2)$ and its radius $r$ is $\sqrt{5^2+(-2)^2-13} = \sqrt{25+4-13} = \sqrt{16} = 4$.
The center of circle $C$ is the intersection of $2x+3y=12$ and $3x-2y=5$. Solving these,we get $x=3$ and $y=2$. So,the center of $C$ is $O(3, 2)$.
The diameter of the first circle is a chord of circle $C$. Let $P$ be a point on the circumference of circle $C$ such that $MP$ is the radius of the first circle,i.e.,$MP = 4$. Since $M$ is the center of the first circle and $MP$ is a radius,$M$ lies on the chord.
The distance $d$ between the centers $O(3, 2)$ and $M(5, -2)$ is $d = \sqrt{(5-3)^2 + (-2-2)^2} = \sqrt{2^2 + (-4)^2} = \sqrt{4+16} = \sqrt{20}$.
In the right-angled triangle formed by the center of $C$,the center of the first circle,and a point on the chord,the radius $R$ of circle $C$ is given by $R^2 = d^2 + r^2 = 20 + 16 = 36$. Thus,$R = 6$.

(D) Let the equation of the circle be $(x-h)^2 + (y-k)^2 = r^2$. Since it passes through $(0,0)$ and $(1,0)$,the radius $r$ is the distance from $(h,k)$ to $(0,0)$,so $r^2 = h^2+k^2$.
Expanding the equation: $x^2 - 2hx + h^2 + y^2 - 2ky + k^2 = h^2+k^2$,which simplifies to $x^2+y^2-2hx-2ky=0$.
Since the circle passes through $(1,0)$,we substitute $x=1, y=0$: $1^2+0^2-2h(1)-2k(0)=0$,which gives $1-2h=0$,so $h=1/2$.
The circle touches $x^2+y^2=9$ (a circle with center $(0,0)$ and radius $3$). The distance between the centers is equal to the difference of the radii (since the smaller circle is inside the larger one): $\sqrt{h^2+k^2} = 3 - r = 3 - \sqrt{h^2+k^2}$.
Thus,$2\sqrt{h^2+k^2} = 3$,so $\sqrt{h^2+k^2} = 3/2$.
Squaring both sides,$h^2+k^2 = 9/4$.
Therefore,$4(h^2+k^2) = 4(9/4) = 9$.

(A) The circle $C_1$ is $x^2+y^2=1$ and $C_2$ is $(x-4)^2+(y-1)^2=r^2$.
The radical axis of two circles is given by $S_1 - S_2 = 0$.
$x^2+y^2-1 - ((x-4)^2+(y-1)^2-r^2) = 0$
$x^2+y^2-1 - (x^2-8x+16+y^2-2y+1-r^2) = 0$
$8x+2y-18+r^2 = 0$,which simplifies to $8x+2y = 18-r^2$.
The line joining the midpoints of the common tangents is the radical axis of the two circles.
This line meets the $x$-axis at $B$. Setting $y=0$ in the radical axis equation,we get $8x = 18-r^2$,so $x = \frac{18-r^2}{8}$.
Thus,$B = \left(\frac{18-r^2}{8}, 0\right)$.
Given $A=(4,1)$ and $AB=\sqrt{5}$,we have $AB^2 = 5$.
$\left(\frac{18-r^2}{8}-4\right)^2 + (0-1)^2 = 5$
$\left(\frac{18-r^2-32}{8}\right)^2 + 1 = 5$
$\left(\frac{-(14+r^2)}{8}\right)^2 = 4$
$\frac{14+r^2}{8} = 2$ (since $r^2 > 0$,the expression inside the square must be positive for the root to be $2$)
$14+r^2 = 16$,which gives $r^2 = 2$.

(B) Let the circle be $x^2+y^2+2gx+2fy+c=0$ $\quad\quad(1)$
The given circles are:
$C_1: x^2+y^2-2x-15=0$ $\quad\quad(2)$
$C_2: x^2+y^2-1=0$ $\quad\quad(3)$
Since $(1)$ is orthogonal to $(2)$,we have $2g_1g_2 + 2f_1f_2 = c_1+c_2$. Here $g_1=g, f_1=f, c_1=c$ and for $(2)$,$g_2=-1, f_2=0, c_2=-15$:
$2(g)(-1) + 2(f)(0) = c-15 \Rightarrow -2g = c-15$ $\quad\quad(4)$
Since $(1)$ is orthogonal to $(3)$,where $g_3=0, f_3=0, c_3=-1$:
$2(g)(0) + 2(f)(0) = c-1 \Rightarrow c=1$
Substituting $c=1$ into $(4)$:
$-2g = 1-15 = -14 \Rightarrow g=7$
Since the circle passes through $(0,1)$:
$0^2+1^2+2g(0)+2f(1)+c=0$ $\Rightarrow 1+2f+1=0$ $\Rightarrow 2f=-2$ $\Rightarrow f=-1$
The circle equation is $x^2+y^2+14x-2y+1=0$.
Centre is $(-g, -f) = (-7, 1)$.
Radius is $\sqrt{g^2+f^2-c} = \sqrt{7^2+(-1)^2-1} = \sqrt{49+1-1} = \sqrt{49} = 7$.
Thus,$(B)$ and $(C)$ are correct.

(C) The given circle equation is $x^2+y^2-4x+6y-12=0$.
Comparing this with the general form $x^2+y^2+2gx+2fy+c=0$,we get $g=-2$,$f=3$,and $c=-12$.
The centre of this circle is $C_1 = (-g, -f) = (2, -3)$ and its radius $r_1$ is $\sqrt{g^2+f^2-c} = \sqrt{(-2)^2+3^2-(-12)} = \sqrt{4+9+12} = \sqrt{25} = 5$.
Since the diameter of this circle is a chord of circle $S$,the length of this chord is equal to the diameter of the first circle,which is $2r_1 = 2(5) = 10$.
Let the centre of circle $S$ be $C_2 = (-3, 2)$. The perpendicular distance $d$ from $C_2$ to the chord is the distance between $(2, -3)$ and $(-3, 2)$.
$d = \sqrt{(-3-2)^2 + (2-(-3))^2} = \sqrt{(-5)^2 + 5^2} = \sqrt{25+25} = \sqrt{50} = 5\sqrt{2}$.
In circle $S$,the radius $R$,the perpendicular distance $d$,and half the chord length $a$ form a right-angled triangle,where $a = \frac{10}{2} = 5$.
Using Pythagoras theorem,$R^2 = d^2 + a^2 = (5\sqrt{2})^2 + 5^2 = 50 + 25 = 75$.
Therefore,$R = \sqrt{75} = 5\sqrt{3}$.

(D) The required circle is concentric with $2x^2+2y^2-8x-12y-9=0$.
Dividing by $2$,we get $x^2+y^2-4x-6y-\frac{9}{2}=0$.
Thus,the centre of the required circle is $(2, 3)$.
The circle passes through the centre of $x^2+y^2+8x+10y-7=0$.
Comparing with $x^2+y^2+2gx+2fy+c=0$,the centre is $(-4, -5)$.
The radius $r$ is the distance between $(2, 3)$ and $(-4, -5)$:
$r = \sqrt{(-4-2)^2 + (-5-3)^2} = \sqrt{(-6)^2 + (-8)^2} = \sqrt{36+64} = 10$.
The equation of the circle is $(x-2)^2 + (y-3)^2 = 10^2$.
$x^2-4x+4 + y^2-6y+9 = 100$.
$x^2+y^2-4x-6y-87=0$.

(D) The given circle is concentric with $2x^2+2y^2-8x-12y-9=0$,which simplifies to $x^2+y^2-4x-6y-4.5=0$.
Thus,the equation of the required circle is of the form $x^2+y^2-4x-6y+k=0$.
This circle passes through the centre of $x^2+y^2+8x+10y-7=0$.
The centre of $x^2+y^2+8x+10y-7=0$ is $(-4, -5)$.
Substituting $(-4, -5)$ into the equation $x^2+y^2-4x-6y+k=0$:
$(-4)^2+(-5)^2-4(-4)-6(-5)+k=0$
$16+25+16+30+k=0$
$87+k=0 \Rightarrow k=-87$.
Therefore,the required equation is $x^2+y^2-4x-6y-87=0$.

(B) For the circle $x^2+y^2+2ax+c=0$,the center is $C_1 = (-a, 0)$ and the radius is $r_1 = \sqrt{a^2-c}$.
For the circle $x^2+y^2+2by+c=0$,the center is $C_2 = (0, -b)$ and the radius is $r_2 = \sqrt{b^2-c}$.
Since the circles touch each other externally,the distance between their centers must be equal to the sum of their radii:
$d(C_1, C_2) = r_1 + r_2$
$\sqrt{(-a-0)^2 + (0-(-b))^2} = \sqrt{a^2-c} + \sqrt{b^2-c}$
$\sqrt{a^2+b^2} = \sqrt{a^2-c} + \sqrt{b^2-c}$
Squaring both sides:
$a^2+b^2 = (a^2-c) + (b^2-c) + 2\sqrt{(a^2-c)(b^2-c)}$
$a^2+b^2 = a^2+b^2-2c + 2\sqrt{(a^2-c)(b^2-c)}$
$2c = 2\sqrt{(a^2-c)(b^2-c)}$
$c = \sqrt{(a^2-c)(b^2-c)}$
Squaring again:
$c^2 = (a^2-c)(b^2-c)$
$c^2 = a^2b^2 - a^2c - b^2c + c^2$
$0 = a^2b^2 - c(a^2+b^2)$
$c(a^2+b^2) = a^2b^2$
Dividing both sides by $a^2b^2c$:
$\frac{a^2+b^2}{a^2b^2} = \frac{1}{c}$
$\frac{1}{b^2} + \frac{1}{a^2} = \frac{1}{c}$

(A) The condition for two circles $x^{2}+y^{2}+2g_{1}x+2f_{1}y+c_{1}=0$ and $x^{2}+y^{2}+2g_{2}x+2f_{2}y+c_{2}=0$ to intersect orthogonally is $2g_{1}g_{2}+2f_{1}f_{2}=c_{1}+c_{2}$.
For the given circles:
Circle $1$: $g_{1}=1, f_{1}=k, c_{1}=6$.
Circle $2$: $g_{2}=0, f_{2}=k, c_{2}=k$.
Substituting these values into the condition:
$2(1)(0) + 2(k)(k) = 6 + k$
$0 + 2k^{2} = 6 + k$
$2k^{2} - k - 6 = 0$
Factoring the quadratic equation:
$(2k+3)(k-2) = 0$
Thus,$k = 2$ or $k = -3/2$.

(D) For the circle $x^2+y^2-6x=0$,the center $C_1 = (3, 0)$ and radius $r_1 = \sqrt{3^2+0^2-0} = 3$.
For the circle $x^2+y^2+6x+2y+1=0$,the center $C_2 = (-3, -1)$ and radius $r_2 = \sqrt{(-3)^2+(-1)^2-1} = \sqrt{9+1-1} = 3$.
The distance between the centers $d = C_1C_2 = \sqrt{(3 - (-3))^2 + (0 - (-1))^2} = \sqrt{6^2 + 1^2} = \sqrt{37}$.
Since $r_1 + r_2 = 3 + 3 = 6$ and $\sqrt{36} < \sqrt{37}$,we have $d > r_1 + r_2$.
Because the distance between the centers is greater than the sum of the radii,the two circles lie outside each other without touching.
Therefore,the number of common tangents that can be drawn is $4$.

(A) The given circle is $x^2+y^2-4x-6y-12=0$. Its centre $C_1$ is $(2, 3)$ and radius $R = \sqrt{2^2+3^2-(-12)} = \sqrt{4+9+12} = \sqrt{25} = 5$.
Let the centre of the required circle be $C_2(h, k)$ and its radius be $r = 3$.
Since the circles touch internally at $P(-1, -1)$,the point $P$ divides the line segment $C_1C_2$ externally in the ratio $R:r = 5:3$.
Using the section formula for external division:
$P = \left(\frac{R h - r x_1}{R - r}, \frac{R k - r y_1}{R - r}\right)$
$-1 = \frac{5h - 3(2)}{5-3} \implies -2 = 5h - 6 \implies 5h = 4 \implies h = \frac{4}{5}$
$-1 = \frac{5k - 3(3)}{5-3} \implies -2 = 5k - 9 \implies 5k = 7 \implies k = \frac{7}{5}$
Thus,the centre is $\left(\frac{4}{5}, \frac{7}{5}\right)$.

(D) Let the required circle be $S: x^{2}+y^{2}+2gx+2fy+c=0$.
For a circle $x^{2}+y^{2}+2g_{i}x+2f_{i}y+c_{i}=0$,the condition for orthogonality is $2gg_{i}+2ff_{i}=c+c_{i}$.
For the given circles:
$S_{1}: x^{2}+y^{2}+6x+0y-1=0 \implies 2g(3)+2f(0)=c-1 \implies 6g=c-1$ $(i)$
$S_{2}: x^{2}+y^{2}+0x-3y+2=0 \implies 2g(0)+2f(-3/2)=c+2 \implies -3f=c+2$ (ii)
$S_{3}: x^{2}+y^{2}+x+y-3=0 \implies 2g(1/2)+2f(1/2)=c-3 \implies g+f=c-3$ (iii)
Subtracting (ii) from $(i)$: $6g+3f=-3 \implies 2g+f=-1$ (iv)
Subtracting (iii) from $(i)$: $5g-f=2$ $(v)$
Adding (iv) and $(v)$: $7g=1 \implies g=1/7$.
Substituting $g$ in (iv): $2(1/7)+f=-1 \implies f=-1-2/7 = -9/7$.
The centre of the circle is $(-g, -f) = (-1/7, 9/7)$.

(C) Let the equation of the required circle be $x^{2}+y^{2}+2gx+2fy+c=0$.
Two circles $x^{2}+y^{2}+2g_{1}x+2f_{1}y+c_{1}=0$ and $x^{2}+y^{2}+2g_{2}x+2f_{2}y+c_{2}=0$ cut orthogonally if $2g_{1}g_{2}+2f_{1}f_{2}=c_{1}+c_{2}$.
For the given circles,the conditions are:
$1) -g-f = c-14 \implies -g-f-c = -14$
$2) 3g-5f = c-10 \implies 3g-5f-c = -10$
$3) -2g+3f = c-27 \implies -2g+3f-c = -27$
Subtracting $(1)$ from $(2)$: $4g-4f = 4 \implies g-f = 1 \implies g = f+1$.
Subtracting $(1)$ from $(3)$: $-g+4f = -13$.
Substituting $g = f+1$: $-(f+1)+4f = -13 \implies 3f = -12 \implies f = -4$.
Then $g = -4+1 = -3$.
From $(1)$: $-(-3)-(-4) = c-14 \implies 3+4 = c-14 \implies c = 21$.
The circle is $x^{2}+y^{2}-6x-8y+21=0$.
The radius $r = \sqrt{g^{2}+f^{2}-c} = \sqrt{(-3)^{2}+(-4)^{2}-21} = \sqrt{9+16-21} = \sqrt{4} = 2$.
The diameter is $2r = 2 \times 2 = 4$.

(C) The radical centre of three circles exists if and only if the radical axes of the circles are not parallel,i.e.,they are not concurrent at infinity.
First,find the radical axis of $S_{1}$ and $S_{2}$ by $S_{1} - S_{2} = 0$:
$(x^{2}+y^{2}-6x-6y+4) - (x^{2}+y^{2}-2x-4y+3) = 0$
$-4x - 2y + 1 = 0 \Rightarrow 4x + 2y - 1 = 0$.
Next,find the radical axis of $S_{2}$ and $S_{3}$ by $S_{2} - S_{3} = 0$:
$(x^{2}+y^{2}-2x-4y+3) - (x^{2}+y^{2}+2kx+2y+1) = 0$
$-(2+2k)x - 6y + 2 = 0 \Rightarrow (2+2k)x + 6y - 2 = 0$.
The radical centre does not exist if these two lines are parallel.
Two lines $a_{1}x + b_{1}y + c_{1} = 0$ and $a_{2}x + b_{2}y + c_{2} = 0$ are parallel if $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}$.
So,$\frac{4}{2+2k} = \frac{2}{6}$.
$24 = 2(2+2k)$ $\Rightarrow 24 = 4 + 4k$ $\Rightarrow 20 = 4k$ $\Rightarrow k = 5$.
Thus,if $k = 5$,the lines are parallel and the radical centre does not exist.
Therefore,$k$ cannot be $5$.

(A) The centers and radii of the given circles $x^{2}+y^{2}=9$ and $x^{2}+y^{2}+2 \alpha x+2 y+1=0$ are $C_{1}(0,0), r_{1}=3$ and $C_{2}(-\alpha, -1), r_{2}=\sqrt{(-\alpha)^{2}+(-1)^{2}-1} = \sqrt{\alpha^{2}} = |\alpha|$.
Since the two circles touch each other internally,the distance between their centers must be equal to the difference of their radii:
$C_{1}C_{2} = |r_{1} - r_{2}|$
$\sqrt{(-\alpha - 0)^{2} + (-1 - 0)^{2}} = |3 - |\alpha||$
$\sqrt{\alpha^{2} + 1} = |3 - |\alpha||$
Squaring both sides:
$\alpha^{2} + 1 = (3 - |\alpha|)^{2}$
$\alpha^{2} + 1 = 9 - 6|\alpha| + \alpha^{2}$
$6|\alpha| = 8$
$|\alpha| = \frac{4}{3}$
$\alpha = \pm \frac{4}{3}$

(D) The given equations of the circles are:
$2 x^{2}+2 y^{2}+4 x+5 y+1=0 \Rightarrow x^{2}+y^{2}+2 x+\frac{5}{2} y+\frac{1}{2}=0 \quad \dots (i)$
$3 x^{2}+3 y^{2}+6 x-7 y+3 k=0 \Rightarrow x^{2}+y^{2}+2 x-\frac{7}{3} y+k=0 \quad \dots (ii)$
Comparing these with the standard form $x^{2}+y^{2}+2 g x+2 f y+c=0$,we get:
For $(i): g_{1}=1, f_{1}=\frac{5}{4}, c_{1}=\frac{1}{2}$
For $(ii): g_{2}=1, f_{2}=-\frac{7}{6}, c_{2}=k$
The condition for two circles to be orthogonal is $2 g_{1} g_{2}+2 f_{1} f_{2}=c_{1}+c_{2}$.
Substituting the values:
$2(1)(1)+2\left(\frac{5}{4}\right)\left(-\frac{7}{6}\right)=\frac{1}{2}+k$
$2-\frac{35}{12}=\frac{1}{2}+k$
$\frac{24-35}{12}=\frac{1}{2}+k$
$-\frac{11}{12}=\frac{1}{2}+k$
$k=-\frac{11}{12}-\frac{6}{12}=-\frac{17}{12}$

(B) The given circle is $x^2+y^2+6x-8y-11=0$. Its center is $C_1 = (-3, 4)$ and radius $r_1 = \sqrt{(-3)^2 + 4^2 - (-11)} = \sqrt{9+16+11} = \sqrt{36} = 6$.
Let the required circle be $x^2+y^2+2gx+2fy+c=0$ with center $C_2 = (-g, -f)$ and radius $r_2 = 3$.
Since the circles touch externally at $(3,0)$,the point $(3,0)$ divides the line segment joining the centers $C_1(-3, 4)$ and $C_2(-g, -f)$ in the ratio $r_1:r_2 = 6:3 = 2:1$ internally.
Using the section formula:
$3 = \frac{2(-g) + 1(-3)}{2+1}$ $\Rightarrow 3 = \frac{-2g-3}{3}$ $\Rightarrow 9 = -2g-3$ $\Rightarrow 2g = -12$ $\Rightarrow g = -6$.
$0 = \frac{2(-f) + 1(4)}{2+1}$ $\Rightarrow 0 = \frac{-2f+4}{3}$ $\Rightarrow 0 = -2f+4$ $\Rightarrow 2f = 4$ $\Rightarrow f = 2$.
The radius of the second circle is $r_2 = \sqrt{g^2+f^2-c} = 3$.
Substituting $g=-6$ and $f=2$: $\sqrt{(-6)^2 + 2^2 - c} = 3$ $\Rightarrow 36+4-c = 9$ $\Rightarrow 40-c = 9$ $\Rightarrow c = 31$.
Finally,$3g-4f+c = 3(-6) - 4(2) + 31 = -18 - 8 + 31 = 5$.

(A) The angle $\theta$ between two circles with radii $r_1, r_2$ and distance between centers $d$ is given by $\cos \theta = \frac{r_1^2+r_2^2-d^2}{2 r_1 r_2}$.
$(A)$ $(x-2)^2+y^2=2$ $(r_1=\sqrt{2}, c_1=(2,0))$ and $(x-2)^2+(y-1)^2=1$ $(r_2=1, c_2=(2,1))$.
$d^2 = (2-2)^2 + (1-0)^2 = 1$.
$\cos \theta = \frac{2+1-1}{2 \times \sqrt{2} \times 1} = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}}$.
$\theta = 45^{\circ}$ or $135^{\circ}$. Thus,$(A)$ matches with $II$.
$(B)$ $x^2+y^2-6x-6y+9=0$ $(r_1=\sqrt{3^2+3^2-9}=3, c_1=(3,3))$ and $x^2+y^2-4x+4y-9=0$ $(r_2=\sqrt{2^2+(-2)^2-(-9)}=\sqrt{17}, c_2=(2,-2))$.
$d^2 = (3-2)^2 + (3-(-2))^2 = 1^2 + 5^2 = 26$.
$\cos \theta = \frac{9+17-26}{2 \times 3 \times \sqrt{17}} = 0$.
$\theta = 90^{\circ}$. Thus,$(B)$ matches with $I$.
$(C)$ $x^2+y^2+4x-14y+28=0$ $(r_1=\sqrt{(-2)^2+7^2-28}=5, c_1=(-2,7))$ and $x^2+y^2+4x-5=0$ $(r_2=\sqrt{(-2)^2+0^2-(-5)}=3, c_2=(-2,0))$.
$d^2 = (-2-(-2))^2 + (7-0)^2 = 49$.
$\cos \theta = \frac{25+9-49}{2 \times 5 \times 3} = \frac{-15}{30} = -\frac{1}{2}$.
$\theta = 120^{\circ}$ or $60^{\circ}$. Thus,$(C)$ matches with $III$.
Therefore,the correct matching is $A-II, B-I, C-III$.

(C) Let $r$ be the radius of the required circle.
From the geometry,the center of the circle lies on the $x$-axis at $(h, 0)$.
The lines $x+y=2$ and $x-y=2$ intersect at $P(2, 0)$.
The distance from the center $(h, 0)$ to the line $x+y-2=0$ is equal to the radius $r$.
$\frac{|h+0-2|}{\sqrt{1^2+1^2}} = r \Rightarrow \frac{|h-2|}{\sqrt{2}} = r$.
Since the circle is to the left of $P(2, 0)$,$h < 2$,so $\frac{2-h}{\sqrt{2}} = r \Rightarrow h = 2 - r\sqrt{2}$.
Also,the circle touches $x^2+y^2=1$ externally,so the distance between centers is $r_1+r_2$.
Distance between $(h, 0)$ and $(0, 0)$ is $h = 1+r$.
Equating the two expressions for $h$:
$1+r = 2 - r\sqrt{2}$
$r(1+\sqrt{2}) = 1$
$r = \frac{1}{\sqrt{2}+1} = \sqrt{2}-1$.
Then $h = 1 + (\sqrt{2}-1) = \sqrt{2}$.
The equation of the circle is $(x-\sqrt{2})^2 + y^2 = r^2 = (\sqrt{2}-1)^2 = 2+1-2\sqrt{2} = 3-2\sqrt{2}$.

(D) For the circle $S_1 \equiv x^2+y^2-14x+6y+33=0$,the center $C_1 = (7, -3)$ and radius $r_1 = \sqrt{7^2 + (-3)^2 - 33} = \sqrt{49+9-33} = 5$.
For the circle $S_2 \equiv x^2+y^2+30x-2y+1=0$,the center $C_2 = (-15, 1)$ and radius $r_2 = \sqrt{(-15)^2 + 1^2 - 1} = \sqrt{225+1-1} = 15$.
The internal center of similitude $A$ divides $C_1C_2$ in the ratio $r_1:r_2 = 5:15 = 1:3$.
$A = \left(\frac{1(-15) + 3(7)}{1+3}, \frac{1(1) + 3(-3)}{1+3}\right) = \left(\frac{-15+21}{4}, \frac{1-9}{4}\right) = \left(\frac{6}{4}, \frac{-8}{4}\right) = \left(\frac{3}{2}, -2\right)$.
The external center of similitude $B$ divides $C_1C_2$ in the ratio $r_1:r_2 = 1:3$ externally.
$B = \left(\frac{1(-15) - 3(7)}{1-3}, \frac{1(1) - 3(-3)}{1-3}\right) = \left(\frac{-15-21}{-2}, \frac{1+9}{-2}\right) = \left(\frac{-36}{-2}, \frac{10}{-2}\right) = (18, -5)$.
The midpoint of $AB$ is $\left(\frac{\frac{3}{2} + 18}{2}, \frac{-2 - 5}{2}\right) = \left(\frac{\frac{39}{2}}{2}, \frac{-7}{2}\right) = \left(\frac{39}{4}, \frac{-7}{2}\right)$.

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

(D) The centre of the circle $C_1: x^2+y^2+2 \alpha x+c=0$ is $O_1(-\alpha, 0)$ and its radius is $r_1 = \sqrt{\alpha^2-c}$.
The centre of the circle $C_2: x^2+y^2+2 \beta x+c=0$ is $O_2(-\beta, 0)$ and its radius is $r_2 = \sqrt{\beta^2-c}$.
For $C_1$ to lie completely inside $C_2$,the distance between centres $d = |O_1O_2| = |-\alpha - (-\beta)| = |\beta - \alpha|$ must satisfy $d + r_1 < r_2$.
Since $r_1 < r_2$,we have $\sqrt{\alpha^2-c} < \sqrt{\beta^2-c}$,which implies $\alpha^2 < \beta^2$ (assuming $c$ is such that radii exist).
Also,$|\beta - \alpha| < \sqrt{\beta^2-c} - \sqrt{\alpha^2-c}$.
Squaring both sides or analyzing the condition for containment,we find that for the circles to be nested with the same constant term $c$,the condition $\alpha \beta > 0$ must hold to ensure the centres are on the same side of the origin and the geometry allows for containment.

(D) The equation of a circle passing through the intersection of a circle $S = 0$ and a line $L = 0$ is given by $S + \lambda L = 0$.
Here,$S = x^2 + y^2 - a^2 = 0$ and $L = x \cos \alpha + y \sin \alpha - p = 0$.
So,the equation of the circle is $(x^2 + y^2 - a^2) + \lambda(x \cos \alpha + y \sin \alpha - p) = 0$.
Since $AB$ is the diameter,the center of this circle must lie on the line $L = 0$.
The center of the circle $(x^2 + y^2 + \lambda x \cos \alpha + \lambda y \sin \alpha - a^2 - \lambda p) = 0$ is $(-\frac{\lambda \cos \alpha}{2}, -\frac{\lambda \sin \alpha}{2})$.
Substituting this into $x \cos \alpha + y \sin \alpha - p = 0$:
$(-\frac{\lambda \cos \alpha}{2}) \cos \alpha + (-\frac{\lambda \sin \alpha}{2}) \sin \alpha - p = 0$
$-\frac{\lambda}{2}(\cos^2 \alpha + \sin^2 \alpha) = p$
$-\frac{\lambda}{2} = p \Rightarrow \lambda = -2p$.
Substituting $\lambda = -2p$ into the equation:
$x^2 + y^2 - a^2 - 2p(x \cos \alpha + y \sin \alpha - p) = 0$.

(C) Let $S_1 = x^2+y^2-8x-6y+21=0$ and $S_2 = x^2+y^2-2x-15=0$.
Using the family of circles equation $S_1 + \lambda S_2 = 0$,we have:
$(x^2+y^2-8x-6y+21) + \lambda(x^2+y^2-2x-15) = 0$.
Since the circle passes through the point $(1, 2)$,substitute $x=1$ and $y=2$ into the equation:
$(1^2+2^2-8(1)-6(2)+21) + \lambda(1^2+2^2-2(1)-15) = 0$.
$(1+4-8-12+21) + \lambda(1+4-2-15) = 0$.
$6 + \lambda(-12) = 0$.
$12\lambda = 6 \Rightarrow \lambda = \frac{1}{2}$.
Substituting $\lambda = \frac{1}{2}$ back into the family equation:
$(x^2+y^2-8x-6y+21) + \frac{1}{2}(x^2+y^2-2x-15) = 0$.
Multiply the entire equation by $2$:
$2(x^2+y^2-8x-6y+21) + (x^2+y^2-2x-15) = 0$.
$2x^2+2y^2-16x-12y+42 + x^2+y^2-2x-15 = 0$.
$3x^2+3y^2-18x-12y+27 = 0$.
$3(x^2+y^2)-18x-12y+27 = 0$.

(A) The equation of the family of circles passing through the intersection of two circles $S_1=0$ and $S_2=0$ is given by $S_1 + \lambda S_2 = 0$.
Given $S_1: x^2+y^2-8x-6y+21=0$ and $S_2: x^2+y^2-2x-15=0$.
The equation is $(x^2+y^2-8x-6y+21) + \lambda(x^2+y^2-2x-15) = 0$.
Since the circle passes through $(1,2)$,substitute $x=1$ and $y=2$:
$(1^2+2^2-8(1)-6(2)+21) + \lambda(1^2+2^2-2(1)-15) = 0$
$(1+4-8-12+21) + \lambda(5-2-15) = 0$
$6 + \lambda(-12) = 0 \Rightarrow \lambda = \frac{1}{2}$.
Substituting $\lambda = \frac{1}{2}$ into the family equation:
$(x^2+y^2-8x-6y+21) + \frac{1}{2}(x^2+y^2-2x-15) = 0$
Multiply by $2$:
$2(x^2+y^2-8x-6y+21) + (x^2+y^2-2x-15) = 0$
$3x^2+3y^2-18x-12y+27 = 0$
Dividing by $3$:
$x^2+y^2-6x-4y+9 = 0$.