Circle

Circle $C_1 \rightarrow$ radius $r_1, O_1(1,2), r_1=3 r^{\prime}$

Circle $\mathrm{C}_2 \rightarrow$ radius $r_2, \mathrm{O}_2(3,-2), r_2=r^{\prime}$

$\because C_1$ and $C_2$ cuts orthogonally

$ \begin{aligned} & r_1^2+r_2^2=0_1 0_2^2 \\ \Rightarrow \quad & \left(3 r^{\prime}\right)^2+r^{\prime 2}=2^2+4^2 \\ \Rightarrow \quad & 10 r^{\prime 2}=20 \Rightarrow r^{\prime}=2 \end{aligned} $

∴ Equation of required circle

$ \begin{aligned} & \Rightarrow(x-1)^2+(y+2)^2=2 \\ & \Rightarrow x^2+y^2-2 x+4 y+3=0 \end{aligned} $

Given, equation of circles

$ x^2+y^2=16\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) $

And $(x-9)^2+y^2=16\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

$ \begin{aligned} C_1 & =(0,0), r_1=4 \\ C_2 & =(9,0), r_2=4 \\ \therefore \quad C_1 C_2 & =\sqrt{(9-0)^2+(0-0)^2} \\ & =\sqrt{(9)^2}=9 \end{aligned} $

And $r_1+r_2=4+4=8$

Since $C_1 C_2>r_1+r_2$

Circles are separate to each other, any tangent to circle (i) with slope ( $m$ ) is

$ \begin{aligned} & y=m x \pm r \sqrt{1+m^2} \\ \Rightarrow \quad & y=m x+4 \sqrt{1+m^2} \\ \Rightarrow \quad & m x-y \pm 4 \sqrt{1+m^2}=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

Equation (iii) to be tangent to circle (ii)

$ \begin{array}{ll} \therefore & \left|\frac{m(9)-0 \pm 4 \sqrt{1+m^2}}{\sqrt{m^2+1}}\right|=4 \\ \Rightarrow & 16\left(m^2+1\right)=\left(9 m \pm 4 \sqrt{1+m^2}\right)^2 \\ \Rightarrow & 16 m^2+16=81 m^2+16\left(1+m^2\right) \\ & \pm 72 m \sqrt{1+m^2} \end{array} $

$ \begin{array}{ll} \Rightarrow & \quad 81 m^2 \pm 72 m \sqrt{1+m^2}=0 \\ \Rightarrow & 9 m\left(9 m \pm 8 \sqrt{1+m^2}\right)=0 \end{array} $

When $9 m=0 \Rightarrow m=0$

Then Eq. (iii) $y= \pm 4$ (horizontal lines)

If, $9 m \pm 8 \sqrt{1+m^2}=0$

$ \begin{aligned} & \Rightarrow \quad(9 m)^2=\left( \pm 8 \sqrt{1+m^2}\right)^2 \\ & \Rightarrow \quad 81 m^2=64\left(1+m^2\right) \\ & \Rightarrow \quad 81 m^2-64 m^2=64 \\ & \Rightarrow \quad 17 m^2=64 \\ & \Rightarrow \quad m= \pm \sqrt{\frac{64}{17}} \\ & \Rightarrow \quad m= \pm \frac{8}{\sqrt{17}} \end{aligned} $

$ \begin{aligned} &\text { Given, equation of circle }\\ &\begin{aligned} & x^2+y^2-4 x-6 y-12=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & \text { centre }(-g,-f)=(2,3) \\ & \text { And radius }=\sqrt{g^2+f^2-c} \\ & \qquad \begin{aligned} & =\sqrt{4+9+12}=\sqrt{25}=5 \end{aligned} \end{aligned} \end{aligned} $

Equation of new circle whose radius is 3

$ \begin{aligned} & (x-h)^2+(y-k)^2=(3)^2 \\ & \Rightarrow \quad(x-h)^2+(y-k)^2=9 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

Since circle (i) and (ii) touches each other internally at $(-1,-1)$

$ \therefore \quad C_1 C_2=\left|r_1-r_2\right| $

where $c_2=(h, k)$

$ \therefore \quad C_1 C_2=|5-3|=2 $

$P$ divides the line segment $C_1 C_2$ in the ratio $r_1: r_2$

$ \quad \begin{aligned}\therefore C_1 P & =r_1 \text { and } C_2 P=r_2 \\ C_1 P & =\sqrt{(-1-2)^2+(-1-3)^2} \\ & =\sqrt{(-3)^2+(-4)^2} \\ & =\sqrt{9+16}=\sqrt{25}=5 \end{aligned} $

and $C_2 P=r_2=3$

$\therefore$ The point of tangency $P$ divides the line segment $C_1 C_2$ externally in the ratio $r_1:-r$

$ \begin{array}{rlrl} \therefore & P =\frac{r_1 C_2-r C_1}{r_1-r} \\ & =\frac{5 C_2-3 C_1}{5-3}=\frac{5 C_2-3 C_1}{2} \\ \Rightarrow & 2 P =5 C_2-3 C_1 \\ \Rightarrow & 5 C_2 =2 P+3 C_1 \\ \Rightarrow & 5(h, k) =2(-1,-1)+3(2,3) \end{array} $

$ \begin{aligned} &\begin{aligned} & \Rightarrow \quad 5(h, k)=(-2,-2)+(6,9) \\ & \Rightarrow \quad 5(h, k)=(4,7) \Rightarrow(h, k)=\left(\frac{4}{5}, \frac{7}{5}\right) \end{aligned}\\ &\therefore \text { Equation of required circle }\\ &\begin{aligned} & \left(x-\frac{4}{5}\right)^2+\left(y-\frac{7}{5}\right)^2=(3)^2 \\ & \Rightarrow \quad\left(x-\frac{4}{5}\right)^2+\left(y-\frac{7}{5}\right)^2=9 \\ & \Rightarrow \quad x^2+\frac{16}{25}-\frac{8}{5} x+y^2+\frac{49}{25}-\frac{14}{5} y=9 \\ & \quad x^2+y^2-\frac{8}{5} x-\frac{14}{5} y+\frac{13}{5}=9 \\ & \Rightarrow \quad 5 x^2+5 y^2-8 x-14 y-32=0 \end{aligned} \end{aligned} $

If $C_1$ and $C_2$ are two circles having no common points, it means they do not intersect and do not touch each other then there will be no common tangents or there will be four common tangents $C_1$ and $C_2$

Equation of the circle touching the $X$-axis and passing through the point $(-1,1)$

$ \begin{array}{ll} & (-1-h)^2+(1-k)^2=k^2 \\ \Rightarrow & 1+h^2+2 h+1+k^2-2 k=k^2 \\ \Rightarrow & 1+2 h+h^2+1-2 k+k^2=k^2 \\ \Rightarrow & 1+2 h+h^2+1-2 k=0 \\ \Rightarrow & h^2+2 h-2 k+2=0 \\ \Rightarrow & x^2+2 x-2 y+2=0 \\ \Rightarrow & 2 y=x^2+2 x+2 \\ \Rightarrow & y=\frac{1}{2}\left(x^2+2 x\right)+1 \\ \Rightarrow & y=\frac{1}{2}(x+1)^2-\frac{1}{2}+1 \\ \Rightarrow & y=\frac{1}{2}(x+1)^2+\frac{1}{2} \\ \Rightarrow & (x+1)^2=2\left(y-\frac{1}{2}\right) \end{array} $

Which represents parabola with focus $(-1,1)$

Given, equation of circles

$ \begin{aligned} & x^2+y^2+2 x-2 y+1=0 \text { and } \\ & x^2+y^2-2 x+2 y-2=0 \end{aligned} $

$\therefore$ Equation of circle passing through the point of intersection of two given circle is

$ \begin{array}{r} \left(x^2+y^2+2 x-2 y+1\right)+\lambda \left(x^2+y^2-2 x+2 y-2\right)=0 \\ \Rightarrow \quad(1+\lambda) x^2+(1+\lambda) y^2+(2-2 \lambda) x +(-2+2 \lambda) y+(1-2 \lambda)=0 \\ \Rightarrow \quad x^2+y^2+\frac{(2-2 \lambda)}{1+\lambda} x+\frac{(-2+2 \lambda)}{1+\lambda} y +\frac{1-2 \lambda}{1+\lambda}=0 \end{array} $

Coordinate of centre of circle

$ =(-g,-f)=\left(-\frac{1-\lambda}{1+\lambda},-\frac{-1+\lambda}{1+\lambda}\right) $

$ \begin{aligned} & \therefore \text { Radius of } \\ & \begin{aligned} &\left(-\frac{1-\lambda}{1+\lambda}\right)^2+\left(-\frac{-1+\lambda}{1+\lambda}\right)^2-\left(\frac{1-2 \lambda}{1+\lambda}\right)=14 \\ & \Rightarrow \quad \frac{(1-\lambda)^2}{(1+\lambda)^2}+\frac{(\lambda-1)^2}{(1+\lambda)^2}-\frac{1-2 \lambda}{1+\lambda}=14 \\ & \Rightarrow \quad(1-\lambda)^2+(\lambda-1)^2-(1+\lambda)(1-2 \lambda) \\ &=14(1+\lambda)^2 \\ & \Rightarrow \quad 2(\lambda-1)^2-\left(1-2 \lambda+\lambda-2 \lambda^2\right) \\ &=14\left(1+\lambda^2+2 \lambda\right) \\ & \Rightarrow \quad 2\left(\lambda^2+1-2 \lambda-\left(1-\lambda-2 \lambda^2\right)\right. \\ &=14\left(1+\lambda^2+2 \lambda\right) \\ & \Rightarrow \quad 4 \lambda^2+1-3 \lambda=14+14 \lambda^2+28 \lambda \\ & \Rightarrow \quad 10 \lambda^2+31 \lambda+13=0 \end{aligned} \end{aligned} $

$\Rightarrow$ This is a quadrate equation is $\lambda$ the existence of real values for $\lambda$ conforms that such circle exists.

Coordinates of centre

$ \begin{array}{ll} & h=\frac{\lambda-1}{1+\lambda}, k=-\frac{\lambda-1}{1+\lambda} \\ \Rightarrow & h=-k \Rightarrow h+k=0 \\ \Rightarrow & x+y=0 \end{array} $

$ \begin{aligned} &\text { Given equation of circle }\\ &x^2+y^2-14 x+6 y+33=0 \ldots \text { (i) } \end{aligned} $

$ \begin{aligned} & \text { Centre of circle }=(-g,-f) \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, =(7,-3) \\ & \text { and radius of circle }=\sqrt{g^2+f^2-c} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\sqrt{49+9-33} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\sqrt{25}=5 \end{aligned} $

Since, circle cuts the $X$-axis, $y=0$

$ \begin{array}{ll} & x^2-14 x+33=0 \\ \Rightarrow & x^2-11 x-3 x+33=0 \\ \Rightarrow & x(x-11)-3(x-11)=0 \\ \Rightarrow & (x-3)(x-11)=0 \\ \Rightarrow & x=3,11 \\ \therefore & A=(3,0) \text { and } B=(11,0) \end{array} $

Since, $C$ is the mid-point of $A B$

$ \begin{aligned} \therefore \quad C & =\left(\frac{3+11}{2}, \frac{0+0}{2}\right) \\ & =(7,0) \end{aligned} $

$\therefore$ Equation of line $L$, which passes through $C$ and having slope $(-1)$ is

$ \begin{aligned} & y-0=(-1)(x-7) \\ & \Rightarrow \quad y=-x+7 \\ & \Rightarrow \quad x+y-7=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \end{aligned} $

Since, $L$ is the radical axis of the circles $s$ and $S^{\prime}\left(S-S^{\prime}=0\right)$

$ \begin{aligned} & \therefore \quad\left(x^2+y^2-14 x+6 y+33\right) -\left(x^2+y^2+2 g^{\prime} x+2 f^{\prime} y+c^{\prime}\right)=0 \\ & \Rightarrow\left(-14-2 g^{\prime}\right) x+\left(6-2 f^{\prime}\right) y+\left(33-c^{\prime}\right)=0 \end{aligned} $

Which is identical with line (ii)

$ \begin{array}{ll} \therefore & \frac{-14-2 g^{\prime}}{1}=\frac{6-2 f^{\prime}}{1}=\frac{33-c^{\prime}}{-7}=k \\ \Rightarrow & -14-2 g^{\prime}=k \text { and } 6-2 f^{\prime}=k \\ \therefore & -14-2 g^{\prime}=6-2 f^{\prime} \\ \Rightarrow & g^{\prime}-f^{\prime}=-10 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{array} $

Since, centre of $S^{\prime}$ lies on (ii)

$ \begin{aligned} \therefore & & -g^{\prime}-f^{\prime} & =7 \\ \Rightarrow & & g^{\prime}+f^{\prime} & =-7 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)\end{aligned} $

Solving Eqs. (iii) and (iv), we get

$ \begin{array}{ll} & g^{\prime}=\frac{-17}{2}, f^{\prime}=\frac{3}{2} \\ \therefore & -14-2\left(-\frac{17}{2}\right)=k \\ \Rightarrow & k=3 \end{array} $

Also, $\frac{33-c^{\prime}}{-7}=k=3$

$ \begin{array}{ll} \Rightarrow & 33-c^{\prime}=-21 \\ \Rightarrow & c^{\prime}=54 \end{array} $

$\therefore$ Equation of required circle is

$ \begin{aligned} & x^2+y^2+2\left(\frac{-17}{2}\right) x+2\left(\frac{3}{2}\right) y+54=0 \\ & \Rightarrow x^2+y^2-17 x+3 y+54=0 \end{aligned} $

Given equation is

$ a x^2+a y^2+2 g x+2 f y-2=0 $

Put $(-1,0)$ into this equation, we get

$ \begin{aligned} & a(-1)^2+a(0)^2+2 g(-1)+2 f(0)-2=0 \\ & \Rightarrow \quad a-2 g-2=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Put $(-1,1)$, we get

$ \begin{aligned} & a(-1)^2+a(1)^2+2 g(-1)+2 f(1)-2=0 \\ & \Rightarrow \quad a+a-2 g+2 f-2=0 \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad 2 a-2 g+2 f-2=0 \\ & \Rightarrow \quad a-g+f-1=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

Put (1, 1), we get

$ \begin{aligned} & a(1)^2+a(1)^2+2 g(1)+2 f(1)-2=0 \\ & \Rightarrow \quad a+g+f-1=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{aligned} $

Subtract Eq. (ii) from Eq. (iii),

$ \begin{aligned} & (a+g+f-1)-(a-g+f-1)=0 \\ & \Rightarrow \quad 2 g=0 \\ & \Rightarrow \quad g=0 \end{aligned} $

Put $g=0$ into Eq. (i), we get

$ \begin{aligned} & & a-2(0)-2 & =0 \\ \Rightarrow & & a & =2 \end{aligned} $

Given circle, $x-2=5 \cos \theta$, $y+1=5 \sin \theta$, where $\theta$ is the perimeter.

$ \begin{aligned} & \Rightarrow(x-2)^2+(y+1)^2=(5 \cos \theta)^2+(5 \sin \theta)^2 \\ & =25\left(\cos ^2 \theta+\sin ^2 \theta\right)=25 \end{aligned} $

$\therefore$ The circle has a centre at $(h, k)=(2,-1)$ and a radius, $r_c=5$

Now, the line equation is $x=1+\frac{r}{2}$ and $y=-2+\frac{\sqrt{3}}{2} r$. This is parametric form.

Let $x_1=1, y_1=-2$

The direction vector of the line is $\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$

So, the slope of the line, $m=\frac{\Delta y}{\Delta x}$

$ \Rightarrow \quad \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3} $

Now, the equation of line is

$ \begin{array}{cc} & y-y_1=m\left(x-x_1\right) \\ \Rightarrow & y-(-2)=\sqrt{3}(x-1) \\ \Rightarrow & y+2=\sqrt{3} x-\sqrt{3} \\ \Rightarrow & \sqrt{3} x-y-(2+\sqrt{3})=0 \end{array} $

Now, calculate the distance from the center of the circle to line,

$ d=\frac{\left|A x_0+B y_0+C\right|}{\sqrt{A^2+B^2}} $

Here, $\left(x_0, y_0\right)=(2,-1)$ and the line is $\sqrt{3} x-y-(2+\sqrt{3})=0$

So, $A=\sqrt{3}, B=-1, C=-(2+\sqrt{3})$

$ \begin{aligned} d & =\frac{|\sqrt{3}(2)+(-1)(-1)-(2+\sqrt{3})|}{\sqrt{(\sqrt{3})^2+(-1)^2}} \\ & =\frac{|2 \sqrt{3}+1-2-\sqrt{3}|}{\sqrt{3+1}} \\ & =\frac{\sqrt{3}-1}{2} \approx 0.366<5 \end{aligned} $

So, the line intersects the circle at two distinct points, meaning it is a secant to the circle.

So, the line is a chord of circle other than diameter.

The equation of the circle is

$ \begin{aligned} & x^2+y^2-6 x+2 y+c=0 \\ \Rightarrow \quad & (x-3)^2+(y+1)^2=10-c \end{aligned} $

So, the center of the circle is $C(3,-1)$ and the radius is $r^2=10-C$

Now, since the equation of tangent is

$ x-2 y=0 $

Slop, $m_t=1 / 2$

Slope of the radius, connecting the centre $(3,-1)$ to the point of tangency

$ P\left(x_p, y_p\right) \text { is } m_r=\frac{y_p-(-1)}{x_p-3}=\frac{y_p+1}{x_p-3} $

Now, since radius is perpendicular to the tangent, so

$ \begin{aligned} & m_t \cdot m_r=-1 \\ \Rightarrow \quad & \frac{1}{2} \cdot \frac{y_p+1}{x_p-3}=-1 \Rightarrow y_p+1=-2 x_p+6 \\ \Rightarrow \quad & y_p=-2 x_p+5 \end{aligned} $

But, $P$ lies on the tangent line, its coordinates satisfy $x_p-2 y_p=0$

$ \begin{array}{ll} \Rightarrow & x_p=2 y_p \\ \text { So, } & y_p=-2 x_p+5 \\ \Rightarrow & -2\left(2 y_p\right)+5=-4 y_p+5 \\ \Rightarrow & 5 y_p=5 \Rightarrow y_p=1 \end{array} $

So, $x_p=2 y_p=2 \times 1=2$

So, the point $P$ is $(2,1)$.

Now, the distance between $(6,3)$ and $(2,1)$ is

$ \begin{aligned} & d=\sqrt{(2-6)^2+(1-3)^2} \\ \Rightarrow & \sqrt{(-4)^2+(-2)^2} \Rightarrow \sqrt{16+4}=\sqrt{20}=2 \sqrt{5} \end{aligned} $

$\therefore$ The distance of the point $(6,3)$ from $P$ is $2 \sqrt{5}$.

The equation of the circle is

$ x^2+y^2-4 x+2 y-4=0 $

Comparing this with the general form

$ x^2+y^2+2 g x+2 f y+c=0 $

We have

$ \begin{aligned} & & 2 g & =-4 \\ \Rightarrow & & g & =-2 \text { and } 2 f=2 \\ \Rightarrow & & f & =1 \end{aligned} $

So, the center of the circle $C$ is

$ (-g,-f)=(2,-1) $

And, the radius is $r=\sqrt{g^2+f^2-c}$

$ \begin{aligned} & =\sqrt{(-2)^2+(1)^2-(-4)} \\ & =\sqrt{4+1+4}=\sqrt{9}=3 \end{aligned} $

The point from which tangents are drawn is $P=(-3,1)$

Now, the circumcircle of $\triangle P A B$ is the circle with $P C$ as its diameter. The center of this circumcircle is the mid-point of $P C$.

Mid-point $M=\left(\frac{-3+2}{2}, \frac{1+(-1)}{2}\right)$

$ \Rightarrow\left(\frac{-1}{2}, \frac{0}{2}\right)=\left(\frac{-1}{2}, 0\right) $

Distance $P C=\sqrt{(2-(-3))^2+(-1-1)^2}$

$ \begin{aligned} & \Rightarrow \quad \sqrt{5^2+(-2)^2} \\ & \Rightarrow \quad \sqrt{25+4}=\sqrt{29} \end{aligned} $

$\therefore$ The radius of the circumcircle

$=\frac{1}{2} \cdot$ distance $P C=\frac{\sqrt{29}}{2}$

Now, the equation of a circle with center ( $h, k$ ) and radius $R$ is

$ (x-h)^2+(y-k)^2=R^2 $

Here, $(h, k)=\left(\frac{-1}{2}, 0\right)$ and $R=\frac{\sqrt{29}}{2}$

So, $\left(x-\left(-\frac{1}{2}\right)\right)^2+(y-0)^2=\left(\frac{\sqrt{29}}{2}\right)^2$

$ \begin{aligned} & \Rightarrow \quad\left(x+\frac{1}{2}\right)^2+y^2=\frac{29}{4} \\ & \Rightarrow \quad x^2+x+\frac{1}{4}+y^2=\frac{29}{4} \\ & \Rightarrow \quad x^2+y^2+x-\frac{28}{4}=0 \\ & \Rightarrow \quad x^2+y^2+x-7=0 \end{aligned} $

Let the equation of circle $C$ be

$ x^2+y^2+2 g x+2 f y+c=0 $

Since, it passes through $(1,1)$ then

$ \begin{array}{r} 1^2+1^2+2 g+2 f+c=0 \\ \Rightarrow \quad 2+2 g+2 f+c=0         \,\,\,...(i)\end{array} $

Now, $S_1: x^2+y^2+2 g x+2 f y+c=0$

and $S_2=x^2+y^2-2 x=0$

So, $S_1-S_2=0$ is the common chord.

$ \Rightarrow(2 g+2 x+2 f y+c=0 $

Now, center of $s_2=x^2+y^2-2 x=0$ is $(1,0)$

Since, the common chord is a diameter of $S_2$, then centre of $S_2$ must lies on the common chord.

$ \begin{array}{ll} \therefore & (2 g+2(l)+2 f(0)+c=0 \\ \Rightarrow & 2 g+2+c=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{array} $

Also, circle $C$ is orthogonal to

$ x^2+y^2+2 y-3=0 $

Here, $g_1=g, f_1=f, c_1=c$ and $g_2=0$,

$ f_2=1, c_2=-3 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$

Using the orthogonality condition

$ \begin{aligned} & 2 g_1 g_2+2 f_1 f_2=c_1+c_2 \\ \Rightarrow \quad & 2 g(0)+2 f(1)=c+(-3) \\ \Rightarrow \quad & 2 f=c-3 \end{aligned} $

From Eq. (ii), $c=-2 g-2$

Put this in Eq. (iii), we get

$ \begin{aligned} & 2 f=c-3=-2 g-2-3 \\ \Rightarrow \quad & 2 f=-2 g-5 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)\end{aligned} $

Substitute $c=-2 g-2$ into Eq. (i), we get

$ \begin{array}{ll} \Rightarrow & 2+2 g+2 f-2 g-2=0 \\ \Rightarrow & 2 f=0 \\ \Rightarrow & f=0 \end{array} $

Put $f=0$ in Eq. (iv), we get

$ \begin{aligned} & & 2(0) & =-2 g-5 \\ \Rightarrow & & 2 g & =-5 \\ \Rightarrow & & g & =\frac{-5}{2} \end{aligned} $

$ \begin{aligned} &\text { The center of circle } C \text { is }\\ &\begin{aligned} & (-g,-f)=\left(-\left(-\frac{5}{2}\right), 0\right) \\ \Rightarrow & \left(\frac{5}{2}, 0\right) \end{aligned} \end{aligned} $

For $S_1: x^2+y^2-2 x+k y+1=0$,

Center $C_1=\left(1,-\frac{k}{2}\right)$

Radius, $r_1=\sqrt{1^2+\left(-\frac{k}{2}\right)^2-1}=\sqrt{\frac{k^2}{4}}=\frac{k}{2}$

For $S_2: x^2+y^2-k x-2 y+1=0$

Center $C_2=\left(\frac{k}{2}, 1\right)$

Radius, $r_2=\sqrt{\left(\frac{k}{2}\right)^2+1^2-1}$

$ \sqrt{\frac{k^2}{4}}=\frac{k}{2} $

Since, angle $\theta$ between two circles is given by $\cos \theta=\frac{d^2-r_1^2-r_2^2}{2 r_1 r_2}$, where $d$ is the distance between the centers.

Given, $\theta=\cos ^{-1}\left(\frac{1}{4}\right)$

$ \begin{aligned} \Rightarrow \quad \cos \theta & =\frac{1}{4} \\ d^2 & =\left(1-\frac{k}{2}\right)^2+\left(\frac{-k}{2}-1\right)^2 \\ & =1-k+\frac{k^2}{4}+1+k+\frac{k^2}{4} \\ & =2+\frac{k^2}{2} \end{aligned} $

Since $k<0, r_1=\frac{-k}{2}$ and $r_2=\frac{-k}{2}$

Substituting into the formula

$ \frac{1}{4}=\frac{\left(2+\frac{k^2}{2}\right)-\left(\frac{-k}{2}\right)^2-\left(\frac{-k}{2}\right)^2}{2\left(\frac{-k}{2}\right)\left(\frac{-k}{2}\right)} $

$ \begin{array}{ll} \Rightarrow & \frac{1}{4}=\frac{2+\frac{k^2}{2}-\frac{k^2}{4}-\frac{k^2}{4}}{2\left(\frac{k^2}{4}\right)} \\ \Rightarrow & \frac{1}{4}=\frac{4}{k^2} \\ \Rightarrow & k^2=16 \\ \Rightarrow & k= \pm 4 \end{array} $

But since $k<0, k=-4$

Now, the equation of the radical axis is

$ \begin{aligned} & S_1-S_2=0 \\ & \left(x^2+y^2-2 x+k y+1\right)-\left(x^2+y^2-k x\right. \\ & -2 y+1)=0 \\ & \Rightarrow-2 x+k y+k x+2 y=0 \\ & \Rightarrow-6 x-2 y=0 \,\,\,\,(\because k=-4)\\ & \Rightarrow 3 x+y=0 \end{aligned} $

The points satisfying this equation are $(-1,3)$ and $(1,-3)$ lies on the radical axis.

$ m=\left(\frac{2+4}{2}, \frac{5+3}{2}\right)=(3,4) $

and $m Q \leq 5$

So, $P$ should be in the shaded region thus, $A B:(y-3)=\frac{3-5}{4-2}(x-4)$

$ \begin{aligned} & \Rightarrow \quad y=-x+4+3 \\ & \Rightarrow \quad x+y-7=0 \end{aligned} $

and equation of the semi-circle

$ \begin{aligned} & (x-3)^2+(y-4)^2=(5)^2 \\ & \Rightarrow \quad x^2-6 x+9+y^2-8 y+16=25 \\ & \Rightarrow \quad x^2+y^2-6 x-8 y=0 \end{aligned} $

Hence, shaded region is

$ x^2+y^2-6 x-8 y \leq 0, x+y-7<0 $

$x^2+y^2-2 x-4 y-4=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

$ \begin{aligned} & c_1=(1,2) \\ & \text { and } r_1=\sqrt{(1)^2+(2)^2-(-4)}=3 \\ & \quad x^2+y^2+2 x+4 y-11=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ & c_2=(-1,-2) \text { and } \\ & r_2=\sqrt{(-1)^2+(-2)^2-(-11)}=4 \\ & c_1 c_2=\sqrt{(-1-1)^2+(-2-2)^2} \\ & \quad=\sqrt{18} < r_1+r_2=7 \end{aligned} $

So, both circles intersect, each others.

Subracting Eq. (i) from Eq. (ii), we get $4 x+8 x-7=0$

Hence, intersects at the points lying on the line $4 x+8 y-7=0$

$ \begin{aligned} & \text { } x^2+y^2-6 x+4 y-12=0 \\ & \Rightarrow\left(x^2-6 x+9\right)+\left(y^2+4 y+4\right)=12+9+4 \\ & \Rightarrow(x-3)^2+(y+2)^2=(5)^2 \end{aligned} $

Centre $=(3,-2)$ and radius $=5$

The line $4 x-3 y+7=0$ touches the circle at the point ( $\alpha, \beta$ )

Foot of perpendicular from $(3,-2)$ to the line

$ \begin{aligned} & 4 x-3 y+7=0 \text { is } \\ & \frac{x-x_0}{a}=\frac{y-y_0}{b}=-\frac{a x_0+b y_0+c}{a^2+b^2} \\ & \Rightarrow \frac{x-3}{4}=\frac{y+2}{-3}=-\frac{4(3)-3(-2)+7}{3^2+4^2}=-1 \end{aligned} $

Thus, $\frac{x-3}{4}=-1 \Rightarrow x=-1$

and $\frac{y+2}{-3}=-1 \Rightarrow y=+1$

so, $(\alpha, \beta)=(-1,1)$

Hence, $\alpha+2 \beta=-1+2=1$

$ \begin{aligned} & \text { } x^2-4 x+y^2+12 y=216 \\ & \Rightarrow(x-2)^2+(y+6)^2=256 \\ & c_1=(2,-6), r_1=16 \\ & \text { and } x^2+6 x+y^2-12 y=-36 \\ & \Rightarrow(x+3)^2+(y-6)^2=9 \\ & \qquad \begin{aligned} \therefore c_2 & =(-3,6), r_2=3 \\ c_1 c_2 & =\sqrt{(2+3)^2+(-6-6)^2} \\ & =\sqrt{169}=13 \end{aligned} \end{aligned} $

and $\quad r_1-r_2=16-3=13$

$\because c_1 c_2=\left|r_1-r_2\right|$ the circle touches internally.

then, the slope of line joining centers

$ =\frac{6-(-6)}{3-2}=-12 / 5 $

$\therefore$ the slope of tangent

$ =\frac{-1}{(-12 / 5)}=5 / 12 $

The four given circles are

$ (x \pm r)^2+(y \pm r)^2=r^2 $

centers are $(r, r),(r,-r),(-r,-r),(-r, r)$ the circle touches all four given centered at the origin $(0,0)$, due to the symmetry.

One circle is internally tangent to the region by the four circles, with radius $\left(r_1\right)=r(\sqrt{2}-1)$

Other circles is externally tangent to the region, with radius $\left(r_2\right)=r(\sqrt{2}+1)$

$ \therefore r_1+r_2=r(\sqrt{2}-1+\sqrt{2}+1)=2 \sqrt{2} r $

Hence, $\frac{r_1+r_2}{r}=\frac{2 \sqrt{2} r}{r}=2 \sqrt{2}$

$\because x^2+y^2+x-3 y-10=0$ and $x^2+y^2+2 x-y-20=0$

So, common chord will be

$ \begin{aligned} & \left(x^2+y^2+x-3 y-10\right) -\left(x^2+y^2+2 x-y-20\right)=0 \\ & \Rightarrow x+2 y-10=0 \Rightarrow x=10-2 y \end{aligned} $

put the ' $x$ ' in first circle

$ \begin{aligned} & (10-2 y)^2+y^2+(10-2 y)-3 y-10=0 \\ & \Rightarrow 100+4 y^2-40 y+y^2-2 y+10 -3 y-10=0 \end{aligned} $

$ \begin{aligned} & \Rightarrow y^2-9 y+20=0 \\ & \Rightarrow(y-5)(y-4)=0 \\ & \text { when } y=4 \Rightarrow x=2 \\ & \text { when } y=5 \Rightarrow x=0 \end{aligned} $

so, intersection points are $(2,4)$ and $(0,3)$ which are end-points of diameters.

So, equation of the circle

$ \begin{array}{ll} & (x-0)(x-2)+(y-4)(y-5)=0 \\ \Rightarrow \quad & x(x-2)+(y-4)(y-5)=0 \\ \Rightarrow \quad & x^2+y^2-2 x-9 y+20=0 \end{array} $

Comparing with

$ \begin{aligned} & x^2+y^2+\alpha x+\beta y+\gamma=0 \\ & \alpha=-2, \beta=-9 \text { and } \gamma=20 \end{aligned} $

Therefore, $\alpha+2 \beta+\gamma$

$ =-2-18+20=0 $

$\therefore \triangle A B C$ is right-angle triangle, right angle at $A$

$ \begin{array}{ll} \therefore & m_{A C} \cdot m_{A B}=-1 \\ & \frac{y-5}{x-4} \cdot \frac{y-2}{x-1}=-1 \\ \Rightarrow & (x-4)(x-1)+(y-5)(y-2)=0 \\ \Rightarrow & x^2+y^2-5 x-7 y+14=0 \end{array} $

$ \begin{aligned} &\text { Here, } r=\left|\frac{4 r+3 r-6}{5}\right|\\ &\begin{array}{lll} 5 r=7 r-6 & -5 r=7 r-6 \\ 6=2 r & \text { or } & r=\frac{6}{12} \\ r=3 & \text { or } & r=\frac{1}{2} \end{array} \end{aligned} $

∴ Centre $(3,3)$ or $\left(\frac{1}{2}, \frac{1}{2}\right)$

∵ Circle lies below the line

∴ Centre $\left(\frac{1}{2}, \frac{1}{2}\right), r=\frac{1}{2}$

∴ Equation of circle,

$ \begin{aligned} & \Rightarrow \quad\left(x-\frac{1}{2}\right)^2+\left(y-\frac{1}{2}\right)^2=\left(\frac{1}{2}\right)^2 \\ & \Rightarrow \quad 4 x^2-4 x+1+4 y^2-4 y+1=1 \\ & \Rightarrow \quad 4 x^2+4 y^2-4 x-4 y+1=0 \end{aligned} $

Equation of circle passing through $x^2+y^2-a^2=0$ and $x \cos \alpha+y \sin \alpha-p=0$, is

$ \begin{aligned} & S+\lambda L=0 \\ & \left(x^2+y^2-a^2\right)+\lambda(x \cos \alpha+y \sin \alpha-p)=0 \\ & x^2+y^2+x(\lambda \cos \alpha)+y(\lambda \sin \alpha)-a^2-\lambda p \\ & =0 \\ & \therefore r=\sqrt{\frac{\lambda^2 \cos ^2 \alpha}{4}+\frac{\lambda^2 \sin ^2 \alpha}{4}+a^2+\lambda p} \\ & \Rightarrow r=\sqrt{\frac{\lambda^2}{4}+\lambda p+a^2} \\ & =\sqrt{\left(\frac{\lambda}{2}+p\right)^2+a^2-p^2} \\ & \quad \min r \Rightarrow \frac{\lambda}{2}+p=0 \\ & \Rightarrow \quad \lambda=-2 p \end{aligned} $

$ \begin{aligned} & L_1: 3 x-4 y+4=0 \\ & \quad L_2: 3 x-4 y-\frac{7}{2}=0 \end{aligned} $

$ \begin{aligned} &\because L_1 \| L_2\\ &\therefore 2 r=\text { Distance between } L_1 \text { and } L_2\\ &\begin{aligned} = & \left|\frac{4+\frac{7}{2}}{\sqrt{3^2+4^2}}\right|=\left|\frac{15}{2 \times 5}\right| \\ 2 r & =\frac{3}{2} \Rightarrow r=\frac{3}{4} \\ \therefore \text { Area }= & \pi r^2=\frac{9 \pi}{16} \end{aligned} \end{aligned} $

Let equation of circle be

$ x^2+y^2+2 g x+2 f y+c=0 $

It passes through $(2,0)$

$ \begin{aligned} \therefore \quad 4+4 g+c & =0 \\ x_{\text {intercept }} & =5 \\ 2 \sqrt{g^2-c} & =5 \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad g^2+4+4 g=\frac{25}{4} \\ & \Rightarrow \quad 4 g^2+16+16 g-25=0 \\ & \Rightarrow \quad 4 g^2+16 g-9=0 \\ & \quad g=\frac{-9}{2}, \frac{1}{2} \end{aligned} $

∵ Centre lies in 1st quadrant $\therefore g<0$

$ \begin{array}{ll} \therefore & g=-\frac{9}{2} \\ \Rightarrow & c=-4-4 g=14 \end{array} $

Equation of circle,

$ x^2+y^2-9 x+2 f y+14=0 $

∵ Centre lies in Ist quadrant

∴ Equation $x^2+y^2-9 x-2 k y+14=0$,

$k \in R^{+}$

Let the centroid of the triangle be at $(h, k)$.

Step 1: Finding the Centroid Coordinates

The centroid's $x$-coordinate ($h$) is the average of all three $x$-coordinates: $ h = \frac{\cos \alpha + \sin \alpha + 1}{3} $ So, $ 3h - 1 = \cos \alpha + \sin \alpha $

The centroid's $y$-coordinate ($k$) is the average of all three $y$-coordinates: $ k = \frac{\sin \alpha - \cos \alpha + 2}{3} $ So, $ 3k - 2 = \sin \alpha - \cos \alpha $

Step 2: Squaring Both Sides

Square both sides of each equation to remove the trigonometric functions:

$ (3h - 1)^2 = (\cos \alpha + \sin \alpha)^2 \quad \cdots (i) $

$ (3k - 2)^2 = (\sin \alpha - \cos \alpha)^2 \quad \cdots (ii) $

Step 3: Add the Two Equations

Add equations $(i)$ and $(ii)$: $ (3h - 1)^2 + (3k - 2)^2 = (\cos \alpha + \sin \alpha)^2 + (\sin \alpha - \cos \alpha)^2 $

Notice that $(\cos \alpha + \sin \alpha)^2 + (\sin \alpha - \cos \alpha)^2 = 2\sin^2 \alpha + 2\cos^2 \alpha = 2 $

So, $ (3h - 1)^2 + (3k - 2)^2 = 2 $

Step 4: Locus Equation

Replace $h$ by $x$ and $k$ by $y$ because the locus is about every possible point $(x, y)$: $(3x - 1)^2 + (3y - 2)^2 - 2 = 0$

Step 5: Expand to Standard Form

Expanding and simplifying:

$ \begin{aligned} & (3x - 1)^2 + (3y - 2)^2 - 2 = 0 \\ & 9x^2 - 6x + 1 + 9y^2 - 12y + 4 - 2 = 0 \\ & 9x^2 + 9y^2 - 6x - 12y + 3 = 0 \\ & 3(x^2 + y^2) - 2x - 4y + 1 = 0 \end{aligned} $

$ \because A B \text { will be the diameter } $

$ \begin{aligned} &\text { ∴ Let equation of circle }\\ &x^2+y^2+2 g x+2 f y=0 \end{aligned} $

( $a, 0$ ) lies on the circle

$ \therefore \quad a^2+2 g a=0 \Rightarrow a=-2 g $

$(0, b)$ lies on the circle

$ \therefore \quad b^2+2 f b \Rightarrow b=-2 f $

∴ Equation of line $A B$

$ \begin{aligned} & \frac{x}{-2 g}+\frac{y}{-2 f}=1\left[\because\left(x_1, y_1\right) \text { lies on } L\right] \\ \Rightarrow \quad & \frac{x}{-g}+\frac{y}{-f}=2 \Rightarrow \frac{x_1}{-g}+\frac{y_1}{-f}=2 \end{aligned} $

∴ Locus is centre $(-g,-f)$ is $\frac{x_1}{x}+\frac{y_1}{y}=2$

Given, $C_1: x^2+y^2=3$,

$ O_1(0,0), r_1=\sqrt{3} $

and $C_2: x^2+y^2-2 x+4 y+4=0$

$ O_2(1,-2), r_2=1 $

Point $P$ divides $O_1 O_2$ externally in $r_1: r_2$

$ \begin{aligned} \therefore & \alpha=\frac{0 \times \sqrt{3}-1(1)}{\sqrt{3}+1}=\frac{-1}{\sqrt{3}+1} \text { and } \\ & \beta=\frac{0 \times \sqrt{3}-(-2)}{\sqrt{3}+1}=\frac{2}{\sqrt{3}+1} \\ \Rightarrow \quad & \frac{\beta}{\alpha}=-2 \end{aligned} $

$ \begin{aligned} &\text { Given, }|x-2|+|y-3|=4\\ &\begin{array}{ll} \therefore & x-2+y-3=4 \\ \Rightarrow & x+y=9 \text { and } x-2-y+3=4 \\ \Rightarrow & x-y=3 \text { and }-x+2+y-3=4 \\ \Rightarrow & x-y=-5 \text { and }-x+2-y+3=4 \\ \Rightarrow & x+y=1 \end{array} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \therefore \text { Centre }\left(\frac{4+0}{2}, \frac{5+1}{2}\right)=(2,3) \\ & \text { and } r=\frac{\sqrt{4^2+4^2}}{2}=2 \sqrt{2} \end{aligned}\\ &\text { ∴ Equation of circle : }\\ &\begin{aligned} & (x-2)^2+(y-3)^2=(2 \sqrt{2})^2 \\ \Rightarrow \quad & x^2+y^2-4 x-6 y+5=0 \end{aligned} \end{aligned} $

$ \begin{gathered} A B=\sqrt{1^2+9}=\sqrt{10} \\ A M=\frac{\sqrt{10}}{2} \end{gathered} $

$ \therefore O A=r=\sqrt{\frac{10}{4}+\frac{10}{4}}=\sqrt{\frac{20}{4}}=\sqrt{5} $

Let $O(\alpha, \beta)$

$ \begin{array}{ll} & O A=O B \\ & (\alpha-1)^2+(\beta-2)^2=(\alpha-2)^2+(\beta+1)^2 \\ \Rightarrow & -2 \alpha+1-4 \beta+4=-4 \alpha+4+2 \beta-1 \\ \Rightarrow & 2 \alpha-6 \beta=0 \\ \Rightarrow & \quad \alpha=3 \beta\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \\ \because & m_{O A} \cdot m_{O B}=-1 \\ \Rightarrow & \quad \frac{\beta-2}{\alpha-1} \cdot \frac{\beta+1}{\alpha-2}=-1 \\ \Rightarrow & (\beta-2)(\beta+1)+(\alpha-1)(\alpha-2)=0 \\ \Rightarrow & \beta^2-\beta-2+\alpha^2-3 \alpha+2=0 \\ \Rightarrow & \beta^2-\beta+9 \beta^2-9 \beta=0 \\ \Rightarrow & 10 \beta^2=10 \beta \\ \Rightarrow & \quad \beta=0,1 \\ & \quad \alpha=0,3 \end{array} $

∴ Centre $(0,0)$ or $(3,1)$

∴ Equation of circle

$ \begin{aligned} & x^2+y^2=5 \text { or }(x-3)^2+(y-1)^2=5 \\ & \Rightarrow x^2+y^2-6 x-2 y+5=0 \end{aligned} $

Given,

$ \begin{aligned} & \quad S_1: 4(x-1)^2+4(y-1)^2=1 \\ & \Rightarrow(x-1)^2+(y-1)^2=\frac{1}{4} \\ & \qquad O_1(1,1), r_1=\frac{1}{2} \\ & S_2: 4(x+1)^2+4(y-1)^2=1 \\ & \Rightarrow(x+1)^2+(y-1)^2=\frac{1}{4} \\ & O_2(-1,1), r_2=1 / 2 \\ & \text { and } S_3: 4(x+1)^2+4(y+1)^2=1 \\ & \Rightarrow(x+1)^2+(y+1)^2=\frac{1}{4} \\ & O_3(-1,-1), r_3=\frac{1}{2} \\ & \because s_1-s_2=0 \\ & \Rightarrow-2 x-2 y+2 x-2 y=0 \\ & \quad x=0 \\ & \quad S_2-s_3=0 \\ & \Rightarrow \quad 2 y=0 \\ & \Rightarrow \quad y=0 \end{aligned} $

∴ Centre of required circle $=(0,0)$

$ \text { radius }=L=\sqrt{s_1} $

$ \begin{aligned} &=\sqrt{1+1-\frac{1}{4}}=\sqrt{\frac{7}{4}}\\ &\text { ∴ Equation of circle }\\ &\begin{aligned} & x^2+y^2=\frac{7}{4} \\ \Rightarrow & 4 x^2+4 y^2=7 \end{aligned} \end{aligned} $

Let the coordinates of the centroid be $G(x, y)$ of the $\triangle A P Q$ is

$ x=\frac{x_A+x_P+x_Q}{3}, y=\frac{y_A+y_P+y_Q}{3} $

Given the points $A(a, 0), P(a \cos \theta, a \sin \theta)$ and $Q(b \sin \theta,-b \cos \theta)$

The coordinates of the centroid are

$ \begin{aligned} & x=\frac{a+a \cos \theta+b \sin \theta}{3} \\ & y=\frac{0+a \sin \theta-b \cos \theta}{3} \end{aligned} $

Let $x=h$ and $y=k$, then,

$ \begin{aligned} & h=\frac{a+a \cos \theta+b \sin \theta}{3}, k=\frac{a \sin \theta-b \cos \theta}{3} \\ & \Rightarrow 3 h-a=a \cos \theta+b \sin \theta \\ & 3 k=a \sin \theta-b \cos \theta \end{aligned} $

Squaring and adding these equations

$ \begin{aligned} (3 h-a)^2+(3 k)^2=(a \cos \theta & +b \sin \theta)^2 +(a \sin \theta-b \cos \theta)^2 \end{aligned} $

$ \begin{aligned} = & a^2 \cos ^2 \theta+b^2 \sin ^2 \theta+2 a b \cos \theta \sin \theta +a^2 \sin ^2 \theta+b^2 \cos ^2 \theta-2 a b \cos \theta \sin \theta \\ = & a^2+b^2 \end{aligned} $

Replacing $h$ with $x$ and $k$ with $y$, we get the locus of the centroid as

$ \begin{aligned} & (3 x-a)^2+(3 y)^2=a^2+b^2 \\ \Rightarrow \quad & \left(x-\frac{a}{3}\right)^2+y^2=\frac{a^2+b^2}{9} \\ \Rightarrow \quad & \left(x-\frac{a}{3}\right)^2+y^2=\left(\frac{\sqrt{a^2+b^2}}{3}\right)^2 \end{aligned} $

So, the locus of the centroid is

$ \left(x-\frac{a}{3}\right)^2+y^2=\left(\frac{\sqrt{a^2+b^2}}{3}\right)^2 $

It is a circle with centre at $\left(\frac{a}{3}, 0\right)$ and radius $\frac{\sqrt{a^2+b^2}}{3}$

Given, lines are $x+2 y-2=0$ and

$ 2 x+3 y-1=0 $

Solving these two lines, we get

$ x=-4 \text { and } y=3 $

Thus, the center of circle is $(-4,3)$.

Since, the general eq. of circle is $(x-h)^2+(y-k)^2=r^2$, where $(h, k)$ is the center and $r$ is the radius.

$ \begin{aligned} & \text { So, }(x-(-4))^2+(y-3)^2=r^2 \\ & \Rightarrow \quad(x+4)^2+(y-3)^2=r^2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

But, the circle passes through the point $(8,8)$,

$ \begin{aligned} & \text { So, }(8+4)^2+(8-3)^2=r^2 \\ & 12^2+5^2=r^2 \\ &\,\,\,\,\,\, \quad 169=r^2 \\ & \Rightarrow \quad r=13 \end{aligned} $

Now, from Eq. (i), we get

$ \begin{aligned} \quad(x+4)^2+(y-3)^2=r^2 & =169 \\ \Rightarrow \quad x^2+8 x+16+y^2-6 y+9 & =169 \\ \Rightarrow \quad x^2+y^2+8 x-6 y-144 & =0 \end{aligned} $

Comparing with the given equation

$ x^2+y^2+p x+q x+r=0 $

We get $p=8, q=-6, r=-144$

$ \begin{aligned} \therefore \quad p^2+q^2+r & =8^2+(-6)^2+(-144) \\ & =64+36-144 \\ & =-44 \end{aligned} $

Given circle is

$ x^2+y^2-2 x+4 y+3=0 $

So, center $=(-g,-f)=\left(\frac{-2}{2}, \frac{4}{2}\right)$

$ =(-1,2) \text { and } c=3 $

Also, the equation of the polar of a point $\left(x_1, y_1\right)$ w.r.t the circle $x^2+y^2+2 g x+2 f y+c=0$ is

$ x x_1+y y_1+g\left(x+x_1\right)+f\left(y+y_1\right)+c=0 $

$ \begin{aligned} &\begin{aligned} & \Rightarrow x x_1+y y_1-\left(x+x_1\right)+2\left(y+y_1\right)+3=0 \\ & \Rightarrow x x_1+y y_1-x-x_1+2 y+2 y_1+3=0 \\ & \Rightarrow\left(x_1-1\right) x+\left(y_1+2\right) y-x_1+2 y_1+3=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned}\\ &\text { Given, polar equation is }\\ &2 x-3 y+1=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

By comparing the coefficients of two equations, we get

$ \begin{aligned} \frac{x_1-1}{2} & =\frac{y_1+2}{-3} \\ & =\frac{-x_1+2 y_1+3}{1} \\ \Rightarrow \quad \frac{x_1-1}{2} & =\frac{y_1+2}{-3} \text { and } \\ \frac{x_1-1}{2} & =\frac{-x_1+2 y_1+3}{1} \\ \Rightarrow-3 x_1+3 & =2 y_1+4 \text { and } x_1-1 \\ & =-2 x_1+4 y_1+6 \\ \Rightarrow 3 x_1+2 y_1 & =-1 \text { and } 3 x_1-4 y_1-7=0 \end{aligned} $

By solving these two equations, we get $x_1=\frac{5}{9}$ and $y_1=\frac{-4}{3}$

$ \text { } \begin{aligned}Now,\,\, 3 x_1-y_1 & =3\left(\frac{5}{9}\right)-\left(\frac{-4}{3}\right) \\ & =\frac{5}{3}+\frac{4}{3}=\frac{9}{3}=3 \\ \therefore \quad 3 x_1-y_1 & =3 \end{aligned} $

Given,

$ \begin{aligned} & \quad s \equiv x^2+y^2+2 g x+2 f y+c=0 \\ & \text { And } s^{\prime} \equiv x^2+y^2-6 x+6 y+2=0 \\ & \Rightarrow \quad\left(x^2-6 x\right)+\left(y^2+6 y\right)=-2 \\ & \Rightarrow \quad\left(x^2-6 x+9\right)+\left(y^2+6 y+9\right) \\ & \quad=-2+9+9 \\ & \Rightarrow \quad(x-3)^2+(y+3)^2=16 \end{aligned} $

Thus, the center of circle $S^{\prime}$ is $(3,-3)$ and radius, $r^{\prime}=4$

$ \begin{aligned} & \text { Now, } S \equiv x^2+y^2+2 g x+2 f y+c=0 \\ & \Rightarrow \quad(x+g)^2+(y+f)^2=g^2+f^2-c \end{aligned} $

So, the center of circle $S$ is $(-g,-f)$ and radius,

$ r=\sqrt{g^2+f^2-c} $

Since, radius of circle $S$ is 1 .

$ \begin{array}{lr} \therefore & \sqrt{g^2+f^2-c}=1 \\ \Rightarrow & g^2+f^2-c=1 \end{array} $

Also, two circle touch externally,

So, distance between $(-g,-f)$ and

$ \begin{aligned} & (3,-3)=r+r^{\prime} \\ & \Rightarrow \quad \sqrt{(-g-3)^2+(-f+3)^2}=4+1 \\ & \Rightarrow \quad(g+3)^2+(f-3)^2=25 \\ & \Rightarrow \quad g^2+6 g+9+f^2-6 f+9=25 \\ & \Rightarrow \quad g^2+f^2+6 g-6 f+18-25=0 \\ & \Rightarrow \quad g^2+f^2+6 g-6 f=7 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Since, point $(-1,-3)$ lies on both circles, it satisfies both circle.

$\therefore$ For circles

$ \begin{aligned} & (-1)^2+(-3)^2+2 g(-1)+2 f(-3)+c=0 \\ & \Rightarrow \quad 1+9-2 g-6 f+c=0 \\ & \Rightarrow \quad-2 g-6 f+c=-10 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

For circle $S^{\prime}$,

$ \begin{aligned} & (-1)^2+(-3)^2-6(-1)+6(-3)+2=0 \\ & \Rightarrow \quad 1+9+6-18+2=0 \\ & \Rightarrow \quad 0=0 \end{aligned} $

Now, the slope of line joining $(-g,-f)$ and $(-1,-3)$ is equal to slope of line joining $(-1,-3)$ and $(3,-3)$

$ \begin{aligned} & & (-f+3) & =\left(\frac{-3+3}{3+1}\right)(-g+1) \\ \Rightarrow & & (-f+3) & =0 \\ \Rightarrow & & f & =3\end{aligned} $

Put $f=3$ into Eq. (i), we get

$ \begin{aligned} & & & g^2+9+6 g-6(3) \\ \Rightarrow & & g^2+9+6 g-18 & =7 \\ \Rightarrow & & g^2+6 g-16 & =0 \\ \Rightarrow & & (g+8)(g-2) & =0 \\ \Rightarrow & & g & =-8 \text { or } g=2 (not \,\,possible)\end{aligned} $

Now, put $g=2, f=3$ into Eq. (ii), we get

$ \begin{aligned} & -2(2)-6(3)+c=-10 \\ \Rightarrow \quad & c=-10+4+18=12 \end{aligned} $

Now, $g+f+c=2+3+12=17$

$ \begin{aligned} & S \equiv x^2+y^2-6 x+8 y+k=0 \\ & \Rightarrow(x-3)^2+(y+4)^2=25-k \end{aligned} $

Thus, the center of the circle $C$ is at $(3,-4)=\left(x_1, y_1\right)$.

Let point $P$ as $\left(x_2, y_2\right)$ and point $Q$ is the point that divides the line segment $C P$ in the ratio $-1: 2$

Also, $3 x+4 y-43=0$

$ \Rightarrow \quad y=\frac{-3}{4} x+\frac{43}{3} $

So, slope of tangent line $=\frac{-3}{4}$

And slope of line $C P=-\left(\frac{-4}{3}\right)=\frac{4}{3}$

Thus, equation of line $C P$ is

$ \begin{array}{lrl} \Rightarrow & y-(-4) & =\frac{4}{3}(x-3) \\ \Rightarrow & y+4 & =\frac{4}{3}(x-3) \\ \Rightarrow & y & =\frac{4}{3} x-8 \end{array} $

Since, $P$ lies on both the lines $3 x+4 y-43=0$ and the line $C P$,

$ \begin{array}{ll} \therefore & 3 x+4\left(\frac{4}{3} x-8\right)-43=0 \\ \Rightarrow & 3 x+\frac{16}{3} x-32-43=0 \end{array} $

$ \begin{aligned} & \Rightarrow \quad \frac{25}{3} x=75 \\ & \Rightarrow \quad x=9 \\ & \text { Now, } y=\frac{4}{3} x-8=\frac{4}{3}(9)-8 \\ & =12-8=4 \end{aligned} $

So, $\quad P=(9,4)$

Using section formula, we find the coordinates of $Q(x, y)$.

$ \begin{aligned} & x=\frac{2 x_1-x_2}{2-1}=\frac{2(3)-9}{1}=-3 \\ & y=\frac{2 y_1-y_2}{2-1}=\frac{2(-4)-4}{1}=-12 \end{aligned} $

So, $Q=(x, y)=(-3,-12)$

The power of point $Q$ w.r.t to circle, $x^2+y^2-6 x+8 y+k=0$ is

$ \begin{aligned} (-3)^2 & +(-12)^2-6(-3)+8(-12)+k \\ & =9+144+18-96+k \\ & =75+k \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Now, the line $3 x+4 y-43=0$ is tangent to the circle, the radius of the circle is the distance from the center $C(3,-4)$ to the line $3 x+4 y-43=0$

$ \begin{aligned}So,\,r & =\frac{\left|A x_0+B y_0+C\right|}{\sqrt{A^2+B^2}} \\ & =\frac{|3(3)+4(-4)-43|}{\sqrt{3^2+4^2}} \\ & =\frac{|9-16-43|}{5}=\frac{50}{5}=10 \end{aligned} $

Now, $r=\sqrt{25-k}$

$ \begin{aligned} & \Rightarrow \quad 10=\sqrt{25-k} \Rightarrow 25-k=100 \\ & \Rightarrow \quad k=-75 \end{aligned} $

Put $k=-75$ in Eq. (i), we get The power of $Q$ w.r.t the circle

$ =75+k=75-75=0 $

$ \begin{array}{r} s_1: x^2+y^2+2 g x+2 f y+c=0 \\ s_2: 2 x^2+2 y^2+3 x+8 y+2 c=0 \\ \Rightarrow \quad x^2+y^2+\frac{3}{2} x+4 y+c=0 \\ s_3: x^2+y^2+2 x+2 y+1=0 \end{array} $

Radical axis $s_1-s_2=0$

$ \begin{array}{ll} \Rightarrow & \left(x^2+y^2+2 g x+2 f y+c\right) \\ & -\left(x^2+y^2+\frac{3}{2} x+4 y+c\right)+0 \\ \Rightarrow & 2 g x+2 f y-\frac{3 x}{2}-4 y=0 \\ \Rightarrow & \left(2 g-\frac{3}{2}\right) x+(2 f-4) y=0 \\ \Rightarrow & (4 g-3) x+(4 f-8) y=0 \end{array} $

The radical axis touches the circle $s_3$, where radius of $s_3$.

$=\sqrt{(-1)^2+(-1)^2-1}=1$ and center is $(-1,-1)$

The perpendicular distance from the center of $s_3$ to the radical axis $=$ radius of $s_3$

$ \begin{aligned} & \text { } \frac{|(4 g-3) \cdot(-1)+(4 f-8)(-1)|}{\sqrt{(4 g-3)^2+(4 f-8)^2}}=1 \\ & \Rightarrow \quad \frac{|-4 g+3-4 f+8|}{\sqrt{(4 g-3)^2+(4 f-8)^2}}=1 \\ & \Rightarrow \quad|-4 g-4 f+11| \\ & \quad=\sqrt{(4 g-3)^2+(4 f-8)^2} \\ & \Rightarrow \quad(-4 g-4 f+11)^2 \\ & \quad=(4 g-3)^2+(4 f-8)^2 \\ & \Rightarrow \quad 16 g^2+16 f^2+121+32 g f-88 g-88 f \\ & \quad=16 g^2-24 g+9+16 f^2-64 f+64 \\ & \Rightarrow \quad 32 g f-64 g-24 f+48=0 \\ & \Rightarrow \quad 8 g f-16 g-6 f+12=0 \\ & \Rightarrow \quad 8 g(f-2)-6(f-2)=0 \\ & \Rightarrow \quad(8 g-6)(f-2)=0 \\ & \Rightarrow \quad(4 g-3)(f-2)=0 \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad 4 g-3=0 \text { or } f-2=0 \\ & \Rightarrow \quad g=\frac{3}{4} \text { or } f=2 \end{aligned} $

Either $g=\frac{3}{4}$ or $f=2$ is the required condition.

$\because x^2+y^2-2 x-4 y-20=0$

After rotation of axes through $\theta=\pi / 4$, it transforms to $a x^2+2 h x y+b y^2+2 g x+2 f y+c=0$ in the given equation

$ a=1, h=0, b=1, g=-1, f=-2, c=-20 $

After rotation of $\theta=\pi / 4$

$ \begin{aligned} a^{\prime} & =a \cos ^2 \theta+b \sin ^2 \theta+2 h \cos \theta \sin \theta \\ & =1\left(\frac{1}{\sqrt{2}}\right)^2+1\left(\frac{1}{\sqrt{2}}\right)^2+0=1 \\ b^{\prime}= & a \sin ^2 \theta+b \cos ^2 \theta-2 h \sin \theta \cos \theta \\ = & 1\left(\frac{1}{\sqrt{2}}\right)^2+1\left(\frac{1}{\sqrt{2}}\right)^2-0=1 \end{aligned} $

also,

$ \begin{aligned} & h^{\prime}=(b-a) \sin \theta \cos \theta+h\left(\cos ^2 \theta-\sin ^2 \theta\right) \\ & =0+0=0 \end{aligned} $

Similarly, $x^{\prime}=x \cos \theta-y \sin \theta=\frac{1}{\sqrt{2}}(x-y)$

$ \begin{aligned} & =\frac{-6 x-2 y}{\sqrt{2}} \\ & =-3 \sqrt{2} x-\sqrt{2} y=2 g x+2 f y \end{aligned} $

then $g=-3 / \sqrt{2}, f=-1 / \sqrt{2}, c=-20$

So,

$ \begin{aligned} & \left|\begin{array}{lll} a & h & g \\ h & b & f \\ g & f & c \end{array}\right|=\left|\begin{array}{ccc} 1 & 0 & -3 / \sqrt{2} \\ 0 & 1 & -1 / \sqrt{2} \\ -3 / \sqrt{2} & -1 / \sqrt{2} & -20 \end{array}\right| \\ & \quad=1\left(-20-\frac{1}{2}\right)-\frac{3}{\sqrt{2}}\left(+\frac{3}{\sqrt{2}}\right) \\ & \quad=-41 / 2-9 / 2=-25 \end{aligned} $

Given equation of circle

$ x^2+y^2+5 k x+2 y+k=0 $

and $2 x^2+2 y^2+2 k x+3 y-1=0$

Equation of common chord

$ 8 k x+y+2 k+1=0 $

Since, this chord is coincide with

$ \begin{aligned} & 4 x+5 y-k=0 \\ & \frac{8 k}{4}=\frac{1}{5}=\frac{2 k+1}{-k} \end{aligned} $

Solving $\frac{8 k}{4}=\frac{1}{5}$

we get $k=\frac{1}{10}$

Solving $8 \frac{k}{4}=\frac{2 k+1}{-k} \Rightarrow 2 k^2+2 k+1=0$

No solution

$\therefore$ No values of $k$ possible

$ \begin{aligned} & \text { } c_1: x^2+y^2-2 x+4 y+1=0 \\ & \Rightarrow(x-1)^2+(y+2)^2=4 \\ & \text { Centre }\left(c_1\right)=(1,-2), r_1=2 \\ & \text { and } c_2: x^2+y^2-4 x-2 y+4=0 \\ & \Rightarrow(x-2)^2+(y-1)^2=1 \\ & \text { Centre }\left(c_2\right)=(2,1), r_2=1 \\ & c_1 c_2=d=\sqrt{(2-1)^2+\left(1+2^2\right.}=\sqrt{10} \end{aligned} $

Let the slope of direct common tangent be $m$ the general line is

$ y=m x+c $

this line has perpendicular distance $r_1$ from $c_1$ and $r_2$ from $c_2$

$ \begin{aligned} & \therefore\left|\frac{m(1)-(-2)+c}{\sqrt{1+m^2}}\right|=2 \\ & \Rightarrow|m+2+c|=2 \sqrt{1+m^2} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & \text { and }\left|\frac{m(2)-1+c}{\sqrt{1+m^2}}\right|=1 \\ & \Rightarrow|2 m+c-1|=\sqrt{1+m^2} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

Put the value of ' $c$ ' from Eq. (i) to Eq. (ii)

$ 2 m-1+\left(-m-2+2 \sqrt{m^2+1}\right)=\sqrt{m^2+1} $

$ \begin{aligned} &\Rightarrow m=3-\sqrt{m^2+1}\\ &\text { Squaring both sides, we get }\\ &\begin{array}{rlrl} & m^2= & m^2+10 & -6 \sqrt{m^2+1} \\ \Rightarrow & & \sqrt{m^2+1} & =5 / 3 \\ \Rightarrow & & m^2+1 & =\frac{25}{9} \\ \Rightarrow & & m^2=16 / 9 \\ \Rightarrow & & m=4 / 3 \end{array} \end{aligned} $

We are told that $(1, a)$ and $(b, 2)$ are conjugate points with respect to the circle

$ x^2+y^2 = 25. $

Step 1. Condition for conjugate points

If $(x_1, y_1)$ and $(x_2, y_2)$ are conjugate points with respect to the circle $x^2+y^2=r^2$,

then the polar of one point passes through the other.

Equation of the polar of $(x_1, y_1)$ w.r.t. $x^2+y^2=r^2$ is:

$ x_1x + y_1y = r^2 $

Step 2. The point $(1,a)$

For the point $(1, a)$, the polar is:

$ 1 \cdot x + a \cdot y = 25 $

That is:

$ x + a y = 25. $

Step 3. The point $(b,2)$ lies on this polar

Since the points $(1,a)$ and $(b,2)$ are conjugate, the point $(b,2)$ lies on the polar of $(1,a)$.

So:

$ b + a(2) = 25 $

$ b + 2a = 25 \quad \text{(1)} $

Step 4. Symmetry of conjugate points

Similarly, the polar of $(b,2)$ is:

$ b x + 2 y = 25. $

And $(1,a)$ must lie on this polar:

$ b(1) + 2(a) = 25 $

Wait—that would give same, yes so far matches (1). Let's check the other way:

Substitute $ (x,y) = (1,a) $:

$ b \cdot 1 + 2 a = 25 $

This is the same as (1). So it’s consistent.

Step 5. We need $4a + 2b = ?$

From (1): $b + 2a = 25.$

Multiply both sides by 2:

$ 2b + 4a = 50 $

So,

$ \boxed{4a + 2b = 50.} $

✅ Final Answer: 50

Option B

Let $(h, k)$ be the pole of the line

$ x+2 b y-5=0 $

with respect to the circle

$ \begin{array}{ll} s \equiv x^2+y^2-4 x-6 y+4=0 \\ \therefore \quad h x+k y-2(x+h)-3(y+k)+4=0 \\ (h-2) x+(k-3) y-2 h-3 k+4=0 \end{array} $

is coincide with $x+2 b y-5=0$

$ \begin{aligned} & \frac{h-2}{1}=\frac{k-3}{2 b}=\frac{2 h+3 k-4}{5} \\ \Rightarrow & \frac{h-2}{1}=\frac{k-3}{2 b} \end{aligned} $

$ 2 b h-k=4 b-3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

and $\frac{h-2}{1}=\frac{2 h+3 k-4}{5}$

$ h-k=2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Solving Eqs. (i) and (ii), we get

$ h=\frac{4 b-5}{2 b-1}, k=\frac{-3}{2 b-1} $

$(h, k)$ lie on line $x+b y+1=0$

$ \begin{aligned} &\begin{aligned} & & \frac{4 b-5}{2 b-1}-\frac{3 b}{2 b-1}+1 & =0 \\ & & 4 b-5-3 b+2 b-1 & =0 \\ & & & =2 \end{aligned}\\ &\therefore \text { Pole of the line from pole }(2,-2) \text { is }\\ &\begin{aligned} & (2-2) x+(-2-3) y-2(2)-3(-2)+4=0 \\ & \Rightarrow \quad-5 y+6=0 \\ & \Rightarrow \quad 5 y-6=0 \end{aligned} \end{aligned} $

The given equations are

$ \begin{aligned} & s: x^2+y^2+4 x+7=0 \\ & s^{\prime}: 2 x^2+2 y^2+3 x+5 y+9=0 \\ & \Rightarrow x^2+y^2+\frac{3}{2} x+\frac{5}{2} y+\frac{9}{2}=0 \\ & s^{\prime \prime}: x^2+y^2+y=0 \end{aligned} $

Radical axis of $s$ and $s^{\prime}: s-s^{\prime}=0$

$ \Rightarrow \quad 4 x-\frac{3}{2} x-\frac{5}{2} y+7-\frac{9}{2}=0 $

$ \Rightarrow \quad \frac{5}{2} x-\frac{5}{2} y+\frac{5}{2}=0 $

$ \Rightarrow \quad x-y+1=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) $

Radical axis of $s$ and $s^{\prime \prime}: s-s^{\prime \prime}=0$

$ \Rightarrow \quad 4 x-y+7=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

From Eqs. (i) and (ii), we have

$ x=-2, y=-1 $

$\therefore$ Radical centre is $P(-2,-1)$

Length of tangent from $P$ to $s^{\prime}$

$ \begin{aligned} & =\sqrt{(-2)^2+(-1)^2+\frac{3}{2}(-2)+\frac{5}{2}(-1)+\frac{9}{2}} \\ & =\sqrt{4+1-3+2}=\sqrt{4}=2 \end{aligned} $

Substitute $x=X+2$ and $y=Y-2$ into the original equation, $x^2+y^2=4$ we get;

$ \begin{aligned} & (X+2)^2+(Y-2)^2=4 \\ \Rightarrow & X^2+4+4 X+Y^2+4-4 Y=4 \\ \Rightarrow & X^2+Y^2+4 X-4 Y+4=0 \\ \therefore & a+b+c=4-4+4=4 \end{aligned} $

Given, point $P(-4,0)$

Equation of circle

$ x^2+y^2-4 x-6 y-12=0 $

Centre $(2,3)$

Centre of circle passing through $P, A, B$ is mid point of $O P$

$ \begin{array}{ll} \therefore\left(\frac{-4+2}{2}, \frac{0+3}{2}\right)=\left(-1, \frac{3}{2}\right) \\ \therefore (g, f)=\left(1, \frac{-3}{2}\right) \end{array} $

Equation of polar is $T=0$

$ \begin{aligned} & \text { i.e., } \alpha x+y(-1)-4 \cdot \frac{1}{2}(x+\alpha) \\ & \quad-6 \cdot \frac{1}{2}(y+(-1))-12=0 \\ & \Rightarrow \quad \alpha x-y-2 x-2 \alpha-3 y+3-12=0 \\ & \Rightarrow \quad(\alpha-2 x-4 y-2 \alpha-9=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

But equation of polar is $y=\beta$ given

So, coefficient of $x$ must be zero.

i.e. $\alpha-2=0 \Rightarrow \alpha=2$

Thus, $0-4 y-2 \times(2)-9=0\,\,\,\,(from Eq. (i))$

$ \begin{aligned} & \Rightarrow \quad-4 y=13 \\ & \Rightarrow \quad y=\frac{-13}{4}=\beta \,\,\,(given)\\ & \therefore 4(\alpha+\beta)=4\left(2-\frac{13}{4}\right)=-5 \end{aligned} $

Here centre of circle $=C=(2,3)$ let $P \equiv(-1,-1)$ and $C P=5$

Let radius $=r$

$ \begin{aligned} & \therefore \sin \frac{\theta}{2}=\frac{r}{C P} \\ & \Rightarrow \quad r^2=(5)^2 \sin ^2 \frac{\theta}{2} \\ & \because \cos \theta=\frac{-7}{25} \,\,\,(given)\end{aligned} $

$ \begin{aligned} & 1-2 \sin ^2 \frac{\theta}{2}=\frac{-7}{25} \Rightarrow 2 \sin ^2 \frac{\theta}{2}=1+\frac{7}{25} \\ & \Rightarrow \quad \sin ^2 \frac{\theta}{2}=\frac{16}{25}=25 \cdot \frac{16}{25}=16 \\ & \therefore \quad r=4\,\,\,\, (\because r>0)\end{aligned} $

$ \begin{aligned} &\text { As we know that, power of a point }\\ &\begin{aligned} & \text { W.r.t a circle }=\left(\sqrt{S_1}\right)^2=S_1 \\ & \therefore 1+36+4-36-a=-16 \\ & \Rightarrow a=5+16=21 \end{aligned} \end{aligned} $

The equation of a circle passing through the intersection of two circles

$ \begin{aligned} & x^2+y^2+2 x+4 y+1=0 \text { and } \\ & x^2+y^2-2 x-4 y-4=0 \text { is } \\ & \left(x^2+y^2+2 x+4 y+1\right) \\ & +\lambda\left(x^2+y^2-2 x-4 y-4\right)=0, \text { where } \lambda \end{aligned} $

$ \begin{aligned} & \text { is a parameter. } \\ & \begin{aligned} \Rightarrow(1+\lambda) x^2+(1+\lambda) y^2+(2-2 \lambda) & +(4-4 \lambda) y +1-4 \lambda=0 \\ \Rightarrow x^2+y^2+\frac{2(1-\lambda)}{(1+\lambda)} x & +\frac{4(1-\lambda)}{(1+\lambda)} y +\frac{1-4 \lambda}{1+\lambda}=0 \ldots \text { (i) } \end{aligned} \end{aligned} $

Applying the orthogonal condition,

$ \begin{aligned} 2\left(\frac{1-\lambda}{1+\lambda}\right) \times 0 & +2 \times \frac{2(1-\lambda)}{(1+\lambda)} \times 0 \\ & =\frac{1-4 \lambda}{1+\lambda}-6 \\ \Rightarrow \quad \frac{1-4 \lambda}{1+\lambda} & =6 \\ \Rightarrow \quad 6+6 \lambda & =1-4 \lambda \Rightarrow \lambda=\frac{-1}{2} \end{aligned} $

Substituting $\lambda=\frac{-1}{2}$ in Eq. (i) and simplifying we get

$ x^2+y^2+6 x+12 y+6=0 $

$\therefore$ Radius of this circle

$ =\sqrt{3^2+6^2-6}=\sqrt{39} $

Let the $x$-intercepts be $(a, 0)$ and

$ (a+4,0) $

Let the $y$-intercepts be $(0, b)$ and

$ (0, b+2 \sqrt{11}) $

Center of circle $=(a+2, b+\sqrt{11})$

Since, the center is in the fourth quadrant $h>0$ and $k<0$

$ \Rightarrow a+2>0 \text { and } b+\sqrt{11}<0 $

Equation of circle $(x-h)^2+(y-k)^2=r^2$

$ \begin{aligned} & \Rightarrow(1-(a+2))^2+(0-(b+\sqrt{11}))^2=r^2 \\ & \Rightarrow(-a-1)^2+(-b-\sqrt{11})^2=r^2 \end{aligned} $

Substituting $(a, 0)$ gives

$ \begin{aligned} & (a-(a+2))^2+(0-(b+\sqrt{11}))^2=r^2 \\ & \Rightarrow(-2)^2+(-b-\sqrt{11})^2=r^2 \end{aligned} $

substituting ( $a+4,0$ ) gives

$ \begin{aligned} & \Rightarrow (a+4-(a+2))^2+(0-(b+\sqrt{11}))^2=r^2 \\ & \Rightarrow 2^2+(-b-\sqrt{11})^2=r^2 \end{aligned} $

Similarly, $(0, b)$ gives

$ \begin{aligned} (0-(a+2))^2+ & (b-(b+\sqrt{11}))^2=r^2 \\ \text { substituting }(0, b+ & 2 \sqrt{11}) \\ (0-(a+2))^2+(b+ & 2 \sqrt{11} -(b+\sqrt{11}))^2=r^2 \\ (-a-2)^2+(\sqrt{11})^2= & r^2 \end{aligned} $

Solving above equation, we get

$ \begin{aligned} & r^2=20 \\ & r=2 \sqrt{5} \end{aligned} $

The inverse point is $\left(\frac{1}{10}, \frac{-1}{5}\right)$

The original point is $(-1,2)$

The circle equation is

$ x^2+y^2-2 x+4 y+c=0 $

Centre of the circle is $(1,-2)$

The original point is $(-1,2)$

$ \begin{aligned} \text { The distance } & =\sqrt{(-1-1)^2+(2-(-2))^2} \\ & =\sqrt{4+16}=\sqrt{20} \end{aligned} $

The centre of the circle is $(1,-2)$

The inverse point is $\left(\frac{1}{10}, \frac{-1}{5}\right)$

$ \begin{aligned} \text { The distance } & =\sqrt{\left(\frac{1}{10}-1\right)^2+\left(\frac{-1}{5}+2\right)^2} \\ & =\frac{9 \sqrt{5}}{10} \end{aligned} $

$ \begin{array}{ll} \Rightarrow & r^2=d_1 d_2=\sqrt{20} \times \frac{9 \sqrt{5}}{10}=9 \\ \Rightarrow & 5-c=9 \Rightarrow c=-4 \end{array} $