System of circles Questions in English

(D) The family of circles passing through the intersection of a circle $S = 0$ and a line $P = 0$ is represented by the equation $S + \lambda P = 0$,where $\lambda$ is a parameter.
Since $\lambda$ can be any real number,$S - \lambda P = 0$ is also a valid representation (by replacing $\lambda$ with $-\lambda$).
Additionally,if we consider the form $\lambda S + P = 0$,it represents the family of circles provided $\lambda \neq 0$ and the line is treated as a degenerate circle.
Thus,all the given forms can represent the family of circles under appropriate conditions.
Therefore,the correct option is $D$.

(B) The equation of a family of circles passing through the intersection of a circle $S = 0$ and a line $L = 0$ is given by $S + \lambda L = 0$.
Given $S = {x^2} + {y^2} - 2x - 6y + 6 = 0$ and $L = 3x + 2y - 5 = 0$,the equation is:
$({x^2} + {y^2} - 2x - 6y + 6) + \lambda (3x + 2y - 5) = 0$
Since the circle passes through the point $(-2, 4)$,we substitute $x = -2$ and $y = 4$ into the equation:
$((-2)^2 + (4)^2 - 2(-2) - 6(4) + 6) + \lambda (3(-2) + 2(4) - 5) = 0$
$(4 + 16 + 4 - 24 + 6) + \lambda (-6 + 8 - 5) = 0$
$6 + \lambda (-3) = 0$
$3\lambda = 6 \implies \lambda = 2$
Substituting $\lambda = 2$ back into the family equation:
$({x^2} + {y^2} - 2x - 6y + 6) + 2(3x + 2y - 5) = 0$
${x^2} + {y^2} - 2x - 6y + 6 + 6x + 4y - 10 = 0$
${x^2} + {y^2} + 4x - 2y - 4 = 0$

(B) Let $P(x_1, y_1)$ be any point on the first circle $x^2 + y^2 + 2gx + 2fy + c_1 = 0$.
Since $P$ lies on the circle,we have $x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c_1 = 0$,which implies $x_1^2 + y_1^2 + 2gx_1 + 2fy_1 = -c_1$.
The length of the tangent from a point $(x_1, y_1)$ to the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is given by $\sqrt{x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c}$.
Substituting the value of $x_1^2 + y_1^2 + 2gx_1 + 2fy_1$ from the first equation,we get the length of the tangent as $\sqrt{-c_1 + c} = \sqrt{c - c_1}$.

(A) The equation of the tangent to the circle $x^2 + y^2 = 5$ at $(1, -2)$ is given by $x(1) + y(-2) = 5$,which simplifies to $x - 2y = 5$,or $x = 2y + 5$.
Substitute $x = 2y + 5$ into the equation of the second circle $x^2 + y^2 - 8x + 6y + 20 = 0$:
$(2y + 5)^2 + y^2 - 8(2y + 5) + 6y + 20 = 0$
Expanding the terms:
$(4y^2 + 20y + 25) + y^2 - 16y - 40 + 6y + 20 = 0$
Combining like terms:
$5y^2 + 10y + 5 = 0$
Dividing by $5$:
$y^2 + 2y + 1 = 0$
$(y + 1)^2 = 0$
This gives $y = -1$. Substituting $y = -1$ into $x = 2y + 5$ gives $x = 2(-1) + 5 = 3$.
Since there is only one point of intersection $(3, -1)$,the line touches the circle.

(C) For the first circle $x^2 + y^2 + 2x + 8y - 23 = 0$:
Center $C_1 = (-1, -4)$ and radius $r_1 = \sqrt{(-1)^2 + (-4)^2 - (-23)} = \sqrt{1 + 16 + 23} = \sqrt{40} = 2\sqrt{10} \approx 6.32$.
For the second circle $x^2 + y^2 - 4x - 10y + 9 = 0$:
Center $C_2 = (2, 5)$ and radius $r_2 = \sqrt{2^2 + 5^2 - 9} = \sqrt{4 + 25 - 9} = \sqrt{20} = 2\sqrt{5} \approx 4.47$.
The distance between the centers $C_1$ and $C_2$ is $d = \sqrt{(2 - (-1))^2 + (5 - (-4))^2} = \sqrt{3^2 + 9^2} = \sqrt{9 + 81} = \sqrt{90} = 3\sqrt{10} \approx 9.49$.
Now,check the relationship between $d$,$r_1 + r_2$,and $|r_1 - r_2|$:
$r_1 + r_2 = 2\sqrt{10} + 2\sqrt{5} \approx 6.32 + 4.47 = 10.79$.
$|r_1 - r_2| = 2\sqrt{10} - 2\sqrt{5} \approx 6.32 - 4.47 = 1.85$.
Since $|r_1 - r_2| < d < r_1 + r_2$ $(1.85 < 9.49 < 10.79)$,the circles intersect at two distinct points.
Therefore,there are $2$ common tangents.

(C) The equation of the first circle is $S_1: x^2 + y^2 - 4 = 0$.
The equation of the second circle is $S_2: x^2 + y^2 - 2x + 6y + a = 0$.
The common chord of the two circles is given by $S_1 - S_2 = 0$,which is $(x^2 + y^2 - 4) - (x^2 + y^2 - 2x + 6y + a) = 0$.
This simplifies to $2x - 6y - 4 - a = 0$.
If a circle bisects the circumference of another circle,the common chord must be a diameter of the second circle.
Therefore,the common chord must pass through the center of the second circle,which is $(1, -3)$.
Substituting $(1, -3)$ into the equation of the common chord: $2(1) - 6(-3) - 4 - a = 0$.
$2 + 18 - 4 - a = 0$.
$16 - a = 0$,which gives $a = 16$.

(C) The equation of a family of circles passing through the intersection of the circle $S \equiv 2x^2 + 2y^2 - 2x - 6y - 25 = 0$ and the line $L \equiv x - y - 1 = 0$ is given by $S + \lambda L = 0$.
$(2x^2 + 2y^2 - 2x - 6y - 25) + \lambda(x - y - 1) = 0$
$2x^2 + 2y^2 + (\lambda - 2)x - (\lambda + 6)y - (25 + \lambda) = 0$
The center of this circle is $\left( -\frac{\lambda - 2}{4}, \frac{\lambda + 6}{4} \right)$.
Since the line $x - y - 1 = 0$ is the diameter,the center must lie on this line:
$-\frac{\lambda - 2}{4} - \frac{\lambda + 6}{4} - 1 = 0$
$-\lambda + 2 - \lambda - 6 - 4 = 0$
$-2\lambda - 8 = 0 \implies \lambda = -4$.
Substituting $\lambda = -4$ into the family equation:
$2x^2 + 2y^2 + (-4 - 2)x - (-4 + 6)y - (25 - 4) = 0$
$2x^2 + 2y^2 - 6x - 2y - 21 = 0$.

(D) For the first circle $x^2 + y^2 - 6x - 2y + 1 = 0$,the center $C_1 = (3, 1)$ and radius $R_1 = \sqrt{3^2 + 1^2 - 1} = \sqrt{9} = 3$.
For the second circle $x^2 + y^2 + 2x - 8y + 13 = 0$,the center $C_2 = (-1, 4)$ and radius $R_2 = \sqrt{(-1)^2 + 4^2 - 13} = \sqrt{1 + 16 - 13} = \sqrt{4} = 2$.
The distance between the centers is $C_1C_2 = \sqrt{(3 - (-1))^2 + (1 - 4)^2} = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = 5$.
Since $R_1 + R_2 = 3 + 2 = 5$,we have $C_1C_2 = R_1 + R_2$.
Therefore,the circles touch each other externally.

(D) Let the centre of the circle be $(h, k)$. Since the circle passes through $(0, 0)$ and $(1, 0)$,the perpendicular bisector of the chord joining $(0, 0)$ and $(1, 0)$ must contain the centre. The midpoint is $(1/2, 0)$ and the line is $x = 1/2$. Thus,$h = 1/2$.
Since the circle passes through $(0, 0)$,its radius $r$ is the distance from $(1/2, k)$ to $(0, 0)$,so $r^2 = (1/2)^2 + k^2 = 1/4 + k^2$.
The circle touches $x^2 + y^2 = 9$,which has centre $(0, 0)$ and radius $R = 3$. The distance between the centres is $\sqrt{(1/2)^2 + k^2} = \sqrt{1/4 + k^2} = r$.
For the circles to touch,the distance between centres must be $|R \pm r|$. Here,the distance between centres is $r$,so $r = |3 \pm r|$.
Case $1$: $r = 3 - r$ $\Rightarrow 2r = 3$ $\Rightarrow r = 3/2$.
Then $r^2 = 9/4$,so $1/4 + k^2 = 9/4$ $\Rightarrow k^2 = 8/4 = 2$ $\Rightarrow k = \pm \sqrt{2}$.
Thus,the centre is $\left( \frac{1}{2}, \pm \sqrt{2} \right)$.

(A) 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 cut orthogonally is $2g_1g_2 + 2f_1f_2 = c_1 + c_2$.
Comparing the given equations with the standard form:
For the first circle: $2g_1 = p \Rightarrow g_1 = \frac{p}{2}$,$2f_1 = 3 \Rightarrow f_1 = \frac{3}{2}$,$c_1 = -5$.
For the second circle: $2g_2 = 5 \Rightarrow g_2 = \frac{5}{2}$,$2f_2 = p \Rightarrow f_2 = \frac{p}{2}$,$c_2 = 7$.
Substituting these values into the condition:
$2\left(\frac{p}{2}\right)\left(\frac{5}{2}\right) + 2\left(\frac{3}{2}\right)\left(\frac{p}{2}\right) = -5 + 7$
$\frac{5p}{2} + \frac{3p}{2} = 2$
$\frac{8p}{2} = 2$
$4p = 2$
$p = \frac{2}{4} = \frac{1}{2}$.

(B) Given equations of the circles are:
$x^2 + y^2 - 6x - 6y + 10 = 0$ $(i)$
$x^2 + y^2 = 2$ $(ii)$
Subtracting equation $(ii)$ from $(i)$:
$(x^2 + y^2 - 6x - 6y + 10) - (x^2 + y^2) = 0 - 2$
$-6x - 6y + 10 = -2$
$-6x - 6y + 12 = 0$
$x + y = 2$ $(iii)$
From $(iii)$,$y = 2 - x$. Substituting this into $(ii)$:
$x^2 + (2 - x)^2 = 2$
$x^2 + 4 - 4x + x^2 = 2$
$2x^2 - 4x + 2 = 0$
$x^2 - 2x + 1 = 0$
$(x - 1)^2 = 0$
$x = 1$
Substituting $x = 1$ into $x + y = 2$,we get $y = 1$.
Thus,the point of contact is $(1, 1)$.

(A) The equation of a coaxial system of circles is given by $S + \lambda L = 0$,where $S = x^2 + y^2 - 9 = 0$ and $L$ is the radical axis.
Since $(2, 3)$ is a limiting point,the circle with radius $0$ passing through $(2, 3)$ is $(x-2)^2 + (y-3)^2 = 0$,which is $x^2 + y^2 - 4x - 6y + 13 = 0$.
The radical axis is $S_1 - S_2 = 0$,which gives $(x^2 + y^2 - 9) - (x^2 + y^2 - 4x - 6y + 13) = 0$,simplifying to $4x + 6y - 22 = 0$ or $2x + 3y - 11 = 0$.
The family of coaxial circles is $(x^2 + y^2 - 9) + k(2x + 3y - 11) = 0$.
The center is $(-k, -\frac{3k}{2})$ and the radius squared is $r^2 = k^2 + \frac{9k^2}{4} + 9 + 11k = \frac{13k^2 + 44k + 36}{4}$.
For limiting points,$r^2 = 0$,so $13k^2 + 44k + 36 = 0$.
Solving for $k$,we get $(13k + 18)(k + 2) = 0$,so $k = -2$ or $k = -\frac{18}{13}$.
For $k = -2$,the center is $(2, 3)$.
For $k = -\frac{18}{13}$,the center is $(\frac{18}{13}, \frac{3}{2} \times \frac{18}{13}) = (\frac{18}{13}, \frac{27}{13})$.

(A) The family of circles passing through the intersection of two circles ${S_1} = 0$ and ${S_2} = 0$ is given by ${S_1} + \lambda {S_2} = 0$ (where $\lambda \neq -1$).
Given ${S_1}: {x^2} + {y^2} - 2x - 4y + 1 = 0$ and ${S_2}: {x^2} + {y^2} - 4x - 2y + 4 = 0$.
The equation is: $({x^2} + {y^2} - 2x - 4y + 1) + \lambda ({x^2} + {y^2} - 4x - 2y + 4) = 0$.
$(1 + \lambda ){x^2} + (1 + \lambda ){y^2} - (2 + 4\lambda )x - (4 + 2\lambda )y + (1 + 4\lambda ) = 0$.
Dividing by $(1 + \lambda )$: ${x^2} + {y^2} - 2\frac{(1 + 2\lambda )}{1 + \lambda }x - 2\frac{(2 + \lambda )}{1 + \lambda }y + \frac{1 + 4\lambda }{1 + \lambda } = 0$.
The centre of this circle is $\left( \frac{1 + 2\lambda }{1 + \lambda }, \frac{2 + \lambda }{1 + \lambda } \right)$.
Since the centre lies on $x + 2y - 3 = 0$,we substitute the coordinates:
$\frac{1 + 2\lambda }{1 + \lambda } + 2\left( \frac{2 + \lambda }{1 + \lambda } \right) - 3 = 0$.
$1 + 2\lambda + 4 + 2\lambda - 3(1 + \lambda ) = 0$.
$5 + 4\lambda - 3 - 3\lambda = 0 \implies \lambda + 2 = 0 \implies \lambda = -2$.
Substituting $\lambda = -2$ into the family equation:
$({x^2} + {y^2} - 2x - 4y + 1) - 2({x^2} + {y^2} - 4x - 2y + 4) = 0$.
$-{x^2} - {y^2} + 6x - 7 = 0 \implies {x^2} + {y^2} - 6x + 7 = 0$.

(A) For the first circle $x^2 + y^2 - 2x - 4y = 0$,the center $C_1 = (1, 2)$ and radius $R_1 = \sqrt{1^2 + 2^2 - 0} = \sqrt{5}$.
For the second circle $x^2 + y^2 - 8y - 4 = 0$,the center $C_2 = (0, 4)$ and radius $R_2 = \sqrt{0^2 + 4^2 - (-4)} = \sqrt{20} = 2\sqrt{5}$.
The distance between the centers $C_1$ and $C_2$ is $C_1C_2 = \sqrt{(1-0)^2 + (2-4)^2} = \sqrt{1^2 + (-2)^2} = \sqrt{1+4} = \sqrt{5}$.
We observe that $|R_2 - R_1| = |2\sqrt{5} - \sqrt{5}| = \sqrt{5}$.
Since $C_1C_2 = |R_2 - R_1|$,the circles touch each other internally.

(B) The equation of a 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 $S_1: x^2 + y^2 + 13x - 3y = 0$ and $S_2: 2x^2 + 2y^2 + 4x - 7y - 25 = 0$.
The required equation is $(x^2 + y^2 + 13x - 3y) + \lambda (2x^2 + 2y^2 + 4x - 7y - 25) = 0$.
Since the circle passes through $(1, 1)$,we substitute $x = 1$ and $y = 1$:
$(1^2 + 1^2 + 13(1) - 3(1)) + \lambda (2(1^2) + 2(1^2) + 4(1) - 7(1) - 25) = 0$
$(1 + 1 + 13 - 3) + \lambda (2 + 2 + 4 - 7 - 25) = 0$
$12 + \lambda (-24) = 0$
$12 = 24\lambda \implies \lambda = \frac{1}{2}$.
Substituting $\lambda = \frac{1}{2}$ into the family equation:
$(x^2 + y^2 + 13x - 3y) + \frac{1}{2}(2x^2 + 2y^2 + 4x - 7y - 25) = 0$
Multiply by $2$:
$2x^2 + 2y^2 + 26x - 6y + 2x^2 + 2y^2 + 4x - 7y - 25 = 0$
$4x^2 + 4y^2 + 30x - 13y - 25 = 0$.

(D) Let the required circle be ${x^2} + {y^2} + 2gx + 2fy + c = 0$.
Two circles ${x^2} + {y^2} + 2g_1x + 2f_1y + c_1 = 0$ and ${x^2} + {y^2} + 2g_2x + 2f_2y + c_2 = 0$ intersect at right angles if $2g_1g_2 + 2f_1f_2 = c_1 + c_2$.
Applying this condition to the given circles:
$(i) \; 2g(1/2) + 2f(1) = c + 3 \Rightarrow g + 2f = c + 3$
$(ii) \; 2g(1) + 2f(2) = c + 5 \Rightarrow 2g + 4f = c + 5$
$(iii) \; 2g(-7/2) + 2f(-4) = c - 9 \Rightarrow -7g - 8f = c - 9$
Subtracting $(i)$ from $(ii)$,we get $g + 2f = 2$.
Substituting $g + 2f = 2$ into $(i)$,we get $2 = c + 3$,so $c = -1$.
Now,$g + 2f = 2$ and $-7g - 8f = -1 - 9 = -10$,which simplifies to $7g + 8f = 10$.
Solving $g + 2f = 2$ and $7g + 8f = 10$:
Multiply the first by $4$: $4g + 8f = 8$.
Subtracting this from $7g + 8f = 10$ gives $3g = 2$,so $g = 2/3$.
Then $2/3 + 2f = 2$ $\Rightarrow 2f = 4/3$ $\Rightarrow f = 2/3$.
Substituting $g=2/3, f=2/3, c=-1$ into the general equation:
${x^2} + {y^2} + 2(2/3)x + 2(2/3)y - 1 = 0$
${x^2} + {y^2} + 4/3x + 4/3y - 1 = 0$
Multiplying by $3$: $3({x^2} + {y^2}) + 4x + 4y - 3 = 0$,which is $3({x^2} + {y^2}) + 4(x + y) - 3 = 0$.

(C) The family of circles passing through the intersection of $S_1: x^2 + y^2 - 1 = 0$ and $S_2: x^2 + y^2 - 2x - 4y + 1 = 0$ is given by $S_2 + \lambda S_1 = 0$.
$(x^2 + y^2 - 2x - 4y + 1) + \lambda(x^2 + y^2 - 1) = 0$
$(1 + \lambda)x^2 + (1 + \lambda)y^2 - 2x - 4y + (1 - \lambda) = 0$
Dividing by $(1 + \lambda)$,we get $x^2 + y^2 - \frac{2}{1 + \lambda}x - \frac{4}{1 + \lambda}y + \frac{1 - \lambda}{1 + \lambda} = 0$.
The centre is $C = \left( \frac{1}{1 + \lambda}, \frac{2}{1 + \lambda} \right)$ and the radius $r = \sqrt{\left( \frac{1}{1 + \lambda} \right)^2 + \left( \frac{2}{1 + \lambda} \right)^2 - \frac{1 - \lambda}{1 + \lambda}} = \sqrt{\frac{1 + 4 - (1 - \lambda)(1 + \lambda)}{(1 + \lambda)^2}} = \sqrt{\frac{5 - (1 - \lambda^2)}{(1 + \lambda)^2}} = \frac{\sqrt{4 + \lambda^2}}{|1 + \lambda|}$.
Since the circle touches the line $x + 2y = 0$,the perpendicular distance from the centre to the line equals the radius:
$\left| \frac{\frac{1}{1 + \lambda} + 2(\frac{2}{1 + \lambda})}{\sqrt{1^2 + 2^2}} \right| = \frac{\sqrt{4 + \lambda^2}}{|1 + \lambda|}$
$\left| \frac{5}{\sqrt{5}(1 + \lambda)} \right| = \frac{\sqrt{4 + \lambda^2}}{|1 + \lambda|}$
$\sqrt{5} = \sqrt{4 + \lambda^2}$ $\Rightarrow 5 = 4 + \lambda^2$ $\Rightarrow \lambda^2 = 1$ $\Rightarrow \lambda = \pm 1$.
For $\lambda = -1$,the equation does not represent a circle. Thus,$\lambda = 1$.
Substituting $\lambda = 1$ into the family equation: $2x^2 + 2y^2 - 2x - 4y = 0$,which simplifies to $x^2 + y^2 - x - 2y = 0$.

(C) Let the equation of the circle be ${x^2} + {y^2} + 2gx + 2fy = 0$ since it passes through the origin $(0, 0)$.
Since the centre $(-g, -f)$ lies on the line $x + y = 4$,we have $-g - f = 4$,or $g + f = -4$ ... $(i)$.
The circle cuts ${x^2} + {y^2} - 4x + 2y + 4 = 0$ orthogonally. The condition for orthogonality is $2g_1g_2 + 2f_1f_2 = c_1 + c_2$.
Here,$g_1 = g, f_1 = f, c_1 = 0$ and $g_2 = -2, f_2 = 1, c_2 = 4$.
So,$2(g)(-2) + 2(f)(1) = 0 + 4$,which simplifies to $-4g + 2f = 4$,or $-2g + f = 2$ ... $(ii)$.
Subtracting $(ii)$ from $(i)$: $(g + f) - (-2g + f) = -4 - 2$,which gives $3g = -6$,so $g = -2$.
Substituting $g = -2$ into $(i)$: $-2 + f = -4$,so $f = -2$.
The equation of the circle is ${x^2} + {y^2} + 2(-2)x + 2(-2)y = 0$,which is ${x^2} + {y^2} - 4x - 4y = 0$.

(C) The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$.
Comparing the given circles with the general form:
For the first circle,$2g_1 = a$,$2f_1 = b$,and $c_1 = c$.
For the second circle,$2g_2 = d$,$2f_2 = e$,and $c_2 = f$.
The condition for two circles to intersect orthogonally is $2g_1g_2 + 2f_1f_2 = c_1 + c_2$.
Substituting the values:
$2(\frac{a}{2})(\frac{d}{2}) + 2(\frac{b}{2})(\frac{e}{2}) = c + f$
$\frac{ad}{2} + \frac{be}{2} = c + f$
Multiplying by $2$ on both sides:
$ad + be = 2(c + f)$.

(A) Let the equations of the circles be:
$S_1: x^2 + y^2 + 4x + 6y - 19 = 0$
$S_2: x^2 + y^2 - 9 = 0$
$S_3: x^2 + y^2 - 2x - 2y - 5 = 0$
The radical axis of $S_1$ and $S_2$ is $S_1 - S_2 = 0$:
$(x^2 + y^2 + 4x + 6y - 19) - (x^2 + y^2 - 9) = 0$
$4x + 6y - 10 = 0 \implies 2x + 3y = 5$ ... $(i)$
The radical axis of $S_2$ and $S_3$ is $S_2 - S_3 = 0$:
$(x^2 + y^2 - 9) - (x^2 + y^2 - 2x - 2y - 5) = 0$
$2x + 2y - 4 = 0 \implies x + y = 2$ ... $(ii)$
Solving equations $(i)$ and $(ii)$:
From $(ii)$,$y = 2 - x$. Substituting in $(i)$:
$2x + 3(2 - x) = 5$
$2x + 6 - 3x = 5$
$-x = -1 \implies x = 1$
Substituting $x = 1$ in $(ii)$,$1 + y = 2 \implies y = 1$.
Thus,the radical centre is $(1, 1)$.

(A) Let the centre of the circle be $(h, k)$ and its radius be $r$.
Since the circle touches $x^2 + y^2 = a^2$ externally,the distance between their centres is equal to the sum of their radii:
$\sqrt{h^2 + k^2} = r + a \implies r = \sqrt{h^2 + k^2} - a$ $(i)$
Since the circle touches $x^2 + y^2 - 4ax = 0$ (which is $(x-2a)^2 + y^2 = (2a)^2$) externally,the distance between their centres is:
$\sqrt{(h-2a)^2 + k^2} = r + 2a$ $(ii)$
Substituting $r$ from $(i)$ into $(ii)$:
$\sqrt{(h-2a)^2 + k^2} = (\sqrt{h^2 + k^2} - a) + 2a$
$\sqrt{(h-2a)^2 + k^2} = \sqrt{h^2 + k^2} + a$
Squaring both sides:
$(h-2a)^2 + k^2 = (h^2 + k^2) + a^2 + 2a\sqrt{h^2 + k^2}$
$h^2 - 4ah + 4a^2 + k^2 = h^2 + k^2 + a^2 + 2a\sqrt{h^2 + k^2}$
$3a^2 - 4ah = 2a\sqrt{h^2 + k^2}$
Dividing by $a$ (assuming $a \neq 0$):
$3a - 4h = 2\sqrt{h^2 + k^2}$
Squaring again:
$(3a - 4h)^2 = 4(h^2 + k^2)$
$9a^2 - 24ah + 16h^2 = 4h^2 + 4k^2$
$12h^2 - 4k^2 - 24ah + 9a^2 = 0$
Replacing $(h, k)$ with $(x, y)$,the locus is $12x^2 - 4y^2 - 24ax + 9a^2 = 0$.

(C) Two circles with radii $r_1$ and $r_2$ and distance between centers $d$ cut orthogonally if $d^2 = r_1^2 + r_2^2$.
Here,$r_1 = r_2 = a$.
The distance between centers $(2, 3)$ and $(5, 6)$ is $d = \sqrt{(5-2)^2 + (6-3)^2} = \sqrt{3^2 + 3^2} = \sqrt{18}$.
Thus,$d^2 = 18$.
Substituting into the condition: $18 = a^2 + a^2$.
$18 = 2a^2$.
$a^2 = 9$.
Since $a$ is a radius,$a = 3$.

(C) The general equation of a circle passing through the origin is $x^2 + y^2 + 2gx + 2fy = 0$.
For a system of co-axial circles,the radical axis of any two circles must be the same.
The given circles are $S_1: x^2 + y^2 - a^2 = 0$ and $S_2: x^2 + y^2 + 2ax - 2a^2 = 0$.
The radical axis of $S_1$ and $S_2$ is $S_1 - S_2 = 0$,which gives $2ax - a^2 = 0$,or $x = \frac{a}{2}$.
Let the required circle be $S_3: x^2 + y^2 + 2gx + 2fy = 0$.
The radical axis of $S_1$ and $S_3$ is $x^2 + y^2 - a^2 - (x^2 + y^2 + 2gx + 2fy) = 0$,which simplifies to $2gx + 2fy + a^2 = 0$.
Since the circles are co-axial,this radical axis must be $x = \frac{a}{2}$.
Comparing $2gx + 2fy + a^2 = 0$ with $x = \frac{a}{2}$ (or $x - \frac{a}{2} = 0$),we get $f = 0$ and $\frac{2g}{1} = \frac{a^2}{-a/2} = -2a$.
Thus,$g = -a$.
Substituting $g = -a$ and $f = 0$ into the equation of $S_3$,we get $x^2 + y^2 - 2ax = 0$.

(C) Let the equation of the family of circles passing through the intersection of $S_1 = x^2 + y^2 - 8x - 2y + 7 = 0$ and $S_2 = x^2 + y^2 - 4x + 10y + 8 = 0$ be $S_1 + \lambda S_2 = 0$.
$(x^2 + y^2 - 8x - 2y + 7) + \lambda(x^2 + y^2 - 4x + 10y + 8) = 0$
$(1 + \lambda)x^2 + (1 + \lambda)y^2 - (8 + 4\lambda)x + (10\lambda - 2)y + (7 + 8\lambda) = 0$
Dividing by $(1 + \lambda)$:
$x^2 + y^2 - \frac{8 + 4\lambda}{1 + \lambda}x + \frac{10\lambda - 2}{1 + \lambda}y + \frac{7 + 8\lambda}{1 + \lambda} = 0$
The centre of this circle is $(\frac{4 + 2\lambda}{1 + \lambda}, -\frac{5\lambda - 1}{1 + \lambda})$.
Since the centre lies on the $y$-axis,the $x$-coordinate of the centre must be $0$.
$\frac{4 + 2\lambda}{1 + \lambda} = 0 \implies 4 + 2\lambda = 0 \implies \lambda = -2$.
Substituting $\lambda = -2$ into the family equation:
$(x^2 + y^2 - 8x - 2y + 7) - 2(x^2 + y^2 - 4x + 10y + 8) = 0$
$x^2 + y^2 - 8x - 2y + 7 - 2x^2 - 2y^2 + 8x - 20y - 16 = 0$
$-x^2 - y^2 - 22y - 9 = 0$
$x^2 + y^2 + 22y + 9 = 0$.

(C) Let $P$ be any point on the circle $x^2 + y^2 = a^2$. Let $PQ$ and $PR$ be the tangents drawn from $P$ to the circle $x^2 + y^2 = a^2 \sin^2 \alpha$,where $Q$ and $R$ are the points of contact.
Let $O$ be the origin $(0, 0)$. The radius of the inner circle is $OQ = a \sin \alpha$.
The distance $OP$ is the radius of the outer circle,so $OP = a$.
In the right-angled triangle $\triangle OQP$,$\sin(\angle OPQ) = \frac{OQ}{OP} = \frac{a \sin \alpha}{a} = \sin \alpha$.
Therefore,$\angle OPQ = \alpha$.
The angle between the tangents is $\angle QPR = 2 \times \angle OPQ = 2\alpha$.

(A) The equation of a 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 $S_1: {x^2} + {y^2} - 8x - 2y + 7 = 0$ and $S_2: {x^2} + {y^2} - 4x + 10y + 8 = 0$.
The equation is $({x^2} + {y^2} - 8x - 2y + 7) + \lambda ({x^2} + {y^2} - 4x + 10y + 8) = 0$.
Since the circle passes through $(3, -3)$,substitute $x = 3$ and $y = -3$ into the equation:
$(9 + 9 - 24 + 6 + 7) + \lambda (9 + 9 - 12 - 30 + 8) = 0$
$7 + \lambda (-16) = 0$
$16\lambda = 7 \Rightarrow \lambda = \frac{7}{16}$.
Substituting $\lambda = \frac{7}{16}$ back into the equation:
$16({x^2} + {y^2} - 8x - 2y + 7) + 7({x^2} + {y^2} - 4x + 10y + 8) = 0$
$16{x^2} + 16{y^2} - 128x - 32y + 112 + 7{x^2} + 7{y^2} - 28x + 70y + 56 = 0$
$23{x^2} + 23{y^2} - 156x + 38y + 168 = 0$.

(C) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$.
Since it intersects the circles $x^2 + y^2 + 2x + 4y + 6 = 0$ and $x^2 + y^2 + 4x + 6y + 2 = 0$ orthogonally,we use the condition $2g_1g_2 + 2f_1f_2 = c_1 + c_2$.
For the first circle: $2g(1) + 2f(2) = c + 6 \implies 2g + 4f - c = 6$.
For the second circle: $2g(2) + 2f(3) = c + 2 \implies 4g + 6f - c = 2$.
Since the circle passes through $(1, 1)$,we have $1^2 + 1^2 + 2g(1) + 2f(1) + c = 0 \implies 2g + 2f + c = -2$.
Solving the system of equations:
$2g + 4f - c = 6$
$4g + 6f - c = 2$
$2g + 2f + c = -2$
Subtracting the first from the second: $2g + 2f = -4 \implies g + f = -2$.
Adding the first and third: $4g + 6f = 4 \implies 2g + 3f = 2$.
Solving these gives $f = 6$ and $g = -8$.
Substituting into the third equation: $2(-8) + 2(6) + c = -2 \implies -16 + 12 + c = -2 \implies c = 2$.
Thus,the equation is $x^2 + y^2 - 16x + 12y + 2 = 0$.

(A) 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 intersect orthogonally is given by the formula:
$2g_1g_2 + 2f_1f_2 = c_1 + c_2$
This is a standard result derived from the Pythagorean theorem applied to the triangle formed by the centers of the circles and their point of intersection.

(A) For two circles with centers $C_1(-g, -f)$ and $C_2(-g', -f')$ and radii $r_1 = \sqrt{g^2 + f^2}$ and $r_2 = \sqrt{g'^2 + f'^2}$ to touch externally,the distance between their centers must be equal to the sum of their radii.
$C_1C_2 = r_1 + r_2$
$\sqrt{(-g + g')^2 + (-f + f')^2} = \sqrt{g^2 + f^2} + \sqrt{g'^2 + f'^2}$
Squaring both sides:
$(g' - g)^2 + (f' - f)^2 = (g^2 + f^2) + (g'^2 + f'^2) + 2\sqrt{g^2 + f^2}\sqrt{g'^2 + f'^2}$
$g'^2 - 2gg' + g^2 + f'^2 - 2ff' + f^2 = g^2 + f^2 + g'^2 + f'^2 + 2\sqrt{(g^2 + f^2)(g'^2 + f'^2)}$
$-2(gg' + ff') = 2\sqrt{(g^2 + f^2)(g'^2 + f'^2)}$
$-(gg' + ff') = \sqrt{(g^2 + f^2)(g'^2 + f'^2)}$
Squaring again:
$(gg' + ff')^2 = (g^2 + f^2)(g'^2 + f'^2)$
$g^2g'^2 + f^2f'^2 + 2gg'ff' = g^2g'^2 + g^2f'^2 + f^2g'^2 + f^2f'^2$
$2gg'ff' = g^2f'^2 + f^2g'^2$
$g^2f'^2 + f^2g'^2 - 2(gf')(fg') = 0$
$(gf' - fg')^2 = 0$
$gf' = fg'$

(B) For the first circle $x^2 + y^2 - 2x - 3 = 0$:
Center $C_1 = (1, 0)$,Radius $r_1 = \sqrt{1^2 + 0^2 - (-3)} = \sqrt{4} = 2$.
For the second circle $x^2 + y^2 - 4x - 6y - 8 = 0$:
Center $C_2 = (2, 3)$,Radius $r_2 = \sqrt{2^2 + 3^2 - (-8)} = \sqrt{4 + 9 + 8} = \sqrt{21}$.
The distance between the centers $C_1$ and $C_2$ is $d = \sqrt{(2-1)^2 + (3-0)^2} = \sqrt{1^2 + 3^2} = \sqrt{10}$.
We check the condition for intersection: $|r_1 - r_2| < d < r_1 + r_2$.
$|2 - \sqrt{21}| \approx |2 - 4.58| = 2.58$.
$d = \sqrt{10} \approx 3.16$.
$r_1 + r_2 = 2 + \sqrt{21} \approx 6.58$.
Since $2.58 < 3.16 < 6.58$,the condition $|r_1 - r_2| < d < r_1 + r_2$ is satisfied.
Therefore,the circles intersect each other.

(A) Let $S_1 = x^2 + y^2 - 6x - 6y + 4 = 0$ and $S_2 = x^2 + y^2 - 2x - 4y + 3 = 0$.
The radical axis is given by $S_1 - S_2 = 0$,which is $(x^2 + y^2 - 6x - 6y + 4) - (x^2 + y^2 - 2x - 4y + 3) = 0$.
Simplifying this,we get $-4x - 2y + 1 = 0$,or $4x + 2y - 1 = 0$.
The equation of the coaxial system is $S_1 + \lambda(4x + 2y - 1) = 0$,which is $x^2 + y^2 - (6 - 4\lambda)x - (6 - 2\lambda)y + (4 - \lambda) = 0$.
The center of these circles is $(3 - 2\lambda, 3 - \lambda)$ and the radius $r$ is given by $r^2 = (3 - 2\lambda)^2 + (3 - \lambda)^2 - (4 - \lambda)$.
For limit points,the radius $r = 0$.
Thus,$(3 - 2\lambda)^2 + (3 - \lambda)^2 - (4 - \lambda) = 0$.
$9 - 12\lambda + 4\lambda^2 + 9 - 6\lambda + \lambda^2 - 4 + \lambda = 0$.
$5\lambda^2 - 17\lambda + 14 = 0$.
Solving for $\lambda$,we get $(5\lambda - 7)(\lambda - 2) = 0$,so $\lambda = 2$ or $\lambda = 7/5$.
For $\lambda = 2$,the center is $(3 - 2(2), 3 - 2) = (-1, 1)$.
For $\lambda = 7/5$,the center is $(3 - 14/5, 3 - 7/5) = (1/5, 8/5)$.
Comparing with the options,the correct limit point is $(-1, 1)$.

(C) Let the required circle be ${x^2} + {y^2} + 2gx + 2fy + c = 0$ $(i)$.
Since it passes through the origin $(0, 0)$,we have $c = 0$.
Given that the centre $(-g, -f)$ lies on $y = x$,we have $-f = -g$,which implies $f = g$.
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 cut orthogonally is $2g_1g_2 + 2f_1f_2 = c_1 + c_2$.
Here,$g_1 = g, f_1 = f, c_1 = 0$ and $g_2 = -2, f_2 = -3, c_2 = 10$.
Substituting these values,we get $2(g)(-2) + 2(f)(-3) = 0 + 10$.
Since $f = g$,this becomes $-4g - 6g = 10$,which gives $-10g = 10$,so $g = -1$.
Thus,$f = -1$.
The equation of the circle is ${x^2} + {y^2} - 2x - 2y = 0$.

(A) Let the vertices of the triangle be $A, B,$ and $C$. Consider the three circles having the sides $AB, BC,$ and $CA$ as their diameters.
The equation of a circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$.
The radical axis of any two of these circles is the common chord of the two circles. For circles described on the sides of a triangle as diameters,the radical axis of any two circles is the altitude of the triangle passing through the vertex common to the two sides.
Since the radical centre is the point of intersection of the radical axes of the three circles taken in pairs,it is the point of intersection of the three altitudes of the triangle.
Therefore,the radical centre is the orthocentre of the triangle.

(D) Let the coaxial system of circles be given by $x^2 + y^2 + 2g_i x + c = 0$ for $i = 1, 2, 3$,where $g_i$ are variables and $c$ is a constant.
Let the fixed point be $(h, k)$. The square of the length of the tangent from $(h, k)$ to the $i$-th circle is $t_i^2 = h^2 + k^2 + 2g_i h + c$.
The centers of the circles are $P(-g_1, 0)$,$Q(-g_2, 0)$,and $R(-g_3, 0)$.
The distances between the centers are $QR = |g_3 - g_2|$,$RP = |g_1 - g_3|$,and $PQ = |g_2 - g_1|$.
Consider the expression $QRt_1^2 + RPt_2^2 + PQt_3^2 = (g_3 - g_2)(h^2 + k^2 + 2g_1 h + c) + (g_1 - g_3)(h^2 + k^2 + 2g_2 h + c) + (g_2 - g_1)(h^2 + k^2 + 2g_3 h + c)$.
Expanding this,we get $(h^2 + k^2 + c)(g_3 - g_2 + g_1 - g_3 + g_2 - g_1) + 2h(g_1 g_3 - g_1 g_2 + g_2 g_1 - g_2 g_3 + g_3 g_2 - g_3 g_1) = 0 + 0 = 0$.

(B) For three co-axial circles with centres $P, Q, R$ and radii $r_1, r_2, r_3$,the radical axis is common.
Let the centres be at $x_1, x_2, x_3$ on the $x$-axis.
The power of a point on the radical axis is constant,so $x_i^2 - r_i^2 = k$ for $i = 1, 2, 3$.
Thus,$r_i^2 = x_i^2 - k$.
The expression $QR r_1^2 + RP r_2^2 + PQ r_3^2$ becomes $(x_3 - x_2)(x_1^2 - k) + (x_1 - x_3)(x_2^2 - k) + (x_2 - x_1)(x_3^2 - k)$.
Since $(x_3 - x_2) + (x_1 - x_3) + (x_2 - x_1) = 0$,the terms involving $k$ cancel out.
We are left with $(x_3 - x_2)x_1^2 + (x_1 - x_3)x_2^2 + (x_2 - x_1)x_3^2$.
This is a cyclic expression equal to $-(x_1 - x_2)(x_2 - x_3)(x_3 - x_1)$.
Given $PQ = |x_1 - x_2|$,$QR = |x_2 - x_3|$,and $RP = |x_3 - x_1|$,the result is $-PQ \cdot QR \cdot RP$.

(A) Let the two circles be ${S_1: x^2 + y^2 + 2gx + 2fy + c = 0}$ and ${S_2: x^2 + y^2 + 2g'x + 2f'y + c' = 0}$.
If ${S_1}$ bisects the circumference of ${S_2}$,then the common chord of the two circles must pass through the center of ${S_2}$.
The equation of the common chord is given by ${S_1 - S_2 = 0}$,which is ${2(g - g')x + 2(f - f')y + (c - c') = 0}$.
The center of ${S_2}$ is ${(-g', -f')}$.
Substituting the center ${(-g', -f')}$ into the equation of the common chord:
${2(g - g')(-g') + 2(f - f')(-f') + (c - c') = 0}$.
Rearranging the terms,we get ${2g'(g - g') + 2f'(f - f') = c - c'}$.

(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 cut orthogonally is $2g_1g_2 + 2f_1f_2 = c_1 + c_2$.
First,rewrite the equations in standard form:
Circle $1$: $x^2 + y^2 + 4x + 6y + 3 = 0$. Here,$2g_1 = 4 \Rightarrow g_1 = 2$,$2f_1 = 6 \Rightarrow f_1 = 3$,and $c_1 = 3$.
Circle $2$: $2(x^2 + y^2) + 6x + 4y + C = 0 \Rightarrow x^2 + y^2 + 3x + 2y + \frac{C}{2} = 0$. Here,$2g_2 = 3 \Rightarrow g_2 = \frac{3}{2}$,$2f_2 = 2 \Rightarrow f_2 = 1$,and $c_2 = \frac{C}{2}$.
Applying the condition: $2(2)(\frac{3}{2}) + 2(3)(1) = 3 + \frac{C}{2}$.
$6 + 6 = 3 + \frac{C}{2}$ $\Rightarrow 12 = 3 + \frac{C}{2}$ $\Rightarrow 9 = \frac{C}{2}$ $\Rightarrow C = 18$.

(A) Let the point of intersection of the two lines be $A$.
The angle subtended by the arc $PQ$ at the centre $C$ is twice the angle subtended by the chord $PQ$ at any point on the circumference (in the alternate segment).
For the line $x + \sqrt{3}y = 1$,the slope is $m_1 = -\frac{1}{\sqrt{3}}$.
For the line $\sqrt{3}x - y = 2$,the slope is $m_2 = \sqrt{3}$.
Since $m_1 \times m_2 = (-\frac{1}{\sqrt{3}}) \times \sqrt{3} = -1$,the two lines are perpendicular to each other.
Therefore,the angle between the lines at their intersection point $A$ is $90^o$.
Since the lines are perpendicular,the chord $PQ$ subtends an angle of $90^o$ at the circumference point $A$. By the circle theorem,the angle subtended by the arc $PQ$ at the centre is $2 \times 90^o = 180^o$.
Thus,the chord $PQ$ is a diameter of the circle.

(D) Given circle: $x^2 + y^2 + 14x + 6y + 2 = 0$.
Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = 7$,$f = 3$,and $c = 2$.
Let the required circle be $x^2 + y^2 + 2g'x + 2f'y + c' = 0$.
The centre of this circle is $(-g', -f') = (0, 2)$,so $g' = 0$ and $f' = -2$.
Since the circles intersect orthogonally,the condition is $2gg' + 2ff' = c + c'$.
Substituting the values: $2(7)(0) + 2(3)(-2) = 2 + c'$.
$0 - 12 = 2 + c' \Rightarrow c' = -14$.
Substituting $g'$,$f'$,and $c'$ into the general equation: $x^2 + y^2 + 2(0)x + 2(-2)y - 14 = 0$.
Thus,the equation is $x^2 + y^2 - 4y - 14 = 0$.

(C) The given circles are $x^2 + y^2 = 4$ and $x^2 + y^2 - 10x + \lambda = 0$.
For the first circle,the center $C_1 = (0, 0)$ and radius $r_1 = \sqrt{4} = 2$.
For the second circle,the center $C_2 = (5, 0)$ and radius $r_2 = \sqrt{5^2 - \lambda} = \sqrt{25 - \lambda}$.
Since the circles touch externally,the distance between their centers must be equal to the sum of their radii: $C_1C_2 = r_1 + r_2$.
The distance $C_1C_2 = \sqrt{(5-0)^2 + (0-0)^2} = 5$.
So,$5 = 2 + \sqrt{25 - \lambda}$.
$3 = \sqrt{25 - \lambda}$.
Squaring both sides,$9 = 25 - \lambda$.
Therefore,$\lambda = 25 - 9 = 16$.

(D) Given the equation of the circle $x^2 + y^2 + 2gx + c = 0$,where $c$ is a constant and $g$ is a parameter.
The centre of the circle is $(-g, 0)$ and the radius is $r = \sqrt{g^2 - c}$.
For a co-axial system,the limiting points are the points where the radius becomes zero,i.e.,$g^2 - c = 0$,which gives $g = \pm \sqrt{c}$.
Since $c > 0$,the limiting points $(\sqrt{c}, 0)$ and $(-\sqrt{c}, 0)$ are real and distinct.
$A$ co-axial system with real and distinct limiting points is known as a non-intersecting type system. However,in the context of standard classification for this specific equation form where $c > 0$,the circles do not intersect each other,thus they are of the non-intersecting type.

(C) 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,rewrite the equations in the form $x^2 + y^2 + 2gx + 2fy + c = 0$ with coefficient $1$ for $x^2$ and $y^2$.
For the first circle: $2x^2 + 2y^2 - 7x = 0 \Rightarrow x^2 + y^2 - 3.5x = 0$.
For the second circle: $x^2 + y^2 - 4y - 7 = 0$.
Subtracting the two equations: $(x^2 + y^2 - 3.5x) - (x^2 + y^2 - 4y - 7) = 0$.
$-3.5x + 4y + 7 = 0$.
Multiply by $-2$ to clear the decimal: $7x - 8y - 14 = 0$.

(B) 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 ${S_1} = {x^2} + {y^2} + 13x - 3y = 0$ and ${S_2} = {x^2} + {y^2} + 2x - 3.5y - 12.5 = 0$ (dividing the second equation by $2$).
The equation is $({x^2} + {y^2} + 13x - 3y) + \lambda ({x^2} + {y^2} + 2x - 3.5y - 12.5) = 0$.
$(1 + \lambda ){x^2} + (1 + \lambda ){y^2} + (13 + 2\lambda )x - (3 + 3.5\lambda )y - 12.5\lambda = 0$.
Dividing by $(1 + \lambda )$,the centre is $\left( { - \frac{{13 + 2\lambda }}{{2(1 + \lambda )}}, \frac{{3 + 3.5\lambda }}{{2(1 + \lambda )}}} \right)$.
Since the centre lies on $13x + 30y = 0$,we have $13\left( { - \frac{{13 + 2\lambda }}{{2(1 + \lambda )}}} \right) + 30\left( { \frac{{3 + 3.5\lambda }}{{2(1 + \lambda )}}} \right) = 0$.
$-169 - 26\lambda + 90 + 105\lambda = 0 \Rightarrow 79\lambda = 79 \Rightarrow \lambda = 1$.
Substituting $\lambda = 1$ into the family equation: $2{x^2} + 2{y^2} + 15x - 6.5y - 12.5 = 0$.
Multiplying by $2$,we get $4{x^2} + 4{y^2} + 30x - 13y - 25 = 0$.

(D) Let the equations of the circles be:
$S_1 \equiv x^2 + y^2 - 16x + 60 = 0$ $(i)$
$S_2 \equiv x^2 + y^2 - 12x + 27 = 0$ $(ii)$
$S_3 \equiv x^2 + y^2 - 12y + 8 = 0$ $(iii)$
The radical axis of circles $(i)$ and $(ii)$ is given by $S_1 - S_2 = 0$:
$(x^2 + y^2 - 16x + 60) - (x^2 + y^2 - 12x + 27) = 0$
$-4x + 33 = 0 \Rightarrow x = \frac{33}{4}$ $(iv)$
The radical axis of circles $(ii)$ and $(iii)$ is given by $S_2 - S_3 = 0$:
$(x^2 + y^2 - 12x + 27) - (x^2 + y^2 - 12y + 8) = 0$
$-12x + 12y + 19 = 0$ $(v)$
Substitute $x = \frac{33}{4}$ into equation $(v)$:
$-12(\frac{33}{4}) + 12y + 19 = 0$
$-99 + 12y + 19 = 0$
$12y = 80 \Rightarrow y = \frac{80}{12} = \frac{20}{3}$
The radical centre is $(\frac{33}{4}, \frac{20}{3})$. Since this is not among the options,the correct answer is $(d)$.

(B) The radical axis of two circles $S_1 = 0$ and $S_2 = 0$ is given by $S_1 - S_2 = 0$.
First,rewrite the equations in the form $x^2 + y^2 + 2gx + 2fy + c = 0$.
The first circle is $x^2 + y^2 - \frac{7}{3}x + \frac{8}{3}y + \frac{11}{3} = 0$.
The second circle is $x^2 + y^2 - 3x - 4y + 5 = 0$.
The radical axis is $\left( -\frac{7}{3} - (-3) \right)x + \left( \frac{8}{3} - (-4) \right)y + \left( \frac{11}{3} - 5 \right) = 0$.
This simplifies to $\left( -\frac{7}{3} + \frac{9}{3} \right)x + \left( \frac{8}{3} + \frac{12}{3} \right)y + \left( \frac{11-15}{3} \right) = 0$.
$\frac{2}{3}x + \frac{20}{3}y - \frac{4}{3} = 0$.
Multiplying by $\frac{3}{2}$,we get $x + 10y - 2 = 0$.

(A) Let the required circle be $x^2 + y^2 + 2gx + 2fy + c = 0$ $... (i)$
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 cut orthogonally is $2g_1g_2 + 2f_1f_2 = c_1 + c_2$.
Applying this to circle $(i)$ and the given circles:
$1$) $2g(1) + 2f(8.5) = c + 4 \implies 2g + 17f = c + 4$ $... (ii)$
$2$) $2g(3.5) + 2f(3) = c + 11 \implies 7g + 6f = c + 11$ $... (iii)$
$3$) $2g(-0.5) + 2f(11) = c + 3 \implies -g + 22f = c + 3$ $... (iv)$
Subtracting $(iii)$ from $(ii)$: $(2g - 7g) + (17f - 6f) = (c - c) + (4 - 11) \implies -5g + 11f = -7$ $... (v)$
Subtracting $(iv)$ from $(iii)$: $(7g - (-g)) + (6f - 22f) = (c - c) + (11 - 3) \implies 8g - 16f = 8 \implies g - 2f = 1 \implies g = 2f + 1$ $... (vi)$
Substituting $(vi)$ into $(v)$: $-5(2f + 1) + 11f = -7 \implies -10f - 5 + 11f = -7 \implies f = -2$.
Substituting $f = -2$ into $(vi)$: $g = 2(-2) + 1 = -3$.
The centre of the circle is $(-g, -f) = (-(-3), -(-2)) = (3, 2)$.

(A) Given circles are:
${S_1} = {x^2} + {y^2} - 3x - 4y + 5 = 0$ .....$(i)$
For the second circle,divide by $2$ to make the coefficients of ${x^2}$ and ${y^2}$ equal to $1$:
${S_2} = {x^2} + {y^2} - 5x - 6y + 6 = 0$ .....$(ii)$
The equation of the radical axis is given by ${S_1} - {S_2} = 0$.
Subtracting $(ii)$ from $(i)$:
$({x^2} + {y^2} - 3x - 4y + 5) - ({x^2} + {y^2} - 5x - 6y + 6) = 0$
$(-3x + 5x) + (-4y + 6y) + (5 - 6) = 0$
$2x + 2y - 1 = 0$.

(A) Given equations of circles are:
$x^2 + y^2 = 25$ $(i)$
$x^2 + y^2 - 8x + 7 = 0$ $(ii)$
Subtracting equation $(ii)$ from equation $(i)$:
$(x^2 + y^2) - (x^2 + y^2 - 8x + 7) = 25 - 0$
$8x - 7 = 25$
$8x = 32$
$x = 4$
Substituting $x = 4$ in equation $(i)$:
$(4)^2 + y^2 = 25$
$16 + y^2 = 25$
$y^2 = 9$
$y = \pm 3$
Thus,the points of intersection are $(4, 3)$ and $(4, -3)$.

(D) The centers of the circles are $C_1(-a, 0)$ and $C_2(0, -b)$.
The radii of the circles are $R_1 = \sqrt{a^2 - c}$ and $R_2 = \sqrt{b^2 - c}$.
The distance between the centers is $C_1C_2 = \sqrt{(-a - 0)^2 + (0 - (-b))^2} = \sqrt{a^2 + b^2}$.
Since the circles touch each other,the distance between their centers must be equal to the sum of their radii: $R_1 + R_2 = C_1C_2$.
$\sqrt{a^2 - c} + \sqrt{b^2 - c} = \sqrt{a^2 + b^2}$.
Squaring both sides: $(a^2 - c) + (b^2 - c) + 2\sqrt{(a^2 - c)(b^2 - c)} = a^2 + b^2$.
$a^2 + b^2 - 2c + 2\sqrt{(a^2 - c)(b^2 - c)} = a^2 + b^2$.
$2\sqrt{(a^2 - c)(b^2 - c)} = 2c$.
$\sqrt{(a^2 - c)(b^2 - c)} = c$.
Squaring again: $(a^2 - c)(b^2 - c) = c^2$.
$a^2b^2 - a^2c - b^2c + c^2 = c^2$.
$a^2b^2 = a^2c + b^2c$.
Dividing both sides by $a^2b^2c$: $\frac{a^2b^2}{a^2b^2c} = \frac{a^2c}{a^2b^2c} + \frac{b^2c}{a^2b^2c}$.
$\frac{1}{c} = \frac{1}{b^2} + \frac{1}{a^2}$.

(A) When the distance between the centers of two circles is equal to the sum of their radii,i.e.,$d = r_1 + r_2$,the circles touch each other externally at a single point.