Geometrical problems regarding circle and its properties Questions in English

(D) The circle $S^{\prime}$ is $x^2+y^2-6x+6y+2=0$. Its center is $C^{\prime}(3, -3)$ and radius $r^{\prime} = \sqrt{3^2+(-3)^2-2} = \sqrt{9+9-2} = \sqrt{16} = 4$.
Since $S$ is a unit circle,its radius $r = 1$.
Let the center of $S$ be $C(h, k)$. Since $S$ and $S^{\prime}$ touch externally at $P(-1, -3)$,the point $P$ divides the line segment $CC^{\prime}$ internally in the ratio $r:r^{\prime} = 1:4$.
Using the section formula:
$-1 = \frac{4h + 1(3)}{1+4} \implies -5 = 4h+3 \implies 4h = -8 \implies h = -2$.
$-3 = \frac{4k + 1(-3)}{1+4} \implies -15 = 4k-3 \implies 4k = -12 \implies k = -3$.
So,the center of $S$ is $C(-2, -3)$.
The equation of $S$ is $(x+2)^2+(y+3)^2 = 1^2$,which simplifies to $x^2+4x+4+y^2+6y+9 = 1$,or $x^2+y^2+4x+6y+12 = 0$.
Comparing this with $x^2+y^2+2gx+2fy+c=0$,we get $2g=4 \implies g=2$,$2f=6 \implies f=3$,and $c=12$.
Therefore,$g+f+c = 2+3+12 = 17$.

(B) To find the angle subtended by the chord $x+y-1=0$ at the center of the circle $x^2+y^2=1$,we follow these steps:
Step $1$: Identify the circle and chord.
The circle $x^2+y^2=1$ has its center at the origin $O(0,0)$ and radius $r=1$.
The chord is given by the line $x+y-1=0$.
Step $2$: Find the perpendicular distance $d$ from the origin to the chord.
The distance $d$ from $(0,0)$ to $x+y-1=0$ is given by $d = \frac{|1(0)+1(0)-1|}{\sqrt{1^2+1^2}} = \frac{1}{\sqrt{2}}$.
Step $3$: Calculate the angle.
Let the chord intersect the circle at points $A$ and $B$. In the isosceles triangle $OAB$,the perpendicular from $O$ to $AB$ bisects the angle $\angle AOB$. Let $\angle AOB = 2\theta$.
In the right-angled triangle formed by the origin,the midpoint of the chord,and one endpoint of the chord,we have $\cos(\theta) = \frac{d}{r} = \frac{1/\sqrt{2}}{1} = \frac{1}{\sqrt{2}}$.
Thus,$\theta = 45^{\circ}$ or $\frac{\pi}{4}$ radians.
The total angle subtended at the center is $\angle AOB = 2\theta = 2 \times 45^{\circ} = 90^{\circ}$ or $\frac{\pi}{2}$ radians.

(A) The equation of the circle is $x^2 + y^2 - 4 x - 6 y + 9 = 0$.
Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -2, f = -3, c = 9$.
Centre $= (-g, -f) = (2, 3)$ and Radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{4 + 9 - 9} = 2$.
The line $6 x + 8 y + \alpha = 0$ intersects the circle at two distinct points if the perpendicular distance from the centre $(2, 3)$ to the line is less than the radius.
Distance $d = \frac{|6(2) + 8(3) + \alpha|}{\sqrt{6^2 + 8^2}} = \frac{|12 + 24 + \alpha|}{10} = \frac{|36 + \alpha|}{10}$.
Condition: $d < r$ $\Rightarrow \frac{|36 + \alpha|}{10} < 2$ $\Rightarrow |36 + \alpha| < 20$.
This implies $-20 < 36 + \alpha < 20$,which simplifies to $-56 < \alpha < -16$.
Given $\alpha$ is an integer multiple of $8$,let $\alpha = 8k$ for some integer $k$.
$-56 < 8k < -16 \Rightarrow -7 < k < -2$.
The possible integer values for $k$ are $-6, -5, -4, -3$.
Thus,there are $4$ possible values for $\alpha$.

(B) The equation of the circle is $x^2+y^2+10x+8y-23=0$. The center $O$ of the circle is $(-5, -4)$.
Since $M$ is the midpoint of the chord $AB$,the line $OM$ is perpendicular to the chord $AB$.
The slope of $OM$ is $m_1 = \frac{-4 - (-5/2)}{-5 - (-7/2)} = \frac{-4 + 2.5}{-5 + 3.5} = \frac{-1.5}{-1.5} = 1$.
Since $OM \perp AB$,the slope of the chord $AB$ is $m_2 = -1/m_1 = -1$.
The equation of the line $AB$ passing through $M\left(\frac{-7}{2}, \frac{-5}{2}\right)$ with slope $-1$ is:
$y - (-5/2) = -1(x - (-7/2))$
$y + 5/2 = -x - 7/2$
$x + y + 6 = 0$
To match the form $ax + by + 1 = 0$,we divide by $-6$:
$-\frac{1}{6}x - \frac{1}{6}y - 1 = 0$
Wait,the given form is $ax + by + 1 = 0$. Let's rewrite $x + y + 6 = 0$ as $\frac{1}{6}x + \frac{1}{6}y + 1 = 0$.
Comparing this with $ax + by + 1 = 0$,we get $a = 1/6$ and $b = 1/6$.
Therefore,$3a + 3b = 3(1/6) + 3(1/6) = 1/2 + 1/2 = 1$.

(D) For the circle $x^2+y^2=4$,the center is $(0,0)$ and radius $r_1=2$. The perpendicular distance from the center $(0,0)$ to the line $2x-2y-3=0$ is $p_1 = \frac{|2(0)-2(0)-3|}{\sqrt{2^2+(-2)^2}} = \frac{3}{\sqrt{8}}$.
The length of the intercept $d_1 = 2\sqrt{r_1^2-p_1^2} = 2\sqrt{4-\frac{9}{8}} = 2\sqrt{\frac{23}{8}}$.
For the circle $x^2+y^2-10x-14y+65=0$,the center is $(5,7)$ and radius $r_2 = \sqrt{5^2+7^2-65} = \sqrt{25+49-65} = \sqrt{9} = 3$.
The perpendicular distance from the center $(5,7)$ to the line $2x-2y-3=0$ is $p_2 = \frac{|2(5)-2(7)-3|}{\sqrt{2^2+(-2)^2}} = \frac{|10-14-3|}{\sqrt{8}} = \frac{|-7|}{\sqrt{8}} = \frac{7}{\sqrt{8}}$.
Wait,checking the calculation for $d_2$: $d_2 = 2\sqrt{r_2^2-p_2^2} = 2\sqrt{9-\frac{49}{8}} = 2\sqrt{\frac{72-49}{8}} = 2\sqrt{\frac{23}{8}}$.
Since $d_1 = 2\sqrt{\frac{23}{8}}$ and $d_2 = 2\sqrt{\frac{23}{8}}$,we have $d_1=d_2$.

(B) The equation of the circle is $x^2 + y^2 - 6x + 8y = 0$.
Comparing this with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get the center $C = (-g, -f) = (3, -4)$ and radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{3^2 + (-4)^2 - 0} = \sqrt{9 + 16} = 5$.
The perpendicular distance $d$ from the center $(3, -4)$ to the line $3x + 4y - 25 = 0$ is given by $d = \frac{|3(3) + 4(-4) - 25|}{\sqrt{3^2 + 4^2}} = \frac{|9 - 16 - 25|}{\sqrt{25}} = \frac{|-32|}{5} = \frac{32}{5}$.
The shortest distance from the line to the circle is $d - r = \frac{32}{5} - 5 = \frac{32 - 25}{5} = \frac{7}{5}$.

(D) The equation of the circle is $x^2+y^2-4x-8y-5=0$. Comparing this with $x^2+y^2+2gx+2fy+c=0$,we get $g=-2, f=-4, c=-5$.
The center of the circle is $(2, 4)$ and the radius $r = \sqrt{g^2+f^2-c} = \sqrt{(-2)^2+(-4)^2-(-5)} = \sqrt{4+16+5} = \sqrt{25} = 5$.
For the line $3x-4y-m=0$ to intersect the circle at two distinct points,the perpendicular distance $d$ from the center $(2, 4)$ to the line must be less than the radius $r$.
$d = \frac{|3(2)-4(4)-m|}{\sqrt{3^2+(-4)^2}} = \frac{|6-16-m|}{\sqrt{9+16}} = \frac{|-10-m|}{5} = \frac{|m+10|}{5}$.
Setting $d < r$,we have $\frac{|m+10|}{5} < 5$,which implies $|m+10| < 25$.
This inequality is equivalent to $-25 < m+10 < 25$.
Subtracting $10$ from all parts,we get $-35 < m < 15$.
The integers $m$ satisfying this are $\{-34, -33, \dots, 14\}$.
The number of such integers is $14 - (-34) + 1 = 14 + 34 + 1 = 49$.

(A) Given the circle equation: $x^2+y^2-6x-4y-12=0$.
Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=-3, f=-2, c=-12$.
Centre of the circle: $(-g, -f) = (3, 2)$.
Radius of the circle: $r = \sqrt{g^2+f^2-c} = \sqrt{(-3)^2+(-2)^2-(-12)} = \sqrt{9+4+12} = \sqrt{25} = 5$.
Since the lines are parallel to the $x$-axis,they are of the form $y = k$.
These lines touch the circle,so the distance from the centre $(3, 2)$ to the line $y = k$ must be equal to the radius $r = 5$.
Therefore,$|k - 2| = 5$.
This gives $k - 2 = 5$ or $k - 2 = -5$.
So,$k = 7$ or $k = -3$.
The equations of the lines are $y = 7$ and $y = -3$.
The pair of straight lines is given by $(y - 7)(y + 3) = 0$.
Expanding this,we get $y^2 + 3y - 7y - 21 = 0$,which simplifies to $y^2 - 4y - 21 = 0$.

(C) Let the radius of the circle be $r$. Since the circle lies in the fourth quadrant and touches the lines $x=0$ and $y=0$,its center must be at $(r, -r)$,where $r > 0$.
The distance from the center $(r, -r)$ to the line $3x+4y-12=0$ must be equal to the radius $r$.
Using the perpendicular distance formula,we have:
$\left| \frac{3(r) + 4(-r) - 12}{\sqrt{3^2 + 4^2}} \right| = r$
$\left| \frac{3r - 4r - 12}{5} \right| = r$
$\left| \frac{-r - 12}{5} \right| = r$
$|r + 12| = 5r$
Since $r > 0$,$r + 12 = 5r$ or $r + 12 = -5r$.
Case $1$: $4r = 12 \Rightarrow r = 3$.
Case $2$: $6r = -12 \Rightarrow r = -2$ (rejected as $r > 0$).
Thus,the radius is $3$ units.

(C) The given equation of the circle is $x^2+y^2-8x-2y-8=0$.
Comparing this with the general form $x^2+y^2+2gx+2fy+c=0$,we get $g=-4$,$f=-1$,and $c=-8$.
The centre of the circle is $(-g, -f) = (4, 1)$ and the radius $r$ is $\sqrt{g^2+f^2-c} = \sqrt{16+1+8} = \sqrt{25} = 5$.
The perpendicular distance $d$ from the centre $(4, 1)$ to the line $x+y+1=0$ is given by:
$d = \left|\frac{4+1+1}{\sqrt{1^2+1^2}}\right| = \frac{6}{\sqrt{2}} = 3\sqrt{2}$.
The length of the chord is given by $2\sqrt{r^2-d^2}$.
Length $= 2\sqrt{5^2 - (3\sqrt{2})^2} = 2\sqrt{25 - 18} = 2\sqrt{7}$ units.

(A) The equation of the circle is $x^2+y^2=16$,which represents a circle centered at the origin $(0,0)$ with radius $r=4$.
Let the two points be $A = (4 \cos \theta, 4 \sin \theta)$ and $B = (4 \cos (\theta+60^{\circ}), 4 \sin (\theta+60^{\circ}))$.
These points represent points on the circle with polar angles $\theta$ and $\theta+60^{\circ}$ respectively.
The angle subtended by the chord $AB$ at the center $O(0,0)$ is $\Delta \phi = (\theta+60^{\circ}) - \theta = 60^{\circ}$.
Since $OA = OB = r = 4$ and the included angle $\angle AOB = 60^{\circ}$,the triangle $\triangle OAB$ is an equilateral triangle.
Therefore,the length of the chord $AB$ is equal to the radius of the circle.
$AB = r = 4$.

(D) The equation of the given circle is $x^2+y^2-6x+2y-28=0$.
Completing the square,we get $(x-3)^2+(y+1)^2 = 28+9+1 = 38$.
Thus,the radius $r$ of the circle is $\sqrt{38}$.
An equilateral triangle inscribed in a circle of radius $r$ has an area given by $A = \frac{3\sqrt{3}}{4} R^2$,where $R$ is the circumradius.
Here,$R = r = \sqrt{38}$.
Therefore,the area is $\frac{3\sqrt{3}}{4} \times (\sqrt{38})^2 = \frac{3\sqrt{3}}{4} \times 38 = \frac{3\sqrt{3} \times 19}{2} = \frac{57\sqrt{3}}{2}$ sq. units.

(A) The equation of the circle is $x^2+y^2-6x+8y-5=0$.
Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=-3, f=4, c=-5$.
Centre $C = (-g, -f) = (3, -4)$.
Radius $r = \sqrt{g^2+f^2-c} = \sqrt{(-3)^2+4^2-(-5)} = \sqrt{9+16+5} = \sqrt{30}$.
The line equation is $2x-y-5=0$.
The perpendicular distance $d$ from the centre $(3, -4)$ to the line is:
$d = \frac{|2(3)-(-4)-5|}{\sqrt{2^2+(-1)^2}} = \frac{|6+4-5|}{\sqrt{5}} = \frac{5}{\sqrt{5}} = \sqrt{5}$.
The length of the chord is given by $2\sqrt{r^2-d^2}$.
Length $= 2\sqrt{(\sqrt{30})^2-(\sqrt{5})^2} = 2\sqrt{30-5} = 2\sqrt{25} = 2 \times 5 = 10$ units.
Thus,the correct option is $A$.

(B) Let the line passing through $M(a, b)$ have an angle of inclination $\theta$.
The parametric equations of the line are $x = a + r \cos \theta$ and $y = b + r \sin \theta$,where $r$ is the distance from $M$.
Substituting these into the circle equation $x^2 + y^2 = k^2$:
$(a + r \cos \theta)^2 + (b + r \sin \theta)^2 = k^2$
$a^2 + 2ar \cos \theta + r^2 \cos^2 \theta + b^2 + 2br \sin \theta + r^2 \sin^2 \theta = k^2$
$r^2 + 2r(a \cos \theta + b \sin \theta) + (a^2 + b^2 - k^2) = 0$
This is a quadratic equation in $r$ whose roots $r_1$ and $r_2$ represent the distances $MC$ and $MD$.
The product of the roots $r_1 \cdot r_2$ is given by the constant term of the quadratic equation.
Therefore,$MC \times MD = a^2 + b^2 - k^2$.

(B) The equations of the sides are:
$y=0$ ... $(i)$
$y=x$ ... (ii)
$2x+3y=10$ ... (iii)
Solving $(i)$ and (iii) gives vertex $A(5,0)$.
Solving $(i)$ and (ii) gives vertex $B(0,0)$.
Solving (ii) and (iii) gives vertex $C(2,2)$.
Let the equation of the circle be $x^2+y^2+2gx+2fy+c=0$ ... (iv).
Since it passes through $B(0,0)$,$c=0$.
Since it passes through $A(5,0)$,$25+10g=0 \Rightarrow g=-5/2$.
Since it passes through $C(2,2)$,$4+4+4g+4f=0 \Rightarrow g+f+2=0$.
Substituting $g=-5/2$,we get $-5/2+f+2=0 \Rightarrow f=1/2$.
The centre of the circle is $(-g, -f) = (5/2, -1/2)$.

(A) The given equation of the circle is $x^2+y^2+2gx+2fy+c=0$ $(c>0)$.
The center of the circle is $(-g, -f)$ and the radius is $r = \sqrt{g^2+f^2-c}$.
Since the circle touches both coordinate axes,the distance from the center to the axes is equal to the radius,so $|-g| = |-f| = r = \sqrt{c}$.
Since the circle lies in the third quadrant,the center must be $(-\sqrt{c}, -\sqrt{c})$.
Thus,the equation of the circle is $(x+\sqrt{c})^2 + (y+\sqrt{c})^2 = c$.
The circle touches the $x$-axis at $(-\sqrt{c}, 0)$ and the $y$-axis at $(0, -\sqrt{c})$.
Let these points be $A(-\sqrt{c}, 0)$ and $B(0, -\sqrt{c})$.
We check if these points lie on the line $x+y+\sqrt{c}=0$:
For point $A$: $-\sqrt{c} + 0 + \sqrt{c} = 0$ (True).
For point $B$: $0 - \sqrt{c} + \sqrt{c} = 0$ (True).
Thus,the chord intercepted by the circle on the line is the segment $AB$.
The length of the chord $AB$ is $\sqrt{(0 - (-\sqrt{c}))^2 + (-\sqrt{c} - 0)^2} = \sqrt{(\sqrt{c})^2 + (-\sqrt{c})^2} = \sqrt{c+c} = \sqrt{2c}$.

(B) The centre of the circle is $(h, k) = (1, 1)$.
The equation of the circle is $(x-1)^2 + (y-1)^2 = r^2$,where $r$ is the radius.
The perpendicular distance $d$ from the centre $(1, 1)$ to the line $x+y+1=0$ is given by $d = \frac{|1+1+1|}{\sqrt{1^2+1^2}} = \frac{3}{\sqrt{2}}$.
The length of the chord is $L = 4\sqrt{2}$,so half the chord length is $a = \frac{L}{2} = 2\sqrt{2}$.
Using the relation $r^2 = d^2 + a^2$,we get $r^2 = (\frac{3}{\sqrt{2}})^2 + (2\sqrt{2})^2 = \frac{9}{2} + 8 = \frac{9+16}{2} = \frac{25}{2} = 12.5$.
The equation of the circle is $(x-1)^2 + (y-1)^2 = 12.5$,which simplifies to $x^2-2x+1 + y^2-2y+1 = 12.5$,or $x^2+y^2-2x-2y-10.5=0$.
Multiplying by $2$,we get $2x^2+2y^2-4x-4y-21=0$.

(C) The equation of the circle is $x^2+y^2-4x+6y-12=0$. Comparing this with $x^2+y^2+2gx+2fy+c=0$,we get $g=-2, f=3, c=-12$.
The center of the circle is $(-g, -f) = (2, -3)$ and the radius $r = \sqrt{g^2+f^2-c} = \sqrt{(-2)^2+3^2-(-12)} = \sqrt{4+9+12} = \sqrt{25} = 5$.
The perpendicular distance $d$ from the center $(2, -3)$ to the line $4x+3y+1=0$ is given by $d = \frac{|4(2)+3(-3)+1|}{\sqrt{4^2+3^2}} = \frac{|8-9+1|}{\sqrt{16+9}} = \frac{0}{5} = 0$.
Since the perpendicular distance $d=0$,the line passes through the center of the circle,meaning the chord is a diameter.
The length of the chord is $2 \times \sqrt{r^2-d^2} = 2 \times \sqrt{5^2-0^2} = 2 \times 5 = 10$.

(C) The circle is $x^2+y^2=61$ with center $O(0,0)$ and radius $r=\sqrt{61}$.
Since $PA=PB=10$,$P$ lies on the perpendicular bisector of chord $AB$. The center $O$ also lies on the perpendicular bisector of chord $AB$. Thus,the line $OP$ is perpendicular to the line $l$.
The slope of $OP$ is $m_{OP} = \frac{6-0}{-5-0} = -\frac{6}{5}$.
Since $l \perp OP$,the slope of line $l$ is $m_l = -\frac{1}{m_{OP}} = \frac{5}{6}$.
Let the equation of line $l$ be $5x-6y+k=0$.
The distance from the center $O(0,0)$ to the chord $AB$ is $d = \frac{|k|}{\sqrt{5^2+(-6)^2}} = \frac{|k|}{\sqrt{61}}$.
In $\triangle OMA$ (where $M$ is the midpoint of $AB$),$OA^2 = OM^2 + AM^2$.
$AM^2 = PA^2 - PM^2$. Since $P(-5,6)$,$OP = \sqrt{(-5)^2+6^2} = \sqrt{61}$.
Let $M$ be the projection of $O$ on $l$. $OM = d = \frac{|k|}{\sqrt{61}}$.
$AM^2 = r^2 - OM^2 = 61 - \frac{k^2}{61}$.
Also $PM^2 = OP^2 - OM^2 = 61 - \frac{k^2}{61}$.
Given $PA=10$,$AM^2 = PA^2 - PM^2 = 100 - (61 - \frac{k^2}{61}) = 39 + \frac{k^2}{61}$.
Equating $AM^2$: $61 - \frac{k^2}{61} = 39 + \frac{k^2}{61} \implies 22 = \frac{2k^2}{61} \implies k^2 = 11 \times 61 = 671$.
Re-evaluating: The line $l$ passes through $M$. $M$ is the projection of $O$ on $l$. $PM = \sqrt{PA^2 - AM^2} = \sqrt{100 - (61 - d^2)} = \sqrt{39+d^2}$.
Since $P, M, O$ are collinear,$PM = |PO - OM| = |\sqrt{61} - d|$.
$39+d^2 = 61 - 2d\sqrt{61} + d^2 \implies 2d\sqrt{61} = 22 \implies d = \frac{11}{\sqrt{61}}$.
Since $d = \frac{|k|}{\sqrt{61}}$,$|k|=11$. Thus $k = \pm 11$. Checking the options,$5x-6y+11=0$ matches.

(A) The equation of the circle is $x^2+y^2=r^2$ with center $(0,0)$ and radius $r$.
The equation of the line is $mx-y+c=0$.
The line intersects the circle at two distinct points if the perpendicular distance from the center $(0,0)$ to the line is less than the radius $r$.
The perpendicular distance $d$ is given by:
$d = \left| \frac{m(0) - (0) + c}{\sqrt{m^2 + (-1)^2}} \right| = \frac{|c|}{\sqrt{m^2+1}}$.
For two distinct points,we must have $d < r$:
$\frac{|c|}{\sqrt{m^2+1}} < r$
$|c| < r \sqrt{m^2+1}$
$-r \sqrt{m^2+1} < c < r \sqrt{m^2+1}$.

(A) The equation of the circle is $x^2 + y^2 - 4x - 6y + c = 0$. The center is $C(2, 3)$ and the radius is $r = \sqrt{2^2 + 3^2 - c} = \sqrt{13 - c}$.
Let $P = (-1, -1)$. The distance $d$ from $P$ to $C$ is $d = \sqrt{(2 - (-1))^2 + (3 - (-1))^2} = \sqrt{3^2 + 4^2} = 5$.
Let $\alpha$ be the angle between the tangent and the line joining the center to the external point. Then $\sin \alpha = \frac{r}{d}$.
The angle between the two tangents is $\theta = 2\alpha$,so $\alpha = \frac{\theta}{2}$.
Given $\cos \theta = -\frac{7}{25}$,we use the identity $\cos \theta = 1 - 2\sin^2 \alpha$.
$-\frac{7}{25} = 1 - 2\sin^2 \alpha \implies 2\sin^2 \alpha = 1 + \frac{7}{25} = \frac{32}{25} \implies \sin^2 \alpha = \frac{16}{25}$.
Thus,$\sin \alpha = \frac{4}{5}$.
Since $\sin \alpha = \frac{r}{d}$,we have $\frac{r}{5} = \frac{4}{5}$,which gives $r = 4$.

(A) Since the circle touches both coordinate axes in the first quadrant,its center is $(r, r)$ and its radius is $r$,where $r > 0$.
The equation of the circle is $(x-r)^2 + (y-r)^2 = r^2$.
The distance from the center $(r, r)$ to the line $4x+3y-6=0$ must be equal to the radius $r$.
Thus,$\frac{|4r+3r-6|}{\sqrt{4^2+3^2}} = r$.
$\frac{|7r-6|}{5} = r$.
This gives two cases: $7r-6 = 5r$ or $7r-6 = -5r$.
Case $1$: $2r = 6 \implies r = 3$. The center is $(3, 3)$ and the equation is $(x-3)^2 + (y-3)^2 = 3^2$,which simplifies to $x^2+y^2-6x-6y+9=0$.
Case $2$: $12r = 6 \implies r = 1/2$. The center is $(1/2, 1/2)$ and the equation is $(x-1/2)^2 + (y-1/2)^2 = (1/2)^2$,which simplifies to $x^2+y^2-x-y+1/4=0$,or $4x^2+4y^2-4x-4y+1=0$.
The line $4x+3y-6=0$ passes through $(1.5, 0)$ and $(0, 2)$. For the center $(3, 3)$,$4(3)+3(3)-6 = 15 > 0$. For the center $(1/2, 1/2)$,$4(1/2)+3(1/2)-6 = 2+1.5-6 = -2.5 < 0$.
Since the circle lies below the line $L=0$ (meaning the center satisfies $4x+3y-6 < 0$),the correct radius is $r = 1/2$.
Thus,the equation is $4x^2+4y^2-4x-4y+1=0$.

(D) The given lines are $L_1: 3x - 4y + 4 = 0$ and $L_2: 6x - 8y - 7 = 0$.
We can rewrite $L_2$ as $3x - 4y - 3.5 = 0$.
Since the lines are parallel,the distance between them is the diameter of the circle.
The distance $d$ between two parallel lines $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$ is given by $d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}$.
Here,$a = 3, b = -4, c_1 = 4, c_2 = -3.5$.
$d = \frac{|4 - (-3.5)|}{\sqrt{3^2 + (-4)^2}} = \frac{7.5}{\sqrt{9 + 16}} = \frac{7.5}{5} = 1.5$.
Since the diameter $D = 1.5 = \frac{3}{2}$,the radius $r = \frac{D}{2} = \frac{3}{4}$.
The area of the circle is $A = \pi r^2 = \pi \left(\frac{3}{4}\right)^2 = \frac{9\pi}{16}$ sq. units.

(A) The given equation of the circle is $x^2+y^2-6x-4y-11=0$.
Comparing this with $x^2+y^2+2gx+2fy+c=0$,we get the centre $C = (-g, -f) = (3, 2)$.
Since $PA$ and $PB$ are tangents to the circle from point $P(1, 8)$,the angles $\angle PAC$ and $\angle PBC$ are $90^{\circ}$.
Thus,the points $P, A, C,$ and $B$ lie on a circle with diameter $PC$.
The centre of the circle passing through $P, A,$ and $B$ is the midpoint of the diameter $PC$.
Midpoint $= \left(\frac{1+3}{2}, \frac{8+2}{2}\right) = \left(\frac{4}{2}, \frac{10}{2}\right) = (2, 5)$.

(C) For a point $A$ outside a circle,if a secant through $A$ intersects the circle at points $C$ and $B$,then the length of the tangent $AT$ from $A$ to the circle is given by the Power of a Point theorem: $AT^2 = AC \cdot AB$.
First,calculate the distances $AC$ and $AB$ using the distance formula:
$AC = \sqrt{(5-3)^2 + (0-1)^2} = \sqrt{2^2 + (-1)^2} = \sqrt{4+1} = \sqrt{5}$
$AB = \sqrt{(7-3)^2 + (-1-1)^2} = \sqrt{4^2 + (-2)^2} = \sqrt{16+4} = \sqrt{20} = 2\sqrt{5}$
Now,substitute these values into the formula:
$AT^2 = AC \cdot AB = \sqrt{5} \cdot 2\sqrt{5} = 2 \cdot 5 = 10$
$AT = \sqrt{10}$ units.

(C) Let the circle pass through points $A(2,2)$ and $B(9,9)$ and touch the $X$-axis at point $P$.
By the power of a point theorem,since $OP$ is a tangent to the circle from the origin $O$ and $OAB$ is a secant line,we have:
$OP^2 = OA \cdot OB$
Calculating the distances $OA$ and $OB$ from the origin $O(0,0)$:
$OA = \sqrt{2^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$
$OB = \sqrt{9^2 + 9^2} = \sqrt{81 + 81} = \sqrt{162} = 9\sqrt{2}$
Substituting these values into the equation:
$OP^2 = (2\sqrt{2}) \cdot (9\sqrt{2}) = 18 \cdot 2 = 36$
$OP = \sqrt{36} = 6$
Thus,$OP = 6$.

(D) The equation of the circle is $S: x^2 + y^2 + 8x - 6y + k = 0$.
The length of the tangent drawn from an external point $(x_1, y_1)$ to the circle $S = 0$ is given by $\sqrt{S_1}$,where $S_1 = x_1^2 + y_1^2 + 8x_1 - 6y_1 + k$.
Given the point is $(-2, 3)$ and the length of the tangent is $4$ units:
$4 = \sqrt{(-2)^2 + (3)^2 + 8(-2) - 6(3) + k}$
$4 = \sqrt{4 + 9 - 16 - 18 + k}$
$4 = \sqrt{k - 21}$
Squaring both sides:
$16 = k - 21$
$k = 16 + 21 = 37$.

(D) The length of the tangent from a point $(x_1, y_1)$ to a circle $S: x^2+y^2+2gx+2fy+c=0$ is given by $\sqrt{S_1} = \sqrt{x_1^2+y_1^2+2gx_1+2fy_1+c}$.
For the first circle $C_1: x^2+y^2+x+y-4=0$,the length of the tangent $L_1$ from $(1,2)$ is:
$L_1 = \sqrt{1^2+2^2+1+2-4} = \sqrt{1+4+1+2-4} = \sqrt{4} = 2$.
For the second circle $C_2: 3x^2+3y^2-x-y-k=0$,we first normalize it to $x^2+y^2-\frac{1}{3}x-\frac{1}{3}y-\frac{k}{3}=0$.
The length of the tangent $L_2$ from $(1,2)$ is:
$L_2 = \sqrt{1^2+2^2-\frac{1}{3}(1)-\frac{1}{3}(2)-\frac{k}{3}} = \sqrt{1+4-\frac{1}{3}-\frac{2}{3}-\frac{k}{3}} = \sqrt{5-1-\frac{k}{3}} = \sqrt{4-\frac{k}{3}}$.
Given the ratio $\frac{L_1}{L_2} = \frac{4}{3}$,we have:
$\frac{2}{\sqrt{4-\frac{k}{3}}} = \frac{4}{3} \Rightarrow \frac{1}{\sqrt{4-\frac{k}{3}}} = \frac{2}{3}$.
Squaring both sides:
$\frac{1}{4-\frac{k}{3}} = \frac{4}{9} \Rightarrow 4-\frac{k}{3} = \frac{9}{4}$.
$\frac{k}{3} = 4 - \frac{9}{4} = \frac{16-9}{4} = \frac{7}{4}$.
$k = 3 \times \frac{7}{4} = \frac{21}{4}$.

(C) Let the point be $P = (f, g)$.
The length of the tangent from a point $(x_1, y_1)$ to a circle $x^2+y^2+2gx+2fy+c=0$ is given by $\sqrt{x_1^2+y_1^2+2gx_1+2fy_1+c}$.
For the circle $S: x^2+y^2-6=0$,the length of the tangent is $L_1 = \sqrt{f^2+g^2-6}$.
For the circle $S': x^2+y^2+3x+3y=0$,the length of the tangent is $L_2 = \sqrt{f^2+g^2+3f+3g}$.
According to the problem,$L_1 = 2L_2$.
Squaring both sides,we get $L_1^2 = 4L_2^2$.
$f^2+g^2-6 = 4(f^2+g^2+3f+3g)$.
$f^2+g^2-6 = 4f^2+4g^2+12f+12g$.
$3f^2+3g^2+12f+12g+6 = 0$.
Dividing by $3$,we get $f^2+g^2+4f+4g+2 = 0$.
Thus,the value of $f^2+g^2+4f+4g+2$ is $0$.

(B) The circle is $x^2+y^2-12x-16y=0$. Its center $C$ is $(6, 8)$ and radius $r = \sqrt{6^2+8^2} = 10$.
Let $P$ be the point of intersection of the tangents. The angle between the tangents is $60^{\circ}$,so the angle between the line joining the center to $P$ and the tangent is $30^{\circ}$.
In $\triangle PAC$,$\sin(30^{\circ}) = \frac{AC}{PC} = \frac{10}{PC} = \frac{1}{2}$,so $PC = 20$.
The distance from the center $(6, 8)$ to the chord $5x-5y+k=0$ is $d = \frac{|5(6)-5(8)+k|}{\sqrt{5^2+(-5)^2}} = \frac{|k-10|}{5\sqrt{2}}$.
In $\triangle PAC$,the distance from $C$ to the chord $AB$ is $d = r \cos(30^{\circ}) = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3}$.
Wait,the angle between the tangents is $60^{\circ}$,so the angle $\angle APC = 30^{\circ}$. In $\triangle PAC$,$\angle PAC = 90^{\circ}$,so $\angle PCA = 60^{\circ}$.
Thus,$d = r \cos(60^{\circ}) = 10 \times \frac{1}{2} = 5$.
Therefore,$\frac{|k-10|}{5\sqrt{2}} = 5 \Rightarrow |k-10| = 25\sqrt{2}$.
$k-10 = \pm 25\sqrt{2} \Rightarrow k = 10 \pm 25\sqrt{2} = 5(2 \pm 5\sqrt{2})$.

(B) The given equation of the circle is $x^2+y^2-4x+6y-12=0$.
Completing the square,we get $(x-2)^2+(y+3)^2 = 12+4+9 = 25 = 5^2$.
Thus,the center is $(2, -3)$ and the radius $r = 5$.
The parametric coordinates are given by $x = 2 + 5\cos\theta$ and $y = -3 + 5\sin\theta$.
For point $P$ with $\theta = \frac{\pi}{3}$,$P = (2 + 5\cos\frac{\pi}{3}, -3 + 5\sin\frac{\pi}{3}) = (2 + \frac{5}{2}, -3 + \frac{5\sqrt{3}}{2}) = (\frac{9}{2}, -3 + \frac{5\sqrt{3}}{2})$.
For point $Q$ with $\theta = \frac{2\pi}{3}$,$Q = (2 + 5\cos\frac{2\pi}{3}, -3 + 5\sin\frac{2\pi}{3}) = (2 - \frac{5}{2}, -3 + \frac{5\sqrt{3}}{2}) = (-\frac{1}{2}, -3 + \frac{5\sqrt{3}}{2})$.
The distance $PQ$ is $\sqrt{(\frac{9}{2} - (-\frac{1}{2}))^2 + (-3 + \frac{5\sqrt{3}}{2} - (-3 + \frac{5\sqrt{3}}{2}))^2} = \sqrt{(\frac{10}{2})^2 + 0^2} = 5$.

(A) Let the diameter be $PR = 2r$. Since $PQ$ and $RS$ are tangents at $P$ and $R$ respectively,$PQ \perp PR$ and $RS \perp PR$.
Let $\angle PRQ = \theta$. In $\triangle PQR$,$\tan \theta = \frac{PQ}{PR}$,so $PR = PQ \cot \theta$.
Since $X$ lies on the circle and $PR$ is the diameter,$\angle PXR = 90^{\circ}$.
In $\triangle PXR$,$\angle XPR = 90^{\circ} - \theta$ and $\angle XRP = \theta$.
In $\triangle PXS$,$\angle XPS = 90^{\circ} - \theta$ and $\angle XSP = \theta$. Thus,$\tan \theta = \frac{RS}{PR}$,so $PR = RS \tan \theta$.
Equating the two expressions for $PR$:
$PQ \cot \theta = RS \tan \theta$
$\tan^2 \theta = \frac{PQ}{RS} \Rightarrow \tan \theta = \sqrt{\frac{PQ}{RS}}$.
Substituting this back into $PR = RS \tan \theta$:
$PR = RS \cdot \sqrt{\frac{PQ}{RS}} = \sqrt{PQ \cdot RS}$.
Since $PR = 2r$,we have $2r = \sqrt{PQ \cdot RS}$.

(C) The equation of the circle is $x^2+y^2-4x-2y-11=0$.
Rewriting this in standard form: $(x-2)^2+(y-1)^2 = 11+4+1 = 16$.
Thus,the center $C$ is $(2,1)$ and the radius $r = \sqrt{16} = 4$.
The point $P$ is $(4,5)$.
The length of the tangent $AP$ from $P(4,5)$ to the circle is given by $\sqrt{S_1} = \sqrt{4^2+5^2-4(4)-2(5)-11} = \sqrt{16+25-16-10-11} = \sqrt{4} = 2$.
The quadrilateral formed is $PACB$,where $A$ and $B$ are points of contact.
This quadrilateral consists of two congruent right-angled triangles $\triangle PAC$ and $\triangle PBC$,both right-angled at $A$ and $B$ respectively.
The area of $\triangle PAC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AC \times AP = \frac{1}{2} \times 4 \times 2 = 4$.
The total area of the quadrilateral $PACB = 2 \times \text{Area}(\triangle PAC) = 2 \times 4 = 8 \text{ sq units}$.

(D) The given lines are $L_1: 3x - 4y + 5 = 0$ and $L_2: 6x - 8y - 9 = 0$.
We can rewrite $L_2$ as $3x - 4y - \frac{9}{2} = 0$.
Since the lines are parallel,the distance between them is equal to the diameter of the circle.
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,$d = \frac{|5 - (-9/2)|}{\sqrt{3^2 + (-4)^2}} = \frac{|5 + 4.5|}{5} = \frac{9.5}{5} = \frac{19/2}{5} = \frac{19}{10}$.
Since the diameter $2r = d$,we have $2r = \frac{19}{10}$,which implies $r = \frac{19}{20}$.
Thus,the correct option is $D$.

(B) Let the centre of the circle be $(h, k)$ and its radius be $r$. Since the circle touches the lines $4x-3y-24=0$ and $4x+3y-42=0$,the perpendicular distance from $(h, k)$ to these lines is equal to $r$.
$r = \left|\frac{4h-3k-24}{5}\right| = \left|\frac{4h+3k-42}{5}\right|$
Also,the circle passes through $(2, 8)$,so $r^2 = (h-2)^2 + (k-8)^2$.
From the distance equality,$4h-3k-24 = \pm(4h+3k-42)$.
Case $1$: $4h-3k-24 = 4h+3k-42$ $\Rightarrow 6k = 18$ $\Rightarrow k = 3$.
Substituting $k=3$ into the radius equation:
$r^2 = \left(\frac{4h-3(3)-24}{5}\right)^2 = \left(\frac{4h-33}{5}\right)^2$
Equating to $(h-2)^2 + (3-8)^2 = (h-2)^2 + 25$:
$\frac{(4h-33)^2}{25} = (h-2)^2 + 25$
$(4h-33)^2 = 25(h^2-4h+4+25) = 25(h^2-4h+29)$
$16h^2 - 264h + 1089 = 25h^2 - 100h + 725$
$9h^2 + 164h - 364 = 0$
$(h-2)(9h+182) = 0$
Since $h \le 8$,we take $h=2$. Thus,the centre is $(2, 3)$ and $r^2 = (2-2)^2 + (3-8)^2 = 25$.
The equation is $(x-2)^2 + (y-3)^2 = 25 \Rightarrow x^2+y^2-4x-6y-12=0$.

(C) The given circle is $x^2+y^2-2x+2y=0$. The center $C$ is $(1, -1)$ and the radius $r$ is $\sqrt{1^2+(-1)^2-0} = \sqrt{2}$.
Let $P = (-1, 1)$. The inverse point $Q$ of $P$ with respect to the circle lies on the line joining the center $C$ and $P$.
The slope of $CP$ is $m = \frac{1 - (-1)}{-1 - 1} = \frac{2}{-2} = -1$.
The equation of the line passing through $C(1, -1)$ and $P(-1, 1)$ is $y - (-1) = -1(x - 1)$,which simplifies to $y + 1 = -x + 1$,or $x + y = 0$.
Since $Q$ must lie on this line,we check the options. None of the options directly represent $x+y=0$. However,the inverse point $Q$ satisfies $CQ \cdot CP = r^2$. Here $CP = \sqrt{(-1-1)^2 + (1-(-1))^2} = \sqrt{4+4} = \sqrt{8} = 2\sqrt{2}$.
$CQ = \frac{r^2}{CP} = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}}$.
Since $Q$ lies on the line $x+y=0$,let $Q = (h, -h)$. Then $CQ^2 = (h-1)^2 + (-h+1)^2 = 2(h-1)^2 = (\frac{1}{\sqrt{2}})^2 = \frac{1}{2}$.
$(h-1)^2 = \frac{1}{4} \implies h-1 = \pm \frac{1}{2} \implies h = \frac{3}{2} \text{ or } h = \frac{1}{2}$.
Since $P$ and $Q$ are on the same side of the center,$Q = (1/2, -1/2)$.
Testing the options for $Q(1/2, -1/2)$: Option $B$ gives $1/2 - (-1/2) + 1 = 2 \neq 0$. Option $C$ gives $1/2 - 1/2 - 2 = -2 \neq 0$. Re-evaluating the question,if the line containing $Q$ is requested,and $Q$ lies on $x+y=0$,there might be a typo in the options. Given the standard form,$x+y=0$ is the line.

(D) Let $O(h, k)$ be the centre of the circle passing through $A(2,3)$,$B(3,-1)$,and $C(-3,2)$.
Since $O$ is the centre,$OA = OB = OC$ (radii of the circle).
$OA^2 = OC^2 \Rightarrow (h-2)^2 + (k-3)^2 = (h+3)^2 + (k-2)^2$
$h^2 - 4h + 4 + k^2 - 6k + 9 = h^2 + 6h + 9 + k^2 - 4k + 4$
$-4h - 6k = 6h - 4k$
$10h + 2k = 0 \Rightarrow k = -5h \quad ... (i)$
$OA^2 = OB^2 \Rightarrow (h-2)^2 + (k-3)^2 = (h-3)^2 + (k+1)^2$
$h^2 - 4h + 4 + k^2 - 6k + 9 = h^2 - 6h + 9 + k^2 + 2k + 1$
$-4h - 6k + 13 = -6h + 2k + 10$
$2h - 8k = -3 \quad ... (ii)$
Substitute $(i)$ into $(ii)$:
$2h - 8(-5h) = -3$
$2h + 40h = -3$ $\Rightarrow 42h = -3$ $\Rightarrow h = -\frac{1}{14}$
$k = -5h = -5(-\frac{1}{14}) = \frac{5}{14}$
Now,calculate $2k - 4h$:
$2(\frac{5}{14}) - 4(-\frac{1}{14}) = \frac{10}{14} + \frac{4}{14} = \frac{14}{14} = 1$

(C) Given the point $P(\alpha, 1)$ lies on the line $y=1$. Let the chord be $PQ$ and its midpoint be $M(h, 0)$ on the $x$-axis.
The equation of the chord of the circle $x^2+y^2-\alpha x-y=0$ with midpoint $(h, k)$ is $T=S_1$.
Here,$T = xh + yk - \frac{\alpha}{2}(x+h) - \frac{1}{2}(y+k)$ and $S_1 = h^2+k^2-\alpha h-k$.
Since the midpoint is $(h, 0)$,we have $T = xh - \frac{\alpha}{2}(x+h) - \frac{1}{2}y$ and $S_1 = h^2-\alpha h$.
So,$xh - \frac{\alpha}{2}x - \frac{\alpha}{2}h - \frac{1}{2}y = h^2-\alpha h$.
Since this chord passes through $P(\alpha, 1)$,we substitute $x=\alpha$ and $y=1$:
$\alpha h - \frac{\alpha^2}{2} - \frac{\alpha h}{2} - \frac{1}{2} = h^2 - \alpha h$.
Multiplying by $2$: $2\alpha h - \alpha^2 - \alpha h - 1 = 2h^2 - 2\alpha h$.
Rearranging terms: $2h^2 - 3\alpha h + \alpha^2 + 1 = 0$.
For two distinct chords to exist,the quadratic equation in $h$ must have two distinct real roots.
Thus,the discriminant $D > 0$:
$(-3\alpha)^2 - 4(2)(\alpha^2+1) > 0$.
$9\alpha^2 - 8\alpha^2 - 8 > 0$.
$\alpha^2 - 8 > 0 \Rightarrow \alpha^2 > 8$.

(C) The equation of the circle is $x^2+y^2=a^2$ and the chord is $x+y=1$.
To find the equation of the pair of lines joining the origin to the intersection points of the circle and the chord,we homogenize the circle equation using the chord equation:
$x^2+y^2=a^2(1)^2$
$x^2+y^2=a^2(x+y)^2$
$x^2+y^2=a^2(x^2+y^2+2xy)$
$(1-a^2)x^2 - 2a^2xy + (1-a^2)y^2 = 0$.
Since the chord subtends a right angle at the origin,the sum of the coefficients of $x^2$ and $y^2$ must be zero:
$(1-a^2) + (1-a^2) = 0$
$2(1-a^2) = 0$
$1-a^2 = 0$
$a^2 = 1$
Since $a$ represents a radius,$a=1$.

(C) The equation of the chord is $y = 2x + 3$.
Substituting this into the circle equation $x^2 + y^2 + 6x - 4y + 4 = 0$:
$x^2 + (2x + 3)^2 + 6x - 4(2x + 3) + 4 = 0$
$x^2 + 4x^2 + 12x + 9 + 6x - 8x - 12 + 4 = 0$
$5x^2 + 10x + 1 = 0$.
The $x$-coordinate of the midpoint $a$ is the average of the roots: $a = \frac{x_1 + x_2}{2} = \frac{-10/5}{2} = -1$.
Since $(a, b)$ lies on the chord $y = 2x + 3$,we have $b = 2a + 3 = 2(-1) + 3 = 1$.
Thus,$2a + 3b = 2(-1) + 3(1) = -2 + 3 = 1$.

(B) The equation of the circle is $x^2+y^2-4x-8y+16=0$.
Given the midpoint of the chord is $(x_1, y_1) = (1,3)$.
The equation of the chord with midpoint $(x_1, y_1)$ is given by $T = S_1$,where $T = xx_1 + yy_1 + g(x+x_1) + f(y+y_1) + c$ and $S_1 = x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c$.
Substituting the values: $x(1) + y(3) - 2(x+1) - 4(y+3) + 16 = 1^2 + 3^2 - 4(1) - 8(3) + 16$.
$x + 3y - 2x - 2 - 4y - 12 + 16 = 1 + 9 - 4 - 24 + 16$.
$-x - y + 2 = -2$.
$-x - y = -4$,which simplifies to $x + y = 4$.
The chord intersects the coordinate axes at $A(0,4)$ and $B(4,0)$.
The triangle formed by the chord and the coordinate axes is a right-angled triangle with base $OB = 4$ and height $OA = 4$.
$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 4 = 8$.

(C) The given equation of the circle is $x^2+y^2-6x-8y+9=0$.
Its centre is $C=(3,4)$ and radius $R=\sqrt{3^2+4^2-9}=\sqrt{9+16-9}=\sqrt{16}=4$.
Let $AB$ be the chord of length $2\sqrt{3}$. The perpendicular distance $CM$ from the centre $C(3,4)$ to the chord $2x+3y+k=0$ is given by $CM = \frac{|2(3)+3(4)+k|}{\sqrt{2^2+3^2}} = \frac{|6+12+k|}{\sqrt{4+9}} = \frac{|18+k|}{\sqrt{13}}$.
In the right-angled triangle $\triangle ACM$,we have $AC^2 = CM^2 + AM^2$,where $AM = \frac{AB}{2} = \frac{2\sqrt{3}}{2} = \sqrt{3}$.
Substituting the values,$4^2 = \left(\frac{|18+k|}{\sqrt{13}}\right)^2 + (\sqrt{3})^2$.
$16 = \frac{(18+k)^2}{13} + 3$.
$13 = \frac{(18+k)^2}{13} \Rightarrow (18+k)^2 = 169$.
Taking the square root,$18+k = \pm 13$.
Case $1$: $18+k = 13 \Rightarrow k = -5$.
Case $2$: $18+k = -13 \Rightarrow k = -31$.
Thus,one of the values of $k$ is $-5$.

(A) Let the centre of the circle be $C(h, k)$. Since $C$ lies in the third quadrant,$h < 0$ and $k < 0$.
The line $AB$ passes through $(1, 2)$ and $(2, 1)$. The equation of line $AB$ is $y - 1 = \frac{2-1}{1-2}(x - 2) \Rightarrow y - 1 = -1(x - 2) \Rightarrow x + y - 3 = 0$.
The distance of $C(h, k)$ from $AB$ is $\frac{|h + k - 3|}{\sqrt{1^2 + 1^2}} = \frac{7}{\sqrt{2}}$.
Since $C$ is in the third quadrant,$h < 0$ and $k < 0$,so $h + k - 3 < 0$. Thus,$-(h + k - 3) = 7 \Rightarrow h + k = -4$.
Since $C$ is equidistant from $A(1, 2)$ and $B(2, 1)$,$CA^2 = CB^2 \Rightarrow (h-1)^2 + (k-2)^2 = (h-2)^2 + (k-1)^2$.
$h^2 - 2h + 1 + k^2 - 4k + 4 = h^2 - 4h + 4 + k^2 - 2k + 1 \Rightarrow 2h = 2k \Rightarrow h = k$.
Substituting $h = k$ into $h + k = -4$,we get $2h = -4 \Rightarrow h = -2, k = -2$. So $C = (-2, -2)$.
The radius $R$ is the distance $CA = \sqrt{(-2-1)^2 + (-2-2)^2} = \sqrt{(-3)^2 + (-4)^2} = \sqrt{9 + 16} = 5$.
The distance of $P(1, -2)$ from $C(-2, -2)$ is $\sqrt{(1 - (-2))^2 + (-2 - (-2))^2} = \sqrt{3^2 + 0^2} = 3$.
Since $3 < 5$,the point $P(1, -2)$ lies inside the circle $S$.

(D) Let the centre of the circle be $O(h, k)$ and radius be $r$. Since $L_1$ is a tangent,the perpendicular distance from $(h, k)$ to $3x - 4y - 8 = 0$ is equal to $r$.
$\frac{|3h - 4k - 8|}{\sqrt{3^2 + (-4)^2}} = r \Rightarrow |3h - 4k - 8| = 5r$. Since $L_1$ is a tangent,$r = 2$,so $|3h - 4k - 8| = 10$.
Also,the distance from $(h, k)$ to $L_2 \equiv x + y - 1 = 0$ is $\sqrt{2}$.
$\frac{|h + k - 1|}{\sqrt{1^2 + 1^2}} = \sqrt{2} \Rightarrow |h + k - 1| = 2$.
Given $h^2 + k^2 = 13$. Solving these equations,we find $(h, k) = (3, 2)$ or $(-2, 3)$.
For $(h, k) = (3, 2)$,the midpoint $(\alpha, \beta)$ of the chord $L_2$ is the projection of the centre $(3, 2)$ on $x + y - 1 = 0$.
$\frac{\alpha - 3}{1} = \frac{\beta - 2}{1} = -\frac{3 + 2 - 1}{1^2 + 1^2} = -\frac{4}{2} = -2$.
$\alpha = 3 - 2 = 1, \beta = 2 - 2 = 0$. Thus,$\alpha + \beta = 1$.
Given $r = 2$,we have $\alpha + \beta + r = 1 + 2 = 3$.

(D) The equation of a chord of a circle $S=0$ with mid-point $(x_1, y_1)$ is given by $T=S_1$.
Here,$S = x^2+y^2-9=0$ and $(x_1, y_1) = (1, 2)$.
$T = x(1) + y(2) - 9 = x+2y-9$.
$S_1 = (1)^2 + (2)^2 - 9 = 1+4-9 = -4$.
Equating $T=S_1$,we get $x+2y-9 = -4$.
Therefore,$x+2y-5=0$.

(A) The given circles are $S_1: x^2+y^2-4x-6y-3=0$ and $S_2: x^2+y^2+8x-4y+11=0$.
Comparing with $x^2+y^2+2gx+2fy+c=0$,we get:
For $S_1$: $g_1 = -2, f_1 = -3, c_1 = -3$.
For $S_2$: $g_2 = 4, f_2 = -2, c_2 = 11$.
The angle $\theta$ between two circles is given by $\cos \theta = \frac{|2g_1g_2 + 2f_1f_2 - c_1 - c_2|}{2\sqrt{g_1^2+f_1^2-c_1}\sqrt{g_2^2+f_2^2-c_2}}$.
Substituting the values:
$2g_1g_2 = 2(-2)(4) = -16$.
$2f_1f_2 = 2(-3)(-2) = 12$.
Numerator: $|-16 + 12 - (-3) - 11| = |-16 + 12 + 3 - 11| = |-12| = 12$.
Denominator: $2\sqrt{(-2)^2+(-3)^2-(-3)} \sqrt{4^2+(-2)^2-11} = 2\sqrt{4+9+3}\sqrt{16+4-11} = 2\sqrt{16}\sqrt{9} = 2(4)(3) = 24$.
Thus,$\cos \theta = \frac{12}{24} = \frac{1}{2}$.
Therefore,$\theta = \frac{\pi}{3}$.