Chord of contact of tangent and common chord Questions in English

(B) The equation of the circle is $x^2 + y^2 - 6x - 8y + 21 = 0$. The center is $C(3, 4)$ and the radius $r = \sqrt{3^2 + 4^2 - 21} = \sqrt{9 + 16 - 21} = \sqrt{4} = 2$.
The equation of the chord of contact $AB$ from the origin $O(0, 0)$ is given by $T = 0$:
$x(0) + y(0) - 3(x + 0) - 4(y + 0) + 21 = 0$
$3x + 4y - 21 = 0$ ... $(i)$
Let $M$ be the intersection of $OC$ and $AB$. $CM$ is the perpendicular distance from $C(3, 4)$ to the line $3x + 4y - 21 = 0$:
$CM = \frac{|3(3) + 4(4) - 21|}{\sqrt{3^2 + 4^2}} = \frac{|9 + 16 - 21|}{\sqrt{25}} = \frac{4}{5}$
In $\triangle AMC$,$\angle AMC = 90^\circ$. By Pythagoras theorem:
$AM = \sqrt{AC^2 - CM^2} = \sqrt{2^2 - (\frac{4}{5})^2} = \sqrt{4 - \frac{16}{25}} = \sqrt{\frac{100 - 16}{25}} = \sqrt{\frac{84}{25}} = \frac{2\sqrt{21}}{5}$
Since $AB = 2AM$,we have:
$AB = 2 \times \frac{2\sqrt{21}}{5} = \frac{4}{5}\sqrt{21}$

(D) The equation of the pair of tangents from a point $(x_1, y_1)$ to a circle $S = 0$ is given by $SS_1 = T^2$.
Here,the point is the origin $(0, 0)$,so $x_1 = 0$ and $y_1 = 0$.
$S = x^2 + y^2 + 2gx + 2fy + c = 0$
$S_1 = 0^2 + 0^2 + 2g(0) + 2f(0) + c = c$
$T = x(0) + y(0) + g(x + 0) + f(y + 0) + c = gx + fy + c$
Substituting these into $SS_1 = T^2$:
$c(x^2 + y^2 + 2gx + 2fy + c) = (gx + fy + c)^2$
$c(x^2 + y^2) + 2gcx + 2fcy + c^2 = (gx + fy)^2 + 2c(gx + fy) + c^2$
$c(x^2 + y^2) + 2gcx + 2fcy + c^2 = (gx + fy)^2 + 2gcx + 2fcy + c^2$
$c(x^2 + y^2) = (gx + fy)^2$.

(A) Let the point be $P(h, k)$ and the circle be $x^2 + y^2 = a^2$. The equation of the chord of contact $AB$ is $xh + yk = a^2$.
The length of the perpendicular $OM$ from the center $O(0, 0)$ to the chord $AB$ is $OM = \frac{a^2}{\sqrt{h^2 + k^2}}$.
In the right-angled triangle $OAM$,$AM = \sqrt{OA^2 - OM^2} = \sqrt{a^2 - \frac{a^4}{h^2 + k^2}} = \frac{a\sqrt{h^2 + k^2 - a^2}}{\sqrt{h^2 + k^2}}$.
Since $AB = 2AM$,we have $AB = \frac{2a\sqrt{h^2 + k^2 - a^2}}{\sqrt{h^2 + k^2}}$.
The length of the perpendicular $PM$ from $P(h, k)$ to the chord $AB$ is $PM = OP - OM = \sqrt{h^2 + k^2} - \frac{a^2}{\sqrt{h^2 + k^2}} = \frac{h^2 + k^2 - a^2}{\sqrt{h^2 + k^2}}$.
The area of triangle $PAB$ is $\frac{1}{2} \times AB \times PM = \frac{1}{2} \times \left( \frac{2a\sqrt{h^2 + k^2 - a^2}}{\sqrt{h^2 + k^2}} \right) \times \left( \frac{h^2 + k^2 - a^2}{\sqrt{h^2 + k^2}} \right) = \frac{a(h^2 + k^2 - a^2)^{3/2}}{h^2 + k^2}$.

(A) Let the point of intersection of the tangents be $P(h, k)$.
The equation of the chord of contact of the tangents from $P(h, k)$ to the circle $x^2 + y^2 = a^2$ is given by $hx + ky = a^2$,or $hx + ky - a^2 = 0$.
This chord touches the circle $x^2 + y^2 - 2ax = 0$,which has center $(a, 0)$ and radius $a$.
The perpendicular distance from the center $(a, 0)$ to the line $hx + ky - a^2 = 0$ must be equal to the radius $a$:
$\frac{|h(a) + k(0) - a^2|}{\sqrt{h^2 + k^2}} = a$
$|ah - a^2| = a\sqrt{h^2 + k^2}$
$|h - a| = \sqrt{h^2 + k^2}$
Squaring both sides:
$(h - a)^2 = h^2 + k^2$
$h^2 - 2ah + a^2 = h^2 + k^2$
$k^2 = a^2 - 2ah = a(a - 2h)$
Replacing $(h, k)$ with $(x, y)$,the locus is $y^2 = a(a - 2x)$.

(C) The length of the tangent from a point $P(\alpha, \beta)$ to a circle $S = 0$ is given by $\sqrt{S(\alpha, \beta)}$.
For the first circle $S_1: x^2 + y^2 - 4x - 5 = 0$,the length of the tangent is $\sqrt{\alpha^2 + \beta^2 - 4\alpha - 5}$.
For the second circle $S_2: x^2 + y^2 + 6x - 2y + 6 = 0$,the length of the tangent is $\sqrt{\alpha^2 + \beta^2 + 6\alpha - 2\beta + 6}$.
Since the lengths of the tangents are equal,we have:
$\alpha^2 + \beta^2 - 4\alpha - 5 = \alpha^2 + \beta^2 + 6\alpha - 2\beta + 6$
Subtracting $\alpha^2 + \beta^2$ from both sides:
$-4\alpha - 5 = 6\alpha - 2\beta + 6$
Rearranging the terms to one side:
$6\alpha + 4\alpha - 2\beta + 6 + 5 = 0$
$10\alpha - 2\beta + 11 = 0$.

(B) The given circles are $S_1: x^2 + y^2 - 9 = 0$ and $S_2: x^2 + y^2 - 12y + 27 = 0$.
The equation of the common tangent for two touching circles is given by $S_1 - S_2 = 0$.
$(x^2 + y^2 - 9) - (x^2 + y^2 - 12y + 27) = 0$
$x^2 + y^2 - 9 - x^2 - y^2 + 12y - 27 = 0$
$12y - 36 = 0$
$12y = 36$
$y = 3$.

(A) Let $S_1 \equiv x^2 + y^2 - 2x + 6y + 6 = 0$ and $S_2 \equiv x^2 + y^2 - 5x + 6y + 15 = 0$.
Since the two circles touch each other,their common tangent is given by the radical axis equation $S_1 - S_2 = 0$.
Subtracting the two equations:
$(x^2 + y^2 - 2x + 6y + 6) - (x^2 + y^2 - 5x + 6y + 15) = 0$
$(-2x + 5x) + (6y - 6y) + (6 - 15) = 0$
$3x - 9 = 0$
$3x = 9$
$x = 3$.
Thus,the equation of the common tangent is $x = 3$.

(A) The equation of the first circle is $S_1 = x^2 + y^2 + 4x + 1 = 0$.
The equation of the second circle is $S_2 = x^2 + y^2 + 6x + 2y + 3 = 0$.
The equation of the common chord is given by $S_1 - S_2 = 0$.
$(x^2 + y^2 + 4x + 1) - (x^2 + y^2 + 6x + 2y + 3) = 0$.
$(4x - 6x) + (0 - 2y) + (1 - 3) = 0$.
$-2x - 2y - 2 = 0$.
Dividing by $-2$,we get $x + y + 1 = 0$.

(D) The equation of the circle is $x^2 + y^2 + x - y - 1 = 0$.
Comparing this with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = 1/2$,$f = -1/2$,and $c = -1$.
The center of the circle is $C(-g, -f) = (-1/2, 1/2)$.
The radius $r$ of the circle is $\sqrt{g^2 + f^2 - c} = \sqrt{(1/2)^2 + (-1/2)^2 - (-1)} = \sqrt{1/4 + 1/4 + 1} = \sqrt{3/2}$.
Let $M(1, 1)$ be the midpoint of the chord. The distance $d$ from the center $C(-1/2, 1/2)$ to the midpoint $M(1, 1)$ is given by the distance formula:
$d = \sqrt{(1 - (-1/2))^2 + (1 - 1/2)^2} = \sqrt{(3/2)^2 + (1/2)^2} = \sqrt{9/4 + 1/4} = \sqrt{10/4} = \sqrt{5/2}$.
For a chord to exist,the distance $d$ from the center to the midpoint must be less than the radius $r$ of the circle.
Here,$d = \sqrt{5/2} \approx 1.58$ and $r = \sqrt{3/2} \approx 1.22$.
Since $d > r$,the point $(1, 1)$ lies outside the circle.
Therefore,no such chord exists.

(B) The circle has radius $a$ and its diameter lies along the $x$-axis with one end at the origin $(0, 0)$. Thus,the center is $(a, 0)$ and the other end of the diameter is $(2a, 0)$.
The equation of the circle is $(x - a)^2 + y^2 = a^2$,which simplifies to $x^2 + y^2 - 2ax = 0$.
The chord $y = mx$ intersects the circle at the origin $O(0, 0)$ and point $B$. Substituting $y = mx$ into the circle equation:
$x^2 + (mx)^2 - 2ax = 0$
$x^2(1 + m^2) - 2ax = 0$
$x(x(1 + m^2) - 2a) = 0$
So,$x = 0$ or $x = \frac{2a}{1 + m^2}$.
For $x = \frac{2a}{1 + m^2}$,$y = m \left( \frac{2a}{1 + m^2} \right) = \frac{2am}{1 + m^2}$.
Thus,$B = \left( \frac{2a}{1 + m^2}, \frac{2am}{1 + m^2} \right)$.
The equation of a circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$.
Using $O(0, 0)$ and $B\left( \frac{2a}{1 + m^2}, \frac{2am}{1 + m^2} \right)$:
$x\left( x - \frac{2a}{1 + m^2} \right) + y\left( y - \frac{2am}{1 + m^2} \right) = 0$
$x^2 + y^2 - \frac{2ax}{1 + m^2} - \frac{2amy}{1 + m^2} = 0$
$(1 + m^2)(x^2 + y^2) - 2a(x + my) = 0$.

(C) Let the mid-point of the chord be $C(h, k)$.
Since the chord subtends a right angle at the origin $O(0, 0)$,the triangle formed by the origin and the endpoints of the chord is an isosceles right-angled triangle.
The distance from the origin to the midpoint $C(h, k)$ is $d = \sqrt{h^2 + k^2}$.
In the right-angled triangle formed by the origin,the midpoint,and one endpoint of the chord,the angle at the origin is $45^\circ$.
Using trigonometry,$\cos(45^\circ) = \frac{d}{r}$,where $r$ is the radius of the circle.
Given $r = 2$,we have $\frac{1}{\sqrt{2}} = \frac{\sqrt{h^2 + k^2}}{2}$.
Squaring both sides,$\frac{1}{2} = \frac{h^2 + k^2}{4}$,which simplifies to $h^2 + k^2 = 2$.
Replacing $(h, k)$ with $(x, y)$,the locus is $x^2 + y^2 = 2$.

(C) The equation of the circle is ${x^2} + {y^2} = 3$,so the radius $r = \sqrt{3}$ and the center is $C(0, 0)$.
Let the chord be $AB$ with length $2$. The perpendicular distance $d$ from the center $C(0, 0)$ to the chord $x - 2y - k = 0$ bisects the chord.
Thus,the distance $d = \sqrt{r^2 - (\text{half-chord length})^2} = \sqrt{(\sqrt{3})^2 - (1)^2} = \sqrt{3 - 1} = \sqrt{2}$.
The perpendicular distance from $(0, 0)$ to $x - 2y - k = 0$ is given by $d = \frac{|0 - 2(0) - k|}{\sqrt{1^2 + (-2)^2}} = \frac{|-k|}{\sqrt{5}}$.
Equating the two expressions for $d$: $\frac{|k|}{\sqrt{5}} = \sqrt{2}$.
$|k| = \sqrt{10}$,which implies $k = \pm \sqrt{10}$.

(A) 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:
$S_1: (x - a)^2 + (y - b)^2 - c^2 = 0$
$S_2: (x - b)^2 + (y - a)^2 - c^2 = 0$
Subtracting $S_2$ from $S_1$:
$((x - a)^2 - (x - b)^2) + ((y - b)^2 - (y - a)^2) = 0$
Expanding the terms:
$(x^2 - 2ax + a^2 - (x^2 - 2bx + b^2)) + (y^2 - 2by + b^2 - (y^2 - 2ay + a^2)) = 0$
$(-2ax + a^2 + 2bx - b^2) + (-2by + b^2 + 2ay - a^2) = 0$
Simplifying the expression:
$2bx - 2ax + 2ay - 2by = 0$
$2(b - a)x - 2(b - a)y = 0$
Dividing by $2(b - a)$ (assuming $a \neq b$):
$x - y = 0$

(C) The equations of the circles are $S_1: x^2 - 2ax + a^2 + y^2 = a^2 \implies x^2 + y^2 - 2ax = 0$ and $S_2: x^2 + y^2 - 2by + b^2 = b^2 \implies x^2 + y^2 - 2by = 0$.
The equation of the common chord is given by $S_1 - S_2 = 0$,which is $-2ax + 2by = 0$,or $ax - by = 0$.
The center of the first circle is $(a, 0)$ and its radius is $r_1 = a$.
The perpendicular distance $p$ from the center $(a, 0)$ to the line $ax - by = 0$ is $p = \frac{|a(a) - b(0)|}{\sqrt{a^2 + (-b)^2}} = \frac{a^2}{\sqrt{a^2 + b^2}}$.
The length of the common chord is $2\sqrt{r_1^2 - p^2} = 2\sqrt{a^2 - \frac{a^4}{a^2 + b^2}}$.
$= 2\sqrt{\frac{a^2(a^2 + b^2) - a^4}{a^2 + b^2}} = 2\sqrt{\frac{a^4 + a^2b^2 - a^4}{a^2 + b^2}} = 2\sqrt{\frac{a^2b^2}{a^2 + b^2}} = \frac{2ab}{\sqrt{a^2 + b^2}}$.

(A) The equations of the circles are $S_1: x^2 + y^2 - 12 = 0$ and $S_2: x^2 + y^2 - 4x + 3y - 2 = 0$.
The equation of the common chord is given by $S_1 - S_2 = 0$.
$(x^2 + y^2 - 12) - (x^2 + y^2 - 4x + 3y - 2) = 0$
$4x - 3y - 10 = 0$.
The center of the first circle $S_1$ is $C_1(0, 0)$ and its radius $R_1 = \sqrt{12} = 2\sqrt{3}$.
The perpendicular distance $p_1$ from the center $(0, 0)$ to the line $4x - 3y - 10 = 0$ is:
$p_1 = \frac{|4(0) - 3(0) - 10|}{\sqrt{4^2 + (-3)^2}} = \frac{|-10|}{\sqrt{16 + 9}} = \frac{10}{5} = 2$.
The length of the common chord is $2\sqrt{R_1^2 - p_1^2}$.
Length $= 2\sqrt{12 - 2^2} = 2\sqrt{12 - 4} = 2\sqrt{8} = 2(2\sqrt{2}) = 4\sqrt{2}$.

(A) The equations of the circles are:
$S_1: (x - a)^2 + (y - b)^2 = c^2$
$S_2: (x - b)^2 + (y - a)^2 = c^2$
Subtracting $S_2$ from $S_1$ gives the equation of the common chord:
$(x - a)^2 - (x - b)^2 + (y - b)^2 - (y - a)^2 = 0$
$(x^2 - 2ax + a^2) - (x^2 - 2bx + b^2) + (y^2 - 2by + b^2) - (y^2 - 2ay + a^2) = 0$
$-2ax + 2bx - 2by + 2ay = 0$
$-2a(x - y) + 2b(x - y) = 0$
$(2b - 2a)(x - y) = 0$
Since $a \neq b$,we have $x - y = 0$,or $y = x$.
Substitute $y = x$ into the first circle equation:
$(x - a)^2 + (x - b)^2 = c^2$
$x^2 - 2ax + a^2 + x^2 - 2bx + b^2 = c^2$
$2x^2 - 2(a + b)x + (a^2 + b^2 - c^2) = 0$
The distance $d$ from the center $(a, b)$ to the line $x - y = 0$ is:
$d = \frac{|a - b|}{\sqrt{1^2 + (-1)^2}} = \frac{|a - b|}{\sqrt{2}}$
The length of the common chord is $2\sqrt{r^2 - d^2}$:
$= 2\sqrt{c^2 - \frac{(a - b)^2}{2}}$
$= 2\sqrt{\frac{2c^2 - (a - b)^2}{2}}$
$= \sqrt{2(2c^2 - (a - b)^2)}$
$= \sqrt{4c^2 - 2(a - b)^2}$

(A) The equation of the chord of contact from the origin $(0, 0)$ to the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is $gx + fy + c = 0$.
The equation of the chord of contact from the point $(g, f)$ is $gx + fy + g(x + g) + f(y + f) + c = 0$,which simplifies to $2gx + 2fy + g^2 + f^2 + c = 0$,or $gx + fy + \frac{g^2 + f^2 + c}{2} = 0$.
The distance $d$ between two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is given by $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$.
Here,$A = g$,$B = f$,$C_1 = c$,and $C_2 = \frac{g^2 + f^2 + c}{2}$.
Thus,$d = \frac{|c - \frac{g^2 + f^2 + c}{2}|}{\sqrt{g^2 + f^2}} = \frac{|\frac{2c - g^2 - f^2 - c}{2}|}{\sqrt{g^2 + f^2}} = \frac{|\frac{c - g^2 - f^2}{2}|}{\sqrt{g^2 + f^2}} = \frac{1}{2} \left( \frac{g^2 + f^2 - c}{\sqrt{g^2 + f^2}} \right)$.

(B) Let the coordinates of point $R$ be $(h, k)$.
The equation of the chord of contact of tangents drawn from $R(h, k)$ to the circle $x^2 + y^2 = a^2$ is given by $hx + ky = a^2$,or $hx + ky - a^2 = 0$.
It is given that this chord of contact is the line $lx + my + n = 0$.
Since both equations represent the same line,their coefficients must be proportional:
$\frac{h}{l} = \frac{k}{m} = \frac{-a^2}{n}$.
From this,we get:
$h = \frac{-a^2l}{n}$ and $k = \frac{-a^2m}{n}$.
Therefore,the coordinates of $R$ are $\left( \frac{-a^2l}{n}, \frac{-a^2m}{n} \right)$.

(B) The equation of the given circle is $S \equiv x^2 + y^2 - 2x + 4y + 1 = 0$. The center is $(1, -2)$ and the radius is $\sqrt{1^2 + (-2)^2 - 1} = 2$.
The chord of contact $BC$ for the point $A(0, 1)$ is given by $T = 0$:
$x(0) + y(1) - (x + 0) + 2(y + 1) + 1 = 0$
$y - x + 2y + 2 + 1 = 0 \implies -x + 3y + 3 = 0$.
The equation of any circle passing through the intersection of the circle $S=0$ and the line $BC=0$ is $S + \lambda(BC) = 0$:
$(x^2 + y^2 - 2x + 4y + 1) + \lambda(-x + 3y + 3) = 0$.
Since this circle passes through $A(0, 1)$,we substitute $x=0, y=1$:
$(0 + 1 - 0 + 4 + 1) + \lambda(0 + 3 + 3) = 0$
$6 + 6\lambda = 0 \implies \lambda = -1$.
Substituting $\lambda = -1$ into the equation:
$(x^2 + y^2 - 2x + 4y + 1) - (-x + 3y + 3) = 0$
$x^2 + y^2 - 2x + 4y + 1 + x - 3y - 3 = 0$
$x^2 + y^2 - x + y - 2 = 0$.

(D) The equation of the common chord is given by ${S_1} - {S_2} = 0$.
Subtracting the two equations: $({x^2} + {y^2} + 5x + 7y + 9) - ({x^2} + {y^2} + 7x + 5y + 9) = 0$.
This simplifies to $-2x + 2y = 0$,or $x - y = 0$.
For the circle ${x^2} + {y^2} + 5x + 7y + 9 = 0$,the center ${C_1}$ is $(-\frac{5}{2}, -\frac{7}{2})$ and the radius $r$ is $\sqrt{(-\frac{5}{2})^2 + (-\frac{7}{2})^2 - 9} = \sqrt{\frac{25}{4} + \frac{49}{4} - \frac{36}{4}} = \sqrt{\frac{38}{4}} = \sqrt{\frac{19}{2}}$.
The perpendicular distance $d$ from the center ${C_1}$ to the line $x - y = 0$ is $d = \frac{|-\frac{5}{2} - (-\frac{7}{2})|}{\sqrt{1^2 + (-1)^2}} = \frac{|1|}{\sqrt{2}} = \frac{1}{\sqrt{2}}$.
The length of the common chord is $2\sqrt{r^2 - d^2} = 2\sqrt{\frac{19}{2} - \frac{1}{2}} = 2\sqrt{\frac{18}{2}} = 2\sqrt{9} = 2 \times 3 = 6$.

(B) The perpendicular distance $d$ from the center $(0, 0)$ of the circle to the line $mx - y + c = 0$ is given by $d = \left| \frac{c}{\sqrt{1 + m^2}} \right|$.
In the right-angled triangle formed by the radius $a$,the distance $d$,and half the chord length $b$,we have $a^2 = d^2 + b^2$.
Thus,$d^2 = a^2 - b^2$.
Substituting the value of $d$,we get $\frac{c^2}{1 + m^2} = a^2 - b^2$.
Therefore,$(1 + m^2)(a^2 - b^2) = c^2$.

(B) The equation of the common chord is obtained by subtracting the two circle equations: $(x^2 + y^2 + 4x + 3y + 2) - (x^2 + y^2 + 2x + 3y + 1) = 0$.
This simplifies to $2x + 1 = 0$,or $x = -1/2$.
For the circle $x^2 + y^2 + 2x + 3y + 1 = 0$,the center $C_1$ is $(-1, -3/2)$ and the radius $r_1$ is $\sqrt{(-1)^2 + (-3/2)^2 - 1} = \sqrt{1 + 9/4 - 1} = 3/2$.
The perpendicular distance $d$ from the center $C_1(-1, -3/2)$ to the line $x + 1/2 = 0$ is $d = |-1 - (-1/2)| = |-1/2| = 1/2$.
The half-length of the common chord is $PM = \sqrt{r_1^2 - d^2} = \sqrt{(3/2)^2 - (1/2)^2} = \sqrt{9/4 - 1/4} = \sqrt{8/4} = \sqrt{2}$.
Therefore,the total length of the common chord $PQ = 2 \times PM = 2\sqrt{2}$.

(B) The given line is $\frac{x}{3} + \frac{y}{4} = 1$,which can be written as $4x + 3y - 12 = 0$.
The circle is ${x^2} + {y^2} = \frac{169}{25}$,so the radius $r = \sqrt{\frac{169}{25}} = \frac{13}{5}$.
The perpendicular distance $d$ from the center $(0, 0)$ to the line $4x + 3y - 12 = 0$ is given by $d = \frac{|4(0) + 3(0) - 12|}{\sqrt{4^2 + 3^2}} = \frac{12}{5}$.
The length of the chord is $2\sqrt{r^2 - d^2} = 2\sqrt{(\frac{13}{5})^2 - (\frac{12}{5})^2} = 2\sqrt{\frac{169 - 144}{25}} = 2\sqrt{\frac{25}{25}} = 2(1) = 2$.

(D) The equations of the circles are:
${S_1 = x^2 + y^2 + 2x - 3y + 6 = 0}$ .....$(i)$
${S_2 = x^2 + y^2 + x - 8y - 13 = 0}$ .....$(ii)$
The equation of the common chord is given by ${S_1 - S_2 = 0}$.
Subtracting equation $(ii)$ from $(i)$:
${(x^2 + y^2 + 2x - 3y + 6) - (x^2 + y^2 + x - 8y - 13) = 0}$
${x + 5y + 19 = 0}$ .....$(iii)$
Now,we check which of the given points satisfies the equation ${x + 5y + 19 = 0}$:
For $(1, -4)$:
${1 + 5(-4) + 19 = 1 - 20 + 19 = 0}$.
Thus,the point $(1, -4)$ lies on the common chord.

(A) The equation of the chord of contact of tangents drawn from a point $(x_1, y_1)$ to the circle $x^2 + y^2 = a^2$ is given by $xx_1 + yy_1 = a^2$.
Given the point $(x_1, y_1) = (5, -3)$ and the circle equation $x^2 + y^2 = 10$,where $a^2 = 10$.
Substituting these values into the formula,we get:
$x(5) + y(-3) = 10$
$5x - 3y = 10$.

(C) Given circle is $x^2 + y^2 - 2x - 6y + 6 = 0$.
Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -1, f = -3, c = 6$.
The centre of this circle is $C(-g, -f) = (1, 3)$ and its radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{(-1)^2 + (-3)^2 - 6} = \sqrt{1 + 9 - 6} = \sqrt{4} = 2$.
Let the required circle have centre $D(2, 1)$.
Since a chord of the required circle is a diameter of the given circle,the length of this chord is equal to the diameter of the given circle,which is $2 \times r = 2 \times 2 = 4$.
Let $AB$ be the chord of the required circle,which is a diameter of the given circle. The length of $AB = 4$,so $AC = CB = 2$.
In the right-angled triangle $\triangle ACD$,$AD$ is the radius of the required circle $(R)$.
$AD^2 = AC^2 + CD^2$.
$CD$ is the distance between $D(2, 1)$ and $C(1, 3)$.
$CD = \sqrt{(2 - 1)^2 + (1 - 3)^2} = \sqrt{1^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5}$.
$R^2 = 2^2 + (\sqrt{5})^2 = 4 + 5 = 9$.
$R = 3$.

(D) Given,the circle is ${x^2} + {y^2} - 2x = 0$ $(i)$ and the line is $y = x$ $(ii)$.
Substituting $y = x$ in $(i)$,we get:
${x^2} + {x^2} - 2x = 0$
$2{x^2} - 2x = 0$
$2x(x - 1) = 0$
So,$x = 0$ or $x = 1$.
Since $y = x$,the points of intersection are $A = (0, 0)$ and $B = (1, 1)$.
The equation of a circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$.
Substituting $A(0, 0)$ and $B(1, 1)$:
$(x - 0)(x - 1) + (y - 0)(y - 1) = 0$
$x(x - 1) + y(y - 1) = 0$
${x^2} - x + {y^2} - y = 0$
${x^2} + {y^2} - x - y = 0$.

(A) The equation of the common chord of two circles $S_1 = 0$ and $S_2 = 0$ is given by $S_1 - S_2 = 0$.
First,write the equations in the form $x^2 + y^2 + 2gx + 2fy + c = 0$.
The first circle is $3x^2 + 3y^2 - 2x + 12y - 9 = 0$. Dividing by $3$,we get $x^2 + y^2 - \frac{2}{3}x + 4y - 3 = 0$ $(S_1)$.
The second circle is $x^2 + y^2 + 6x + 2y - 15 = 0$ $(S_2)$.
The common chord is $S_1 - S_2 = 0$:
$(x^2 + y^2 - \frac{2}{3}x + 4y - 3) - (x^2 + y^2 + 6x + 2y - 15) = 0$
$(-\frac{2}{3} - 6)x + (4 - 2)y + (-3 + 15) = 0$
$-\frac{20}{3}x + 2y + 12 = 0$
Multiplying by $-3$,we get $20x - 6y - 36 = 0$,which simplifies to $10x - 3y - 18 = 0$.

(D) The equation of the common chord $PQ$ is obtained by subtracting the two circle equations: $(x^2 + y^2 + 2ax + cy + a) - (x^2 + y^2 - 3ax + dy - 1) = 0$.
This simplifies to $5ax + (c - d)y + (a + 1) = 0$.....$(i)$
The given equation of the line passing through $P$ and $Q$ is $5x + by - a = 0$.....$(ii)$
Since both equations represent the same line,their coefficients must be proportional:
$\frac{5a}{5} = \frac{c - d}{b} = \frac{a + 1}{-a}$
From the first and third parts: $a = \frac{a + 1}{-a}$
$-a^2 = a + 1$
$a^2 + a + 1 = 0$
For the quadratic equation $a^2 + a + 1 = 0$,the discriminant $D = b^2 - 4ac = 1^2 - 4(1)(1) = -3$.
Since $D < 0$,there are no real values of $a$ that satisfy this condition.

(D) The equation of the circle is $x^2 + y^2 = 25$.
For a point $(x_1, y_1) = (3, 2)$,we check its position relative to the circle by calculating $S_1 = x_1^2 + y_1^2 - 25$.
$S_1 = 3^2 + 2^2 - 25 = 9 + 4 - 25 = -12$.
Since $S_1 < 0$,the point $(3, 2)$ lies inside the circle.
$A$ chord of contact is only defined for points lying outside the circle.
Therefore,no such chord of contact exists,and consequently,triangle $OAB$ cannot be formed.

(C) Let the point be $P(4, 3)$ and the circle be $x^2 + y^2 = 9$. The radius $r = 3$.
The equation of the chord of contact $AB$ is $T = 0$,which is $4x + 3y = 9$.
The distance from the origin $O(0, 0)$ to the chord $AB$ is $OQ = \frac{|4(0) + 3(0) - 9|}{\sqrt{4^2 + 3^2}} = \frac{9}{5}$.
In $\triangle OAQ$,$AQ = \sqrt{OA^2 - OQ^2} = \sqrt{3^2 - (\frac{9}{5})^2} = \sqrt{9 - \frac{81}{25}} = \sqrt{\frac{225 - 81}{25}} = \sqrt{\frac{144}{25}} = \frac{12}{5}$.
The length of the chord of contact $AB = 2 \times AQ = 2 \times \frac{12}{5} = \frac{24}{5}$.
The distance $PQ$ is the distance from $P(4, 3)$ to the line $4x + 3y - 9 = 0$,which is $PQ = \frac{|4(4) + 3(3) - 9|}{\sqrt{4^2 + 3^2}} = \frac{|16 + 9 - 9|}{5} = \frac{16}{5}$.
The area of $\triangle PAB = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AB \times PQ = \frac{1}{2} \times \frac{24}{5} \times \frac{16}{5} = \frac{192}{25}$.

(A) Let $(h, k)$ be the coordinates of the centre of circle ${C_2}$.
Its equation is $(x - h)^2 + (y - k)^2 = 5^2$.
The equation of ${C_1}$ is $x^2 + y^2 = 4^2$. The equation of the common chord of ${C_1}$ and ${C_2}$ is $2hx + 2ky = h^2 + k^2 - 9$ $(i)$.
Let $p$ be the length of the perpendicular from the centre $(0, 0)$ of ${C_1}$ to $(i)$.
Then $p = \frac{|h^2 + k^2 - 9|}{\sqrt{4h^2 + 4k^2}}$.
The length of the common chord is $2\sqrt{4^2 - p^2}$,which is of maximum length if $p = 0$.
This implies $h^2 + k^2 - 9 = 0$ $(ii)$.
The slope of $(i)$ is given as $\frac{3}{4}$.
Thus,$-\frac{h}{k} = \frac{3}{4} \Rightarrow k = -\frac{4h}{3}$ $(iii)$.
Substituting $(iii)$ into $(ii)$,we get $h^2 + (-\frac{4h}{3})^2 = 9$ $\Rightarrow h^2 + \frac{16h^2}{9} = 9$ $\Rightarrow \frac{25h^2}{9} = 9$ $\Rightarrow h^2 = \frac{81}{25}$ $\Rightarrow h = \pm \frac{9}{5}$.
If $h = \frac{9}{5}$,then $k = -\frac{4}{3} \times \frac{9}{5} = -\frac{12}{5}$.
If $h = -\frac{9}{5}$,then $k = -\frac{4}{3} \times (-\frac{9}{5}) = \frac{12}{5}$.
Thus,the centres are $\left( -\frac{9}{5}, \frac{12}{5} \right)$ and $\left( \frac{9}{5}, -\frac{12}{5} \right)$.

(D) The equation of the common chord (line $L$) of two circles ${S_1} = 0$ and ${S_2} = 0$ is given by ${S_1} - {S_2} = 0$.
Given circles are:
${S_1}: {x^2} + {y^2} - 25 = 0$
${S_2}: {x^2} + {y^2} - 8x + 7 = 0$
Subtracting the two equations:
$({x^2} + {y^2} - 25) - ({x^2} + {y^2} - 8x + 7) = 0$
$8x - 32 = 0$
$x = 4$
This is the equation of the line $L$.
The second circle is ${x^2} + {y^2} - 8x + 7 = 0$. Comparing this with the general form ${x^2} + {y^2} + 2gx + 2fy + c = 0$,we get $2g = -8$,so $g = -4$,and $f = 0$.
The centre of the second circle is $(-g, -f) = (4, 0)$.
The length of the perpendicular from the point $(4, 0)$ to the line $x - 4 = 0$ is given by:
$d = \frac{|4 - 4|}{\sqrt{1^2 + 0^2}} = \frac{0}{1} = 0$.
Thus,the length of the perpendicular is $0$.

(D) The equation of the pair of tangents from a point $(x_1, y_1)$ to a circle $S = 0$ is given by $SS_1 = T^2$.
Here,$S = x^2 + y^2 - 2x + 4y + 3 = 0$.
For point $(6, -5)$,$S_1 = (6)^2 + (-5)^2 - 2(6) + 4(-5) + 3 = 36 + 25 - 12 - 20 + 3 = 32$.
The tangent $T$ at $(x, y)$ relative to $(6, -5)$ is $xx_1 + yy_1 - (x + x_1) + 2(y + y_1) + 3 = 0$.
Substituting $(6, -5)$,we get $T = 6x - 5y - (x + 6) + 2(y - 5) + 3 = 5x - 3y - 13 = 0$.
Now,$SS_1 = T^2$ becomes $(x^2 + y^2 - 2x + 4y + 3)(32) = (5x - 3y - 13)^2$.
Expanding this,$32x^2 + 32y^2 - 64x + 128y + 96 = 25x^2 + 9y^2 - 30xy - 130x + 78y + 169$.
Rearranging terms: $7x^2 + 23y^2 + 30xy + 66x + 50y - 73 = 0$.

(D) The equation of the chord of contact from a point $(x_1, y_1)$ to the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is given by $xx_1 + yy_1 + g(x + x_1) + f(y + y_1) + c = 0$.
Given the point $(x_1, y_1) = (2, -3)$ and the circle $x^2 + y^2 + 4x - 6y - 12 = 0$,we have $g = 2$,$f = -3$,and $c = -12$.
Substituting these values into the formula:
$x(2) + y(-3) + 2(x + 2) - 3(y - 3) - 12 = 0$
$2x - 3y + 2x + 4 - 3y + 9 - 12 = 0$
$4x - 6y + 1 = 0$

(B) Let $S = x^2 + y^2 - 2x + 4y = 0$. For the point $(x_1, y_1) = (0, 1)$,we have $S_1 = 0^2 + 1^2 - 2(0) + 4(1) = 5$.
The equation of the tangent $T$ at $(x_1, y_1)$ is given by $x x_1 + y y_1 - (x + x_1) + 2(y + y_1) = 0$.
Substituting $(0, 1)$,we get $T = x(0) + y(1) - (x + 0) + 2(y + 1) = -x + 3y + 2$.
The equation of the pair of tangents is given by $SS_1 = T^2$.
$5(x^2 + y^2 - 2x + 4y) = (-x + 3y + 2)^2$.
$5x^2 + 5y^2 - 10x + 20y = x^2 + 9y^2 + 4 - 6xy - 4x + 12y$.
Rearranging the terms,we get $4x^2 - 4y^2 + 6xy - 6x + 8y - 4 = 0$.

(B) The equation of the common chord $PQ$ is obtained by subtracting the two circle equations: $(x^2 + y^2 + 2ax + cy + a) - (x^2 + y^2 - 3ax + dy - 1) = 0$.
This simplifies to $5ax + (c - d)y + (a + 1) = 0$ ...... $(i)$
We are given that the line $PQ$ is $5x + 6y - a = 0$ ...... (ii)
Comparing the coefficients of $(i)$ and (ii),we have: $\frac{5a}{5} = \frac{c - d}{6} = \frac{a + 1}{-a}$.
From the first and third parts: $a = \frac{a + 1}{-a} \implies -a^2 = a + 1 \implies a^2 + a + 1 = 0$.
The discriminant of this quadratic equation is $D = 1^2 - 4(1)(1) = -3$.
Since $D < 0$,there are no real values of $a$ that satisfy this condition.

(D) Given circles are $S_1: x^2 + y^2 - 4x - 4y = 0$ and $S_2: x^2 + y^2 - 16 = 0$.
The equation of the common chord is $S_1 - S_2 = 0$,which gives $-4x - 4y + 16 = 0$,or $x + y - 4 = 0$.
Thus,the equation of the common chord is $x + y = 4$,or $\frac{x + y}{4} = 1$.
To find the pair of lines joining the origin to the intersection points of the circle $S_2 = 0$ and the chord,we homogenize the circle equation using the chord equation:
$x^2 + y^2 - 16(1)^2 = 0$
$x^2 + y^2 - 16\left(\frac{x + y}{4}\right)^2 = 0$
$x^2 + y^2 - 16\left(\frac{x^2 + y^2 + 2xy}{16}\right) = 0$
$x^2 + y^2 - (x^2 + y^2 + 2xy) = 0$
$-2xy = 0$,which implies $xy = 0$.
The lines are $x = 0$ (y-axis) and $y = 0$ (x-axis).
The angle between the lines $x = 0$ and $y = 0$ is $\frac{\pi}{2}$.

(A) The equations of the circles are:
$S_1: x^2 + y^2 + \frac{7}{2}x - \frac{5}{2}y + 1 = 0$ (Dividing by $2$)
$S_2: x^2 + y^2 - 4x + 8y - 18 = 0$
The equation of the common chord is $S_1 - S_2 = 0$:
$(x^2 + y^2 + \frac{7}{2}x - \frac{5}{2}y + 1) - (x^2 + y^2 - 4x + 8y - 18) = 0$
$\frac{15}{2}x - \frac{21}{2}y + 19 = 0 \Rightarrow 15x - 21y + 38 = 0$
The center of circle $S_2$ is $C_2(2, -4)$ and its radius $r = \sqrt{2^2 + (-4)^2 - (-18)} = \sqrt{4 + 16 + 18} = \sqrt{38}$.
The perpendicular distance $d$ from $C_2(2, -4)$ to the line $15x - 21y + 38 = 0$ is:
$d = \frac{|15(2) - 21(-4) + 38|}{\sqrt{15^2 + (-21)^2}} = \frac{|30 + 84 + 38|}{\sqrt{225 + 441}} = \frac{152}{\sqrt{666}}$.
The length of the common chord is $2\sqrt{r^2 - d^2}$:
$2\sqrt{38 - \frac{152^2}{666}} = 2\sqrt{38 - \frac{23104}{666}} = 2\sqrt{\frac{25308 - 23104}{666}} = 2\sqrt{\frac{2204}{666}} = 2\sqrt{\frac{1102}{333}}$.

(C) The equation of the common chord is given by subtracting the two circle equations:
$[(x - a)^{2} + y^{2} - c^{2}] - [x^{2} + (y - b)^{2} - c^{2}] = 0$
$x^{2} - 2ax + a^{2} + y^{2} - c^{2} - x^{2} - y^{2} + 2by - b^{2} + c^{2} = 0$
$-2ax + 2by + a^{2} - b^{2} = 0$
$2ax - 2by - a^{2} + b^{2} = 0 \dots (1)$
The distance $p$ from the center of the first circle $(a, 0)$ to the line $(1)$ is:
$p = \frac{|2a(a) - 2b(0) - a^{2} + b^{2}|}{\sqrt{(2a)^{2} + (-2b)^{2}}} = \frac{|2a^{2} - a^{2} + b^{2}|}{\sqrt{4a^{2} + 4b^{2}}} = \frac{a^{2} + b^{2}}{2\sqrt{a^{2} + b^{2}}} = \frac{1}{2}\sqrt{a^{2} + b^{2}}$
The length of the common chord is $2\sqrt{c^{2} - p^{2}}$:
$L = 2\sqrt{c^{2} - \left(\frac{\sqrt{a^{2} + b^{2}}}{2}\right)^{2}}$
$L = 2\sqrt{c^{2} - \frac{a^{2} + b^{2}}{4}}$
$L = \sqrt{4c^{2} - a^{2} - b^{2}}$

(C) Let the center of circle $C_2$ be $(h, k)$. Its equation is $(x - h)^{2} + (y - k)^{2} = 5^{2} = 25$.
The equation of $C_1$ is $x^{2} + y^{2} = 16$.
The equation of the common chord is given by $S_1 - S_2 = 0$,which is $2hx + 2ky = h^{2} + k^{2} - 9$.
The length of the common chord is maximum when the chord passes through the center of $C_1$,i.e.,$(0, 0)$.
Thus,$h^{2} + k^{2} - 9 = 0$,or $h^{2} + k^{2} = 9$.
The slope of the common chord $2hx + 2ky = 9$ is $-h/k = 3/4$,which implies $k = -4h/3$.
Substituting $k$ into $h^{2} + k^{2} = 9$,we get $h^{2} + (-4h/3)^{2} = 9$,so $h^{2} + 16h^{2}/9 = 9$,which gives $25h^{2}/9 = 9$,so $h^{2} = 81/25$.
Thus,$h = \pm 9/5$. If $h = 9/5$,then $k = -4(9/5)/3 = -12/5$. If $h = -9/5$,then $k = 12/5$.
Therefore,the possible centers are $(9/5, -12/5)$ and $(-9/5, 12/5)$.

(B) The equation of the circle is $x^{2} + y^{2} - 6x - 8y + 21 = 0$. The center is $C(3, 4)$ and the radius $r = \sqrt{3^{2} + 4^{2} - 21} = \sqrt{9 + 16 - 21} = \sqrt{4} = 2$.
The equation of the chord of contact $AB$ from the origin $(0, 0)$ is given by $T = 0$,which is $x(0) + y(0) - 3(x + 0) - 4(y + 0) + 21 = 0$,simplifying to $3x + 4y - 21 = 0$.
The perpendicular distance $CM$ from the center $C(3, 4)$ to the line $3x + 4y - 21 = 0$ is $CM = \frac{|3(3) + 4(4) - 21|}{\sqrt{3^{2} + 4^{2}}} = \frac{|9 + 16 - 21|}{5} = \frac{4}{5}$.
In the right-angled triangle $\triangle AMC$,$AM = \sqrt{AC^{2} - CM^{2}} = \sqrt{2^{2} - (\frac{4}{5})^{2}} = \sqrt{4 - \frac{16}{25}} = \sqrt{\frac{100 - 16}{25}} = \sqrt{\frac{84}{25}} = \frac{2\sqrt{21}}{5}$.
Since $AB = 2AM$,we have $AB = 2 \times \frac{2\sqrt{21}}{5} = \frac{4}{5}\sqrt{21}$.

(A) The equation of the line is $y = mx + c$,which can be written as $\frac{y - mx}{c} = 1$.
The equation of the circle is $x^{2} + y^{2} = a^{2}$.
To find the joint equation of the lines passing through the origin and the intersection points,we homogenize the circle equation using the line equation:
$x^{2} + y^{2} = a^{2} \left( \frac{y - mx}{c} \right)^{2}$
$c^{2}(x^{2} + y^{2}) = a^{2}(y^{2} - 2mxy + m^{2}x^{2})$
$x^{2}(c^{2} - a^{2}m^{2}) + 2a^{2}mxy + y^{2}(c^{2} - a^{2}) = 0$.
For these lines to be at right angles,the sum of the coefficients of $x^{2}$ and $y^{2}$ must be zero:
$(c^{2} - a^{2}m^{2}) + (c^{2} - a^{2}) = 0$
$2c^{2} - a^{2}(1 + m^{2}) = 0$
$2c^{2} = a^{2}(1 + m^{2})$.

(D) The equation of the chord of contact for a point $(x_1, y_1)$ with respect to the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is given by $xx_1 + yy_1 + g(x + x_1) + f(y + y_1) + c = 0$.
For the point $(0, 0)$,the chord of contact is $g(x + 0) + f(y + 0) + c = 0$,which simplifies to $gx + fy + c = 0$ (Equation $1$).
For the point $(g, f)$,the chord of contact is $xg + yf + g(x + g) + f(y + f) + c = 0$,which simplifies to $2gx + 2fy + g^2 + f^2 + c = 0$,or $gx + fy + \frac{1}{2}(g^2 + f^2 + c) = 0$ (Equation $2$).
The distance between two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is given by $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$.
Here,$A = g$,$B = f$,$C_1 = c$,and $C_2 = \frac{1}{2}(g^2 + f^2 + c)$.
Distance $d = \frac{|\frac{1}{2}(g^2 + f^2 + c) - c|}{\sqrt{g^2 + f^2}} = \frac{|\frac{1}{2}(g^2 + f^2 - c)|}{\sqrt{g^2 + f^2}} = \frac{g^2 + f^2 - c}{2 \sqrt{g^2 + f^2}}$.

(B) The given equations of the circles are:
$C_1: x^2 + y^2 - 6x - 16 = 0$
$C_2: x^2 + y^2 - 8y - 9 = 0$
The equation of the common chord is given by $C_1 - C_2 = 0$:
$(x^2 + y^2 - 6x - 16) - (x^2 + y^2 - 8y - 9) = 0$
$-6x + 8y - 7 = 0$ or $6x - 8y + 7 = 0$
For circle $C_1$,the center is $(3, 0)$ and the radius $r_1 = \sqrt{3^2 + 0^2 - (-16)} = \sqrt{9 + 16} = 5$.
The perpendicular distance $d$ from the center $(3, 0)$ to the common chord $6x - 8y + 7 = 0$ is:
$d = \frac{|6(3) - 8(0) + 7|}{\sqrt{6^2 + (-8)^2}} = \frac{|18 + 7|}{\sqrt{36 + 64}} = \frac{25}{10} = \frac{5}{2}$
The length of the common chord is $2 \sqrt{r_1^2 - d^2}$:
$= 2 \sqrt{5^2 - (\frac{5}{2})^2} = 2 \sqrt{25 - \frac{25}{4}} = 2 \sqrt{\frac{100 - 25}{4}} = 2 \sqrt{\frac{75}{4}} = 2 \times \frac{5 \sqrt{3}}{2} = 5 \sqrt{3}$

(C) The circle $C_1$ has radius $r_1 = 1$ and touches both axes in the first quadrant,so its center is $(1, 1)$. Its equation is $(x-1)^2 + (y-1)^2 = 1$.
Let the second circle $C_2$ have radius $r$ and center $(r, r)$. Its equation is $(x-r)^2 + (y-r)^2 = r^2$.
The common chord of two intersecting circles is a line perpendicular to the line joining their centers. The length of the common chord is maximized when the common chord passes through the center of the smaller circle or when the circles are positioned such that the chord is the diameter of the smaller circle.
For two circles $(x-h_1)^2 + (y-k_1)^2 = r_1^2$ and $(x-h_2)^2 + (y-k_2)^2 = r_2^2$,the common chord is the radical axis: $2x(h_2-h_1) + 2y(k_2-k_1) + h_1^2 + k_1^2 - r_1^2 - h_2^2 - k_2^2 + r_2^2 = 0$.
Here,$h_1=1, k_1=1, r_1=1$ and $h_2=r, k_2=r, r_2=r$.
The radical axis equation is $2x(r-1) + 2y(r-1) + 1 + 1 - 1 - r^2 - r^2 + r^2 = 0$,which simplifies to $2(r-1)(x+y) + 1 - r^2 = 0$.
The length of the common chord is maximized when the radical axis passes through the center of the circle with the smaller radius. Setting $x=1, y=1$ in the radical axis equation: $2(r-1)(2) + 1 - r^2 = 0 \implies 4r - 4 + 1 - r^2 = 0 \implies r^2 - 4r + 3 = 0$.
Solving for $r$: $(r-3)(r-1) = 0$. Since $r \neq 1$ (as the circles would be identical),we get $r = 3$.

(A) The equation of the chord of contact of tangents drawn from a point $(x_1, y_1)$ to the circle $x^2 + y^2 = r^2$ is given by $xx_1 + yy_1 = r^2$.
Here,the point is $(x_1, y_1) = (5, -3)$ and the circle is $x^2 + y^2 = 10$,so $r^2 = 10$.
Substituting these values into the formula:
$x(5) + y(-3) = 10$
$5x - 3y = 10$
Thus,the equation of the chord of contact is $5x - 3y = 10$.

(D) The equation of the chord of contact for a circle $x^2 + y^2 + 2gx + 2fy + c = 0$ from an external point $(x_1, y_1)$ is given by $T = 0$,where $T = xx_1 + yy_1 + g(x + x_1) + f(y + y_1) + c = 0$.
Given the circle $x^2 + y^2 + 4x + 6y - 12 = 0$,we have $g = 2$,$f = 3$,and $c = -12$.
The point is $(x_1, y_1) = (2, 3)$.
Substituting these values into the formula:
$x(2) + y(3) + 2(x + 2) + 3(y + 3) - 12 = 0$
$2x + 3y + 2x + 4 + 3y + 9 - 12 = 0$
$4x + 6y + 1 = 0$
$4x + 6y = -1$
Since none of the provided options match this result,the correct answer is none of these.

(C) Let the two given circles be $C_1: x^2 + y^2 + 5x - 8y + 1 = 0$ and $C_2: x^2 + y^2 - 3x + 7y - 25 = 0$.
The equation of the common chord is given by $C_1 - C_2 = 0$.
$(x^2 + y^2 + 5x - 8y + 1) - (x^2 + y^2 - 3x + 7y - 25) = 0$
$8x - 15y + 26 = 0$.
The given circle is $x^2 + y^2 - 2x = 0$. Comparing this with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -1, f = 0$.
The center of this circle is $(-g, -f) = (1, 0)$.
The perpendicular distance $d$ from the point $(x_0, y_0) = (1, 0)$ to the line $ax + by + c = 0$ is given by $d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}$.
Substituting the values: $d = \frac{|8(1) - 15(0) + 26|}{\sqrt{8^2 + (-15)^2}} = \frac{|8 + 26|}{\sqrt{64 + 225}} = \frac{34}{\sqrt{289}} = \frac{34}{17} = 2$.