Circle

To find the equation of the circle passing through the origin and cutting the given circles orthogonally, we use the condition for orthogonality between circles.

The condition for two circles to be orthogonal is expressed as:

$ 2g_1g_2 + 2f_1f_2 = c_1 + c_2 $

Applying this condition to the circles in question, we have:

For the circle $x^2 + y^2 + 6x - 15 = 0$:

Center: $(-3, 0)$

Radius term: $-15$

For the circle $x^2 + y^2 - 8y - 10 = 0$:

Center: $(0, 4)$

Radius term: $-10$

Let's determine $g$ and $f$ for the required circle $x^2 + y^2 + 2gx + 2fy = 0$.

For the first condition of orthogonality with $x^2 + y^2 + 6x - 15 = 0$, we get:

$ 2g(-3) + 2f(0) = 0 - 15 $

Solving for $g$:

$ -6g = -15 \quad \Rightarrow \quad g = \frac{15}{6} = \frac{5}{2} $

For the second condition of orthogonality with $x^2 + y^2 - 8y - 10 = 0$, we have:

$ 2g(0) + 2f(4) = -10 $

Solving for $f$:

$ 8f = -10 \quad \Rightarrow \quad f = -\frac{5}{4} $

The equation of the required circle is then:

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

Multiplying the entire equation by 2 to eliminate fractions, we obtain:

$ 2x^2 + 2y^2 - 10x + 5y = 0 $

This is the equation of the required circle.

Equation of circle passing through 3 point is

$ \begin{array}{l} \left|\begin{array}{cccc} x^{2}+y^{2} & x & y & 1 \\\\ x_{1}^{2}+y_{1}^{2} & x_{1} & y_{1} & 1 \\\\ x_{2}^{2}+y_{2}^{2} & x_{2} & y_{2} & 1 \\\\ x_{3}^{2}+y_{3}^{2} & x_{3} & y_{3} & 1 \end{array}\right|=0 \\\\ \left|\begin{array}{cccc} 1+k^{2} & 1 & k & 1 \\\\ 2 & -1 & 1 & 1 \\\\ 1 & 0 & -1 & 1 \\\\ 1 & 1 & 0 & 1 \end{array}\right|=0 \end{array} $

$ C_{1}=C_{1}-C_{4}, C_{4}=C_{4}-C_{2} $

$ \left|\begin{array}{cccc} K^{2} & 1 & K & 0 \\\\ 1 & -1 & 1 & 2 \\\\ 0 & 0 & -1 & 1 \\\\ 0 & 1 & 0 & 0 \end{array}\right|=0 $

$ R_{2}=R_{2}+R_{3} $, we get

$ \left|\begin{array}{cccc} K^{2} & 1 & K & 0 \\\\ 1 & -1 & 0 & 3 \\\\ 0 & 0 & -1 & 1 \\\\ 0 & 1 & 0 & 0 \end{array}\right|=0 $

$ \Rightarrow K^{2}(3)-1(K)=0 $

$ \Rightarrow 3 K^{2}-K=0 ; K=0, K=\frac{1}{3} $

If lines are perpendicular

then, $ m_{1} m_{2}=-1=-1 \times \frac{-1}{a}=-1 $

$ a=-1 $

If this lines are tangent mean $ \perp $ distance from centre $ = $ Radius of circle

Centre $ \equiv(-1,1) $ radius $ =\sqrt{1+1-1}=1 $

Perpendicular distance $ =r $

$ \Rightarrow\left|\frac{-1+1+K}{\sqrt{2}}\right|=1 $

$ |K|=\sqrt{2}, K=+\sqrt{2} \quad\{\because K > 1\} $

Perpendicular distance $ =r $

$ \Rightarrow \quad\left|\frac{-1+a+b}{\sqrt{1+a^{2}}}\right|=1 $

$ \left|\frac{-1-1+b}{\sqrt{2}}\right|=1 $ $ (\because a=-1) $

$ \Rightarrow \quad|2-b|=\sqrt{2} $

$ \Rightarrow \quad 2-b=-\sqrt{2} $ and $ 2-b=+\sqrt{2} $

$ \Rightarrow \quad b=2+\sqrt{2} $ and $ b=2-\sqrt{2} $

$ b-K=2+\sqrt{2}-\sqrt{2}=2 $

Given, $ c_{1} \equiv x^{2}+y^{2}+2 x-6 y+1=0 $

$ c_{2} \equiv x^{2}+y^{2}-4 x+2 y+4=0 $

$ c_{1} \equiv(-1,3), r_{1}=\sqrt{1+9-1}=3 $

$ c_{2} \equiv(2,-1), r_{2}=\sqrt{4+1-4}=1 $

$ \therefore $ Internal similitude ( $ P $ ) centre divide $ c_{1} $ and $ c_{2} $ in ratio $ r_{1}: r_{2} $

$ \begin{aligned} \Rightarrow \quad P & \equiv(h, k) \\\\ h & =\frac{-1+6}{3+1} \Rightarrow h=\frac{5}{4} \Rightarrow 4 h=5 \end{aligned} $

Given equation

$ \begin{array}{l} x^{2}+y^{2}-4 x-8 y+16=0 \text { and } \\\\ x^{2}+y^{2}-6 x-16 y+64=0 \text { can be } \end{array} $

written as

$ S_{1} \equiv(x-2)^{2}+(y-4)^{2}=2^{2} $

$ \Rightarrow \quad $ Centre $ \equiv(2,4) $ and Radius $ =2 $

$ S_{2} \equiv(x-3)^{2}+(y-8)^{2}=3^{2} $

$ \Rightarrow \quad $ Centre $ \equiv(3,8) $ and Radius $ =3 $

Equation of tangent to $ S_{1} $

$ \begin{array}{l} \Rightarrow \quad(y-4)=m(x-2)+2 \sqrt{1+m^{2}} \\\\ \Rightarrow \quad y-m x=4-2 m+2 \sqrt{1+m^{2}} . . \end{array} $

Equation of tangent to $ S_{2} $

$ \begin{array}{l} \Rightarrow \quad(y-8)=m(x-3)+3 \sqrt{1+m^{2}} \\\\ \Rightarrow \quad y-m x=8-3 m+3 \sqrt{1+m^{2}} \ldots \end{array} $

If tangent is common, then Eq. (i) and

(ii) will be same

$ 4-2 m+2 \sqrt{1+m^{2}}=8-3 m+3 \sqrt{1+m^{2}} $

$ \Rightarrow \quad m-4=\sqrt{1+m^{2}} $

On squaring both sides, we get

$ \begin{array}{l} \Rightarrow \quad m^{2}+16-8 m=1+m^{2} \\\\ \Rightarrow \quad 8 m=15 \Rightarrow m=\frac{15}{8} \end{array} $

Angle between 2 circle is $ \cos \theta=\frac{r_{1}^{2}+r_{2}^{2}-d^{2}}{2 r_{1} r_{2}} $

$ c_{1} \equiv(-1,3), r_{1}=\sqrt{1+9+6}=4 $

$ c_{2} \equiv(3,1), r_{2}=\sqrt{9+1-K}=\sqrt{10-K} $

$ c_{1} c_{2}=d=\sqrt{4^{2}+2^{2}}=\sqrt{20} $

$ \frac{-5}{24}=\frac{16+10-K-20}{2 \cdot 4 \cdot \sqrt{10-K}} $

$ \frac{-5}{3}=\frac{6-K}{\sqrt{10-K}} $

$ 5 \sqrt{10-K}=3 K-18 $

$ 250-25 K=9 K^{2}+324-108 K $

$ 9 K^{2}-83 K+74=0 $

$ 9 K^{2}-74 K-9 K+74=0 $

$ K=\frac{74}{9}, K=1 $

$ c=(p, q) $, constant $ \tan =c $

$ c_{1}=(1,2), r_{1}=\sqrt{1+4-4}=1 $

$ c_{2}=(-1,2), r_{2}=\sqrt{1+4-1}=2 $

$ c_{3}=(2,1), r_{3}=\sqrt{4+1+11}=4 $

$ 2 g_{1} g_{2}+2 f_{1} f_{2}=c_{1}+c_{2} $

[ $ \because $ Condition for orthogonal]

$ \begin{array}{l} 2 p+4 q=c+4 \\\\ -2 p+4 q=c+1 \\\\ 4 p+2 q=c-11 \\\\ \text { btracting Eq: (i) fro } \\\\ 4 p=3 \Rightarrow p=\frac{3}{4} \end{array} $

On subtracting Eq. (i) from (ii), we get

On subtracting Eqs. (iii) from (ii), we get

$ -6 p+2 q=12 \Rightarrow-6 \times \frac{3}{4}+2 q=12 $

$ 2 q=12+\frac{9}{2} \Rightarrow 2 q=\frac{33}{2} \Rightarrow q=\frac{33}{4} $

$ \therefore \quad p+q=\frac{33}{4}+\frac{3}{4}=\frac{36}{4}=9 $

Given, circle

$ \begin{aligned} x^2+y^2-2 x-2 y-1 & =0, P\left(\frac{\pi}{4}\right) \text { and } Q\left(\frac{\pi}{3}\right) \\\\ (x-1)^2+(y-1)^2 & =1^2 \end{aligned} $

Center, $(h, k)=(1,1)$ and radius, $r=1$

$ \begin{aligned} \therefore \quad P\left(\frac{\pi}{4}\right) & =\left(h+r \cos \frac{\pi}{4}, y+r \sin \frac{\pi}{4}\right) \\\\ & =\left(1+\frac{1}{\sqrt{2}}, 1+\frac{1}{\sqrt{2}}\right) \\\\ \text { and } \quad Q\left(\frac{\pi}{3}\right) & =\left(h+r \cos \frac{\pi}{3}, y+r \sin \frac{\pi}{3}\right) \\\\ & =\left(1+\frac{1}{2}, 1+\frac{\sqrt{3}}{2}\right) \end{aligned} $

Slope of described tangent $=$ Slope of chord $P Q$

$ \begin{aligned} & =\frac{\left(1+\frac{\sqrt{3}}{2}\right)-\left(1+\frac{1}{\sqrt{2}}\right)}{\left(1+\frac{1}{2}\right)-\left(1+\frac{1}{\sqrt{2}}\right)}=\frac{\sqrt{6}-2}{\sqrt{2}-2} \times \frac{\sqrt{2}+2}{\sqrt{2}+2} \\\\ & =\frac{\sqrt{12}-2 \sqrt{2}+2 \sqrt{6}-4}{2-4} \end{aligned} $

$\begin{aligned} & =-\frac{1}{2}(2 \sqrt{3}-2 \sqrt{2}+2 \sqrt{6}-4) \\\\ & =2+\sqrt{2}-\sqrt{3}-\sqrt{6}\end{aligned}$

Let $S: x^2+y^2+2 g x+2 f y+c=0$

$\because$ The power of the point $(2,0)$ with respect to circle $S$ is -4

$ \begin{array}{lr} \therefore(2)^2+(0)^2+2 g(2)+2 f(0)+c=-4 \\\\ \Rightarrow \quad 4 g+c+8=0 \quad \ldots \text { (i) } \end{array} $

$\because$ The length of tangent drawn from the point $(1,1)$ to $S$ is 2

$ \begin{array}{lr} \therefore & \sqrt{(1)^2+(l)^2+2 g(l)+2 f(l)+c}=2 \\\\ \Rightarrow & 2 g+2 f+c+2=4 \\\\ \Rightarrow & 2 g+2 f+c-2=0 \quad \ldots \text { (ii) } \end{array} $

$\because$ The circle $S$ is passing through $(-1,-1)$.

$ \begin{array}{ll} \therefore(-1)^2+(-1)^2+2 g(-1)+2 f(-1)+c=0 \\\\ \Rightarrow & -2 g-2 f+c+2=0 \\\\ \Rightarrow & 2 g+2 f-c-2=0 \ldots \text { (iii) } \end{array} $

On solving Eqs. (i), (ii) and (iii), we get

$ \begin{aligned} & g=-2 f=3 \text { and } c=0 \\\\ & \therefore \quad S=x^2+y^2-4 x+6 y=0 \\\\ & \text { Radius of } S=\sqrt{g^2+f^2-c} \\\\ & =\sqrt{(-2)^2+(3)^2-0} \\\\ & =\sqrt{4+9}=\sqrt{13} \end{aligned} $

Given, the pole of the line $x-5 y-7=0$ with respect to the circle

$ S \equiv x^2+y^2-2 x+4 y+1=0 \text { is } P(a, b) $

$\therefore$ Equation of polar $T=0$

$ \begin{aligned} & a x+b y-(a+x)+2(b+y)+1=0 \\\\ & (a-1) x+(b+2) y+(2 b-a+1)=0 \end{aligned} $

This equation will be same as

$ x-5 y-7=0 $

$ \begin{array}{ll} \therefore & \frac{a-1}{1}=\frac{b+2}{-5}=\frac{2 b-a+1}{-7} \\\\ \Rightarrow & a=0 \text { and } b=3 \\\\ \therefore & P(a, b)=P(0,3) \end{array} $

Center of circle $S=0$ is $C(1,-2)$

$ \begin{aligned} & P C=\sqrt{(1-0)^2+(-2-3)^2}=\sqrt{26} \\\\ & \sqrt{a^3+b^3-1}=\sqrt{0+27-1}=\sqrt{26} \end{aligned} $

Hence, $\quad P C=\sqrt{a^3+b^3-1}$

We have,

$ \begin{array}{ll} & \\\\ S_1 \equiv x^2+y^2+2 x+2 y+1=0 \\\\ \therefore S_1 \equiv(x+1)^2+(y+1)^2=1 \\\\ \text { and } S_2 \equiv x^2+y^2-2 x-2 y+1=0 \\\\ \therefore S_2 \equiv(x-1)^2+(y-1)^2=1 \end{array} $

From the diagram it is clear that transverse common tangents are $x=0$ and $y=0$ (i.e. $Y$-axis and $X$-axis)

The combined equation of lines $x=0$ and $y=0$ is $x y=0$

Let required circle

$ \begin{aligned} & S: x^2+y^2+2 g x+2 f y+c=0 \\\\ & S_1: x^2+y^2+4 x-5=0 \\\\ & S_2: x^2+y^2-4 y+3=0 \end{aligned} $

Here, $g_1=2, f_1=0, g_2=0, f_2=-2$, $c_1=-5$ and $c_2=3$

$\because$ Circle $S$ is passing throught the point (1, 1)

$ \begin{array}{lr} \therefore 1^2+1^2+2 g(1)+2 f(1)+c=0 \\\\ \Rightarrow 2 g+2 f+c+2=0 \quad \ldots \text { (i) } \end{array} $

$\because S$ and $S_1$ are orthogonal circles

$ \begin{array}{lc} \therefore & 2 g g_1+2 f f_1=c+c_1 \\\\ \Rightarrow & 2 g(2)+2 f(0)=c+(-5) \\\\ \Rightarrow & 4 g-c+5=0 \quad \ldots \text { (ii) } \end{array} $

$\because S$ and $S_2$ are orthogonal circles.

$ \begin{array}{lc} \therefore & 2 g g_2+2 f f_2=c+c_2 \\\\ \Rightarrow & 2 g(0)+2 f(-2)=c+3 \\\\ \Rightarrow & -4 f-c-3=0 \quad \ldots \text { (iii) } \end{array} $

On solving Eqs. (i), (ii) and(iii), we get

$ g=-\frac{3}{4}, f=-\frac{5}{4} $

and $\quad c=2$

$\therefore$ Center of $S=(-g,-f)=\left(\frac{3}{4}, \frac{5}{4}\right)$

Given,

$ \begin{aligned} & S_1 \equiv x^2+y^2-6 x+5=0 \\\\ & S_2 \equiv x^2+y^2+4 y-5=0 \end{aligned} $

Here, $C_1=(3,0), r_1=\sqrt{(-3)^2+0^2-5}=2$

$ \begin{aligned} & C_2=(0,-2), r_2=\sqrt{0^2+2^2-(-5)}=3 \\\\ & C_1 C_2=\sqrt{(0-3)^2+(-2-0)^2}=\sqrt{13} \end{aligned} $

Area of $\Delta A C_1 C_2=\frac{1}{2} r_1 r_2=\frac{1}{2} A M \times C_1 C_2$

$ \Rightarrow \frac{1}{2}(2)(3)=\frac{1}{2}(A M)(\sqrt{13}) \Rightarrow A M=\frac{6}{\sqrt{13}} $

$\therefore$ Required length of common tangent

$ =A B=2(A M)=2\left(\frac{6}{\sqrt{13}}\right)=\frac{12}{\sqrt{13}} $

Let coordinates of $C$ be $(h, k)$.

$ A=(2,3), B=(3,2) $

Coordinates of centroid

$ \begin{aligned} =\left(\frac{2+h+3}{3}, \frac{k+3+2}{3}\right) & =\left(\frac{h+5}{3}, \frac{k+5}{3}\right) \\\\ \therefore \quad\left(\frac{h+5}{3}\right)^2+\left(\frac{k+5}{3}\right)^2 & =25 \quad \text { [given] } \\\\ (h+5)^2+(k+5)^2 & =(15)^2 \\\\ (x+5)^2+(Y+5)^2 & =(15)^2 \end{aligned} $

which is the equation of circle with radius 15 units.

$\Rightarrow$ Diameter $=30$ units

Three points on circle are

$ (1,1),(-2,2),(2,-2) $

Let equation of circle be

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

$\because$ Above points lie on the circle

$ \begin{aligned} & \therefore \quad 1+1+2 g+2 f+c=0 \\\\ & \Rightarrow \quad 2 g+2 f+c+2=0 \quad \ldots \text { (i) } \\\\ & (-2)^2+(2)^2+2(-2) g+2 \times 2 \times f+c=0 \\\\ & 8-4 g+4 f+c=0 \\\\ & \quad 4 g-4 f-c-8=0 \quad \ldots \text { (ii) } \end{aligned} $

and

$ \begin{gathered} (2)^2+(-2)^2+2 \times 2 \times g+2(-2) f+c=0 \\\\ 4 g-4 f+c+8=0 \quad \ldots \text { (iii) } \end{gathered} $

On solving Eqs. (i), (ii) and (iii), we get

$ c=-8, g=\frac{3}{2}, f=\frac{3}{2} $

$\therefore$ Centre of circle $\equiv\left(\frac{-3}{2}, \frac{-3}{2}\right)$

Now, perpendicular distance from centre of the circle to the line

$ \begin{aligned} & =\left|\frac{3\left(-\frac{3}{2}\right)-4\left(-\frac{3}{2}\right)+1}{\sqrt{3^2+(-4)^2}}\right| \\\\ & =\frac{\left|-\frac{9}{2}+\frac{12}{2}+1\right|}{5}=\frac{1}{2} \end{aligned} $

Given circle is

$ \begin{aligned} & x^2+y^2-4 x+6 y+4=0 \\\\ & (x-2)^2+(y+3)^2=(3)^2 \\\\ & \text { Centre }=(2,-3) \text { and radius }=r=3 \end{aligned} $

$\because$ Perpendicular distance between centre and line $=r$

$ \begin{aligned} \therefore & \left|\frac{4 \times 2-3(-3)+p}{\sqrt{4^2+(3)^2}}\right|=3 \\\\ \Rightarrow \quad & 17+p= \pm 15 \\\\ & p=-2 \quad (\because p+3>0) \end{aligned} $

So, equation of line is

$ 4 x-3 y-2=0 \quad \ldots \text { (i) } $

Let line touches the circle at $(h, k)$.

Then, equation of tangent at $(h, k)$ is

$ \begin{aligned} & h x+k y-2(x+h)+3(y+k)+4=0 \\\\ & (h-2) x+(k+3) y-2 h+3 k+4=0 \quad \ldots \text { (ii) } \end{aligned} $

On comparing Eq. (ii) with Eq. (i), we get

$ \begin{aligned} \frac{4}{h-2} & =\frac{-3}{k+3} \\\\ 4 k+3 h & =-6 \quad \ldots \text { (iii) } \end{aligned} $

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

$ \Rightarrow \quad 7 k-6 h=-6 \quad \ldots \text { (iv) } $

On solving Eqs. (iii) and (iv), we get

$ h=\frac{-2}{5}, k=\frac{-6}{5} $

$ \text { Hence, } h-2 k=\frac{-2}{5}+\frac{12}{5}=2 $

We know that inverse of any point $P(x, y)$ with respect to circle centred at $(h, k)$ is $Q\left(x^{\prime}, y^{\prime}\right)$, where

$ x^{\prime}=\alpha(x-h)+h \Rightarrow y^{\prime}=\alpha(y-k)+k $

and $\alpha=\frac{r^2}{(x-h)^2+(y-k)^2}$

$r=$ radius of circle

Here, $h=2, k=-2$ and $r=2$,

$ \begin{aligned} & x=3, y=3 \\\\ & \Rightarrow \alpha=\frac{2^2}{(3-2)^2+(3+2)^2}=\frac{4}{26}=\frac{2}{13} \end{aligned} $

Now,

$ \begin{aligned} & x^{\prime}=a=\frac{2}{13}(3-2)+(2)=\frac{28}{13} \\\\ & y^{\prime}=b=\frac{2}{13}(3+2)-2=-\frac{16}{13} \end{aligned} $

Hence, $a+5 b=\frac{28}{13}+5\left(-\frac{16}{13}\right)$

$ =\frac{-}{-52} \frac{52}{13}=-4 $

We have,

$ \begin{aligned} & S_1: x^2+y^2-4 x+6 y+4=0 \\\\ & C_1:(2-3), r_1=3 \\\\ & S_2: x^2+y^2+2 x-2 y-2=0 \\\\ & C_2:(-1,1), r_2=2 \\\\ & \text { Here, } C_1 C_2=r_1+r_2 \end{aligned} $

So, the point $P$ divides $C_1 C_2$ in the ratio of $3: 2$

$ P\left(\frac{1}{5}, \frac{-3}{5}\right) $

Slope of line passing through $C_1$ and $C_2$ is $\frac{1+3}{-1-2}=\frac{4}{-3} \Rightarrow P:\left(\frac{1}{5},-\frac{3}{5}\right)$ and here only one transverse common tangent which is perpendicular to $C_1 C_2$

$\Rightarrow$ slope of tangent $=-\frac{1}{-\frac{4}{3}}=\frac{3}{4}$

$\because$ Tangent is also passes through $\left(\frac{1}{5},-\frac{3}{5}\right)$

So, equation of tangent is

$ y+\frac{3}{5}=\frac{3}{4}\left(x-\frac{1}{5}\right) \Rightarrow 3 x-4 y-3=0 $

On comparing with

$ a x+b y+c=0, \frac{a}{c}=-1 $

We have,

$ S \equiv x^2+y^2+2 g x+2 f y+6=0 $

cuts another circle

$ \begin{aligned} & x^2+y^2-6 x-6 y-6=0 \text { orthogonally. } \\\\ & \Rightarrow 2 g_1 g_2+2 f_1 f_2=c_1+c_2 \\\\ & 2 g(-3)+2 f(-3)=6-6 \Rightarrow g+f=0 \end{aligned} $

and angle between $S=0$ and

$ x^2+y^2+6 x+6 y+2=0 \text { is } 60^{\circ} . $

The condition for angle $\theta$ between two circles is given by

$ \begin{aligned} \cos \theta & =\frac{\left|2\left(g_1 g_2+f_1 f_2\right)-\left(c_1+c_2\right)\right|}{2 \sqrt{r_1^2 r_2^2}} \\\\ \cos 60^{\circ} & =\frac{2(3 g+3 f)-(6+2)}{2 \sqrt{4^2 \cdot r^2}} \end{aligned} $

(Let radius of $\dot{s}=0$ is $r$ )

$ \begin{aligned} & \frac{1}{2}=\frac{|2 \times 3 \times 0-8|}{2 \times 4 \times r}(\because g+f=0) \\\\ & 4 r=g \Rightarrow r=2 \end{aligned} $

We have,

$ \begin{aligned} & C_1: x^2+y^2-2 x-8 y+8=0 \,\,\,\,\,\,\,\,\,\,\,.......(i) \\\\ & \text { Centre }(1,4), r_1=3 \\\\ & C_2: x^2+y^2-8 x+15=0 \,\,\,\,\,\,\,\,\,\,\,.......(ii) \\\\ & \text { Centre }(4,0), r_2=1 \\\\ & \text { As } \frac{C_1 P}{C_2 P}=\frac{C_1 A}{C_2 C}=\frac{3}{1} \\\\ & \Rightarrow P_x=\frac{12-1}{2}=\frac{11}{2}\,\,\,\,\,\,\,\,\,\,\, \text { (external division) } \\\\ & P_y=\frac{0-4}{2}=-2 \end{aligned} $

Let slope of tangent $A P$ be $m$.

Then, equation of $A P$ is :

$ \begin{aligned} & y+2=m\left(x-\frac{11}{2}\right) \\\\ & 2 m x-2 y-(4+11 m)=0 \end{aligned} $

$\because$ Perpendicular distance of $C_2$ from $A P$ is 1 .

$ \therefore\left|\frac{2 m \times 4-2 \times 0-(4+11 m)}{\sqrt{(2 m)^2+(-2)^2}}\right|=1 $

$ \begin{aligned} & (3 m+4)^2=4 m^2+4 \\\\ & 5 m^2+24 m+12=0 \end{aligned} $

So, $m_1+m_2=-\frac{24}{5}$ where $m_1$ and $m_2$ are slope of $A P$ and $B P$.

$A(a, 3), B(b, 5), C(a, b)$ circumcentre $O(1,1)$

We know that, $O A^2=O B^2=O C^2$

On equating II and III, we get

$ \begin{aligned} & (a-1)^2=16 \Rightarrow a-1= \pm 4 \\ & \Rightarrow \quad a=5,-3 \end{aligned} $

On equating I and III, we get

$ (b-1)^2=4 \Rightarrow b-1= \pm 2 \Rightarrow b=3,-1 $

Sum of absolute distinct values

$ =5+3+1=9 $

$ \begin{aligned} &\text { Passes through }(-6,3)\\ &\begin{aligned} & \text { Radius }=a \\ & c(-a, a) \end{aligned} \end{aligned} $

Equation will be

$ \begin{aligned} (x+a)^2+(y-a)^2 & =a^2 \\ \Rightarrow \quad x^2+y^2+2 a x-2 a y+a^2 & =0 \end{aligned} $

At $(-6,3)$,

$ \begin{aligned} 36+9-12 a-6 a+a^2 & =0 \\ \Rightarrow \quad a^2-18 a+45 & =0 \end{aligned} $

$⇒ a=15,3 $

Equation of circle will be

$ \begin{aligned} & x^2+y^2+30 x-30 y+225=0 \\ & x^2+y^2+6 x-6 y+9=0 \end{aligned} $

$x^2+y^2-10 x=0$

Tangent at $(9,3)$

$ \begin{aligned} & T=0 \Rightarrow 9 x+3 y-5(x+9)=0 \\ & -4 x+3 y=45 \end{aligned} $

$ \begin{aligned} &\text { Normal at }(9,3)\\ &\begin{aligned} & 3 x-4 y=\lambda \Rightarrow 3 \times 9-4 \times 3=\lambda \\ & \Rightarrow \lambda=15 \Rightarrow 3 x-4 y=15 \\ & \text { Area of triangle }=\frac{1}{2} \times\left(\frac{45}{4}-5\right) \times 3=\frac{75}{8} \end{aligned} \end{aligned} $

Given, circles

$ \begin{aligned} & x^2+y^2-4=0 \text { and } \\ & (x-3)^2+(y-4)^2=49 \\ & c_1\left(0,0, r_1=2 \text { and } c_2(3,4), r_2=7\right. \\ & \text { Now, } c_1 c_2=\sqrt{9+16}=5 \\ & \left|r_1-r_2\right|=5 \\ & c_1 c_2=\left|r_1-r_2\right| \end{aligned} $

⇒ Circles touch internally.

Given, circle

$ x^2+y^2+2 x+8 y-23=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Equation of new circle touching

Eq. (i) at point $(1,2)$ is

$ \begin{gathered} x^2+y^2+2 x+8 y-23+\lambda(x-1)^2 \left.+(y-2)^2\right]=0 \\ \Rightarrow(1+\lambda) x^2+(1+\lambda) y^2+(2-2 \lambda) x +(8-4 \lambda) y-23+5 \lambda=0 \\ \Rightarrow x^2+y^2-2 x\left(\frac{\lambda-1}{\lambda+1}\right)-4 y\left(\frac{\lambda-2}{\lambda+1}\right) +\frac{5 \lambda-23}{\lambda+1}=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \text { } \end{gathered} $

$ \begin{aligned} &\text { Given that } r=\sqrt{10}\\ &\begin{aligned} & \Rightarrow\left(\frac{\lambda-1}{\lambda+1}\right)^2+4\left(\frac{\lambda-2}{\lambda+1}\right)^2-\frac{5 \lambda-23}{\lambda+1}=10 \\ & \Rightarrow \lambda^2-2 \lambda+1+4 \lambda^2-16 \lambda+16 -\left(5 \lambda^2-18 \lambda-23\right)=10(\lambda+1)^2 \\ & \Rightarrow \quad 10 \lambda^2+20 \lambda-30=0 \\ & \Rightarrow \quad \lambda^2+2 \lambda-3=0 \end{aligned} \end{aligned} $

$ \Rightarrow \quad \lambda=-3,1 $

At $\lambda=1 \Rightarrow$ Eq. (ii)becomes

$ \begin{array}{lr} \Rightarrow & x^2+y^2+2 y-9=0 \\ \Rightarrow & |a+b+c|=7 \end{array} $

At $\lambda=-3 \Rightarrow$ Eq. (ii) becomes

$ \begin{aligned} & \Rightarrow x^2+y^2-4 x-10 y+19=0 \\ & \Rightarrow \quad|a+b+c|=5 \end{aligned} $

$x^2+y^2-4 x-6 y+10=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Let the circle passing through origin centred on line $x-y=0$

$ x^2+y^2+2 g x+2 f y=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

∵ Line (i) and circle (ii) cuts orthogonally

$ \begin{aligned} & \left\{\begin{array}{l} g=f \\ \text { and } c=0 \end{array}\right. \\ & -4 g-6 g=0+10 \Rightarrow g=-1 \end{aligned} $

Now, radius of (ii) circle

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

Diameter $=2 r=2 \sqrt{2}$

$ \begin{aligned} &\text { Equation of required circle is }\\ &\begin{aligned} & \begin{array}{l} x^2+y^2+6 x+4 y-12+\lambda \\ \quad\left(x^2+y^2-4 x-6 y-12\right)=0 \end{array} \\ & \Rightarrow x^2+y^2+\frac{(6-4 \lambda) x}{\lambda+1}+\frac{(4-6 \lambda) y}{\lambda+1} -\frac{12(\lambda+1)}{\lambda+1}=0 \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { Radius }=\sqrt{13} \\ & \begin{array}{l} \Rightarrow \sqrt{\frac{(2 \lambda-3)^2}{(\lambda+1)^2}+\frac{(3 \lambda-2)^2}{(\lambda+1)^2}+12}=\sqrt{13} \\ \Rightarrow \quad(\lambda+1)^2=13 \lambda^2+13-24 \lambda \\ \Rightarrow 12 \lambda^2-26 \lambda+12=0 \\ \Rightarrow \quad 6 \lambda^2-13 \lambda+6=0 \\ \quad \lambda=\frac{3}{2}, \frac{2}{3} \end{array} \\ & \text { At } \lambda=\frac{3}{2} \Rightarrow x^2+y^2-2 y-12=0 \\ & \text { At } \lambda=\frac{2}{3} \Rightarrow x^2+y^2+2 x-12=0 \end{aligned} $

Let, coordinates of $P=(x, y)$

So, distance of $P$ from

$ (0,2)=\sqrt{(x-0)^2+(y-2)^2} $

and distance of $P$ from

$ (-1,0)=\sqrt{(x+1)^2+(y-0)^2} $

According to the question,

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

On squaring both sides, we get

$ \begin{aligned} & 2\left(x^2+y^2-4 y+4\right)=x^2+y^2+2 x+1 \\ & \Rightarrow x^2+y^2-2 x-8 y+7=0 \end{aligned} $

which is the equation of a circle.

$ \begin{aligned} \text { Centre of circle } & =\left(\frac{-(-2)}{2}, \frac{-(-8)}{2}\right) \\ & =(1,4) \end{aligned} $

and radius $(r)=\sqrt{(1)^2+(4)^2-7}=\sqrt{10}$

Hence, a circle with centre $(1,4)$ and radius $\sqrt{10}$ units.

$(3,4),(3,2)$ and $(1,4)$

$ \begin{gathered} x=a+r \cos \theta, y=b+r \sin \theta, \\ \cos \theta=\frac{x-a}{r}, \sin \theta=\frac{y-b}{r} \\ \because \quad \sin ^2 \theta+\cos ^2 \theta=1 \\ \left(\frac{x-a}{r}\right)^2+\left(\frac{y-b}{r}\right)^2=1 \end{gathered} $

Equation of circle $(x-a)^2+(y-b)^2=r^2$

$ x^2+y^2-2 a x-2 b y+a^2+b^2=r^2 $

Putting the point $(3,4),(3,2),(1,4)$ respectively.

$ \begin{aligned} & (3,4) \rightarrow 9+16-6 a-8 b+a^2+b^2=r^2 \\ & a^2+b^2-6 a-8 b+25=r^2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & (3,2) \rightarrow 9+4-6 a-4 b+a^2+b^2=r^2 \\ & a^2+b^2-6 a-4 b+13=r^2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ & (1,4) \rightarrow 1+16-2 a-8 b+a^2+b^2=r^2 \\ & a^2+b^2-2 a-8 b+17=r^2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{aligned} $

Subtracting Eq. (ii) from Eq. (i),

$ \begin{array}{ll} & -4 b+12=0 \\ \Rightarrow & b=3 \end{array} $

Subtracting Eq. (iii) from Eq. (i),

$ \begin{array}{r} -4 a+8=0 \\\Rightarrow \,\,\,\,\,\,\,\,\, a=2 \end{array} $

Putting values of $a$ and $b$ in Eq. (i),

$ \begin{array}{lc} & 4+9-12-24+25=r^2 \\ \Rightarrow & r^2=2 \Rightarrow r=\sqrt{2} \\ \therefore & b^a r^a=3^2(\sqrt{2})^2=9 \times 2=18 \end{array} $

Given, equation of circle is

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

On differentiating Eq. (i) w.r.t. $x$ on both sides, we get

$ \begin{aligned} & 2 x+2 y \frac{d y}{d x}=0 \\ \Rightarrow \quad & 2 y \frac{d y}{d x}=-2 x \Rightarrow \frac{d y}{d x}=\frac{-x}{y} \end{aligned} $

∴ Slope of tangent to $x^2+y^2=4$ at

$P(\sqrt{3}, 1)$ is given by $\left(\frac{d y}{d x}\right)_{(\sqrt{3}, 1)}$.

So, $\left(\frac{d y}{d x}\right)_{(\sqrt{3}, 1)}=\frac{-\sqrt{3}}{1}=-\sqrt{3}$

So, slope of line $L=\frac{-1}{-\sqrt{3}}=\frac{1}{\sqrt{3}}$

$ \left[\because \text { slope of line }=\frac{-1}{d y / d x}\right] $

Let equation of line $L$ be $x-\sqrt{3} y+c=0$

As, $L$ is tangent to $(x-3)^2+y^2=1$.

So, perpendicular distance of $L$ from its centre should be equal to its radius i.e. 1. Centre $(3,0)$

$ \therefore \quad \frac{|3+c|}{\sqrt{4}}=1 \Rightarrow c=-1 \text { or }-5 $

So, equation of line $L$ is $x-\sqrt{3} y=1$ or $x-\sqrt{3} y=5$

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

$ \begin{aligned} & P(6,-5) \\ & (x-1)^2+(y+2)^2=2 \end{aligned} $

Center $(1,-2)$

Radius $=\sqrt{2}$

Equation of tangent to given circle

$ \begin{aligned} y+2 & =m(x-1) \pm \sqrt{2} \sqrt{1+m^2} \\ (\because y-k & \left.=m(x-h) \pm a \sqrt{1+m^2}\right) \end{aligned} $

$m$ is slope of the tangent.

Tangent is passing through $(6,-5)$

$ \begin{aligned} -5+2 & =m(6-1) \pm \sqrt{2} \sqrt{1+m^2} \\ -3 & =5 m \pm \sqrt{2} \sqrt{1+m^2} \\ -3-5 m & = \pm \sqrt{2} \sqrt{1+m^2} \end{aligned} $

Squaring on both sides,

$ \begin{array}{rr} & 9+25 m^2+30 m=2\left(1+m^2\right) \\ \Rightarrow & 23 m^2+30 m+7=0 \\ \Rightarrow & 23 m^2+23 m+7 m+7=0 \\ \Rightarrow & (23 m+7)(m+1)=0 \\ \Rightarrow & m_1=-1, m_2=-\frac{7}{23} \end{array} $

$ \begin{array}{ll} & \therefore \quad \tan \theta=\left|\frac{m_1-m_2}{1+m_1 m_2}\right|=\left|\frac{-1+\frac{7}{23}}{1+\frac{7}{23}}\right| \\ \Rightarrow & \tan \theta=\left|\frac{-16}{30}\right| \Rightarrow \tan \theta=\frac{16}{30}=\frac{8}{15} \\ \Rightarrow & \tan \theta=\frac{8}{15} \Rightarrow \cot \theta=\frac{15}{8} \end{array} $

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

Centre $\left(c_1\right)=(2,3)$

and $r_1=\sqrt{(2)^2+(3)^2-k}=\sqrt{13-k}$

and $x^2+y^2+8 x-4 y+11=0$

Centre $\left(c_2\right)=(-4,2)$

and $r_2=\sqrt{(-4)^2+(2)^2-11}=\sqrt{9}=3$

Now,

$ d=c_1 c_2=\sqrt{(2+4)^2+(3-2)^2}=\sqrt{37} $

As, we know

$ \cos \theta=\frac{r_1^2+r_2^2-d^2}{2 r_1 \cdot r_2} $

$ \begin{aligned} & \Rightarrow \cos \pi / 2=\frac{13-k+9-37}{2(\sqrt{13-k}) \times 3} \\ & \Rightarrow \quad 0=22-37-k \Rightarrow k=-15 \end{aligned} $

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

Centres of four circles are $\left(\lambda_{,}{ }^A \lambda\right),\left(-\lambda_{,}^B, \lambda\right),\left(-\lambda_{,}{ }^C-\lambda\right)$ and $(\lambda,-\lambda)$. Radius of circles are $\lambda$.

Radius of possible circles are, $O P$ or $O Q$

$ \begin{aligned} O Q & =O A+A Q=O A+\lambda \\ \text { and } O A & =\sqrt{(\lambda-0)^2+(\lambda-0)^2} \\ & =\lambda \sqrt{2} \end{aligned} $

So, $O Q=\lambda(\sqrt{2}+1)$

and $O P=\lambda \sqrt{2}-\lambda=\lambda(\sqrt{2}-1)$

From the given options

Then, radius $=(\sqrt{2}-1) \lambda$

$ \begin{aligned} & \text { } S_1: x^2+y^2-1=0 \\ & S_2: x^2+y^2-2 x-3=0 \\ & S_3: x^2+y^2-2 y-3=0 \end{aligned} $

For radical circle,

$ \begin{aligned} & S_1-S_2=0 \\ & \Rightarrow-1+2 x+3=0 \Rightarrow x=-1 \\ & \text { and } S_2-S_3=0 \Rightarrow-2 x-3+2 y+3=0 \\ & \Rightarrow x=y=-1 \end{aligned} $

So, centre of radical circle $=C(\alpha, \beta)$

$ =(-1,-1) $

and $r_1=1, r_2=\sqrt{(1)^2-(-3)}=2$

and $r_3=\sqrt{(+1)^2-(-3)}=2$

So, $r=r_1+r_2+r_3=5$

Therefore, equation of circle

$ (x+1)^2+(y+1)^2=25 $

Equation of $H F$, which is mid-parallel to $A D$ and $B C$

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

Equation of $G E$, which is mid-parallel to $A B$ and $C D$

$ x-y+3=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

On solving Eqs. (i) and (ii), we get

$ x=(1 / 2) \text { and } y=(7 / 2) $

∴ Centre $=(1 / 2,7 / 2)$

$ \text { and radius }=\frac{1}{2} \times \frac{|-6-(-2)|}{\sqrt{1^2+1^2}}=\frac{2 \sqrt{2}}{2}=\sqrt{2} $

∴ Equation of circle

$ \begin{aligned} \left(x-\frac{1}{2}\right)^2+\left(y-\frac{7}{2}\right)^2=(\sqrt{2})^2 \\ \Rightarrow \quad 2 x^2+2 y^2-2 x-14 y+21 & =0 \end{aligned} $

We have,

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

which touch the positive $X$-axis and positive $Y$-axis.

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

and equation of circles

$ \Rightarrow \quad(x-r)^2+(y-r)^2=r^2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

$\because P(2,4)$ point lies on the circle,

$ \begin{aligned} & (2-r)^2+(4-r)^2=r^2 \\ & \Rightarrow 4+r^2-4 r+16+r^2-8 r=r^2 \\ & \Rightarrow \quad r^2-12 r+20=0 \\ & \Rightarrow \quad r=10,2 \end{aligned} $

Hence, the difference of their areas

$ \begin{aligned} & =\pi\left(r_1^2-r_2^2\right)=\pi\left(10^2-2^2\right) \\ & =\pi(100-4)=96 \pi \end{aligned} $

We have,

$ \begin{array}{r} 2 x-3 y+3=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ 2 x+y+1=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ 6 x+4 y+1=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{array} $

represent the sides of a triangle. On solving Eqs. (i) and (ii), we get Point $A\left(-\frac{3}{4}, \frac{1}{2}\right)$.

Similarly, on solving Eq. (ii) with Eq. (iii) and Eq. (i) with Eq. (iii), we get point $B\left(-\frac{3}{2}, 2\right)$ and point $C\left(\frac{8}{13}, \frac{-15}{26}\right)$ respectively.

Let $E$ and $F$ is the mid-point of $A C$ and $A B$.

$ \therefore E \equiv\left(-\frac{7}{104}, \frac{-1}{26}\right) \text { and } F \equiv\left(\frac{-9}{8}, \frac{5}{4}\right) $

Now, equation of line passing through $E$ and perpendicular to $A C$ is

$ 312 x+208 y=-29\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv) $

and equation of line passing through $F$ and perpendicular to $A B$ is

$ 8 x-16 y=-29\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(v) $

$\because A B$ and $A C$ represent the chord of the circle and we know that perpendicular line from two chords of a circle intersect at the centre of circle.

On solving Eqs. (iv) and (v), we get

$ x=\frac{-9}{8}, y=\frac{10}{8} $

So, centre of circle is $O\left(\frac{-9}{8}, \frac{10}{8}\right)$.

$ \begin{aligned} \text { Distance }(A O) & =\sqrt{\left(\frac{-9}{8}+\frac{3}{4}\right)^2+\left(\frac{10}{8}-\frac{1}{2}\right)^2} \\ & =\text { radius }(r) \\ \Rightarrow \quad r & =\frac{3}{8} \sqrt{5} \end{aligned} $

Hence, required equation of circle with centre $\left(\frac{-9}{8}, \frac{10}{8}\right)$ and radius $\frac{3}{8} \sqrt{5}$ is

$ \begin{array}{r} \quad\left(x+\frac{9}{8}\right)^2+\left(y-\frac{10}{8}\right)^2=\left(\frac{3}{8} \sqrt{5}\right)^2 \\ \Rightarrow \quad 8 x^2+8 y^2+18 x-20 y+17=0 \end{array} $

Wrong information

As, $S^{\prime}=x^2+y^2+4 x+4=0$

represent the circle whose centre is ( $-2,0$ ) and radius is zero.

52. (a) We know that two circles

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

and $x^2+y^2+2 g^{\prime} x+2 f^{\prime} y+c^{\prime}=0$ cuts orthogonally, if $2 g g^{\prime}+2 f f^{\prime}=c+c^{\prime}$

So, $2 g(-2)+2 f(-2)=4+(-4)$

$ \Rightarrow \quad g+f=0 $

We also know that if angle of intersection of two circles is $\theta$, then

$ \cos \theta=\left|\frac{r_1^2+r_2^2-d^2}{2 r_1 r_2}\right| \text {, where } d $

represents the distance between centre of these circles.

For the circle $x^2+y^2+2 g x+2 f y+4=0$, the centre is ( $-g,-f$ ) and radius $r_1$ is

$ \sqrt{g^2+f^2-4}=\sqrt{2 f^2-4} $

$ \,\,\,\, \,\,\,\, \,\,\,\, \,\,\,\, \,\,\,\, \,\,\,\, \,\,\,\, \,\,\,\, \,\,\,\, \,\,\,\, \,\,\,\, \,\,\,\, \,\,\,\, \,\,\,\, \,\,\,\, \,\,\,\, \,\,\,\, [\because g+f=0] $

For the circle $x^2+y^2+4 x+4 y+4=0$, the centre is $(-2,-2)$ and radius $r_2$ is $\sqrt{(-2)^2+(-2)^2-4}=2$

Now, $d=\sqrt{(-g+2)^2+(-f+2)^2}$

$ \begin{array}{r} \Rightarrow d^2=(g-2)^2+(f-2)^2 \\ \Rightarrow d^2=g^2+4-4 g+f^2+4-4 f=2 f^2+8 \\ \quad[\because g+f=0] \end{array} $

$ \begin{aligned} & \therefore \quad \cos 60^{\circ}=\left|\frac{\left(2 f^2-4\right)+4-\left(2 f^2+8\right)}{2 \sqrt{2 f^2-4}(2)}\right| \\ & \Rightarrow \quad \frac{1}{2}=\left|\frac{-8}{4 \sqrt{2 f^2-4}}\right| \\ & \Rightarrow \quad 2 f^2-4=16 \Rightarrow 2 f^2=20 \end{aligned} $

$ \begin{aligned}Thus,\,\, r_1 & =\sqrt{2 f^2-4} \\ & =\sqrt{20-4}=\sqrt{16}=4 \end{aligned} $

We know that two circles

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

and $x^2+y^2+2 g^{\prime} x+2 f^{\prime} y+c^{\prime}=0$ cuts orthogonally, if $2 g g^{\prime}+2 f f^{\prime}=c+c^{\prime}$

$ \begin{aligned} & \text { So, } 2 g(2)+2 f(2)=c+7 \\ & \Rightarrow \quad 4 g+4 f=c+7 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & \quad 2 g(-2)+2 f(2)=c+7 \\ & \Rightarrow \quad-4 g+4 f=c+7 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ & 2 g(-2)+2 f(-2)=c+7 \\ & \Rightarrow \quad 4 g-4 f=c+7 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{aligned} $

on solving the Eqs. (i), (ii) and (iii), we get

$ g=0, f=0 \text { and } c=-7 $

Now, the equation of circle $s=0$ becomes

$ \begin{aligned} x^2+y^2-7 & =0 \\ \Rightarrow \quad 2 x+2 y \frac{d y}{d x}-0 & =0 \Rightarrow \frac{d y}{d x}=-\frac{x}{y} \end{aligned} $

$ \begin{aligned} &\Rightarrow \quad\left(\frac{d y}{d x}\right)_{(\sqrt{3}, 2)}=-\frac{\sqrt{3}}{2}\\ &\text { ∴ The required equation be }\\ &\begin{array}{ll} & y-2=-\frac{\sqrt{3}}{2}(x-\sqrt{3}) \\ \Rightarrow & \sqrt{3} x+2 y-7=0 \end{array} \end{aligned} $

$C(2,3)$ is centre

$x+y+1=0$ is chord

As, $\angle A C B=60^{\circ}$

$\therefore \triangle A B C$ is equilateral

$ \Rightarrow A C=A B=B C=\text { radius of circle }=r(\text { let }) $

Draw $C M \perp A B$

$ \Rightarrow \quad A M=M B $

Now, $C M=\left|\frac{2+3+1}{\sqrt{1^2+1^2}}\right|=\frac{6}{\sqrt{2}}=3 \sqrt{2}$

$\triangle A C M$ is right angled triangle, then

$ \begin{aligned} A C^2 & =C M^2+A M^2 \\ r^2 & =18+\left(\frac{r}{2}\right)^2 \Rightarrow \frac{3}{4} r^2=18 \\ r^2 & =24 \end{aligned} $

Equation of circle $(x-2)^2+(y-3)^2=24$

$ \Rightarrow x^2+y^2-4 x-6 y-11=0 $

$X$-intercept $=6$

$ \begin{aligned} \Rightarrow & & 2 \sqrt{g^2-c} & =6 \\ \Rightarrow & & c & =g^2-9 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

$Y$-intercept $=8$

$ \begin{aligned} \Rightarrow & & 2 \sqrt{f^2-c} & =8 \\ \Rightarrow & & c & =f^2-16\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \end{aligned} $

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

$ \begin{array}{r} f^2-g^2=7 \\& (f+g)(f-g)=7 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{array} $

$\because g x+f y=1$ passes through $(1,-1)$

Then,

$ f-g=-1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)$

From Eq. (iii), we get

$ f+g=-7 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(v)$

On solving Eqs. (iv) and (v), we get

$ f=-4, g=-3 $

From Eq. (i), we get

$ \begin{aligned} c & =0 \\ \text { Radius } & =\sqrt{g^2+f^2-c} \\ & =\sqrt{\left(-3^2\right)+(-4)^2-0}=5 \end{aligned} $

Let equation of circle be

$ x^2+y^2+2 g x+2 f y+c=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Circle passes through $(3,1)$ and $(-2,4)$

$ \begin{array}{rr} \therefore \quad 6 g+2 f+c+10=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ -4 g+8 f+c+20=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{array} $

Also, centre $(-g,-f)$ lies on the line

$ \begin{array}{ll} & x-y+1=0 \\ \therefore & -g+f+1=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)\end{array} $

On solving Eqs. (ii) and (iii), we get

$ \begin{array}{r} 10 g-6 f-10=0 \\&or\,\, g-\frac{3}{5} f-1=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(v)\end{array} $

or On solving Eqs. (iv) and (v), we get

$ f=0 \text { and } g=1 $

Using Eq. (ii), we get

$ c=-16 $

Equation of circle is

$ \begin{aligned} x^2+y^2+2 x-16 & =0 \\ (x+1)^2+y^2 & =(\sqrt{17})^2 \end{aligned} $

Parametric form

$ \begin{gathered} x+1=\sqrt{17} \cos \theta, y=\sqrt{17} \sin \theta \\or\,\, x=-1+\sqrt{17} \cos \theta \\ y=\sqrt{17} \sin \theta \end{gathered} $

$S \equiv x^2+y^2-8 x+10 y+5=0$

$ P(1,1) \text { and } Q(1,-1) $

Equation of polar of $P(1,1)$

$ \begin{array}{} S_1=0 \\ 3 x-6 y+6=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{array} $

Equation of chord with $Q$ as mid-point is

$ \begin{aligned} S_1 & =S_{11} \\ -3 x+4 y+7 & =0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

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

$ x=11, y=\frac{13}{2} $

∴ Point of intersection $\left(11, \frac{13}{2}\right)$

Given, circles

$ x^2+y^2-2 x+2 y+1=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

and  $ x^2+y^2+2 x-2 y+k=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

On comparing with

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

$ \begin{aligned}and\,\, & x^2+y^2+2 g_1 x+2 f_1 y+c_1=0 \\ & g=-1, f=1, c=1 \end{aligned} $

and  $ g_1=1, f_1=-1, c_1=k $

Angle between circles (i) and (ii) is $\frac{\pi}{3}$.

$ \begin{array}{rlr} \therefore \cos \frac{\pi}{3} & =\frac{\left(c+c_1\right)-2\left(g g_1+f f_1\right)}{2 \sqrt{g^2+f^2-c} \sqrt{g_1^2+f_1^2-c_1}} \\ \Rightarrow & \frac{1}{2} =\frac{1+k-2(-1-1)}{2 \sqrt{1+1-1} \sqrt{1+1-k}} \\ \Rightarrow & \sqrt{2-k}=k+5 \\ \Rightarrow & 2-k=k^2+10 k+25 \\ \Rightarrow & k^2+11 k+23=0 \\ \Rightarrow & k =\frac{-11 \pm \sqrt{121-92}}{2} \\ & =\frac{-11 \pm \sqrt{29}}{2} \end{array} $

$\therefore k$ is an irrational number.

Given, line

$ x-y+1=0 $

and circle $x^2+y^2+2 x+2 y+1=0$

For intersection substituting $y=x+1$ in Eq. (ii), we have

$ \begin{aligned} &y=x+1 \text { in Eq. (ii), we have }\\ &\begin{array}{lr} x^2+(x+1)^2+2 x+2(x+1)+1=0 \\ x^2+x^2+2 x+1+2 x+2 x+2+1=0 \\ \Rightarrow 2 x^2+6 x+4=0 \\ \Rightarrow x^2+3 x+2=0 \\ \Rightarrow (x+1)(x+2)=0 \\ \Rightarrow x=-1, x=-2 \end{array} \end{aligned} $

∴ For $x=-1, y=0$

and For $x=-2, y=-1$

$A(-1,0)$ and $B(-2,-1)$ are diameter of a circle.

Equation of circle is

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

On comparing with

$ \begin{aligned} & x^2+y^2+2 g x+2 f y+c=0 \\ & g=\frac{3}{2}, f=\frac{1}{2}, c=1 \\ & g+f=\frac{3}{2}+\frac{1}{2}=2=2 c \end{aligned} $

The given two diameters are,

$ \begin{aligned} & x-y=2 \\ & 2 x+3 y=14 \end{aligned} $

Now, solving these equations, we get

$ x=4 \text { and } y=2 $

So, centre of circle is $(4,2)$.

Now, equation of circle is

$ (x-4)^2+(y-2)^2=r^2 $

where, ' $r$ ' is radius.

Since, it also passes through $(1,-2)$, then

$ \begin{array}{rlrl} (1-4)^2+(-2-2)^2 & =r^2 \\ \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, 9+16 & =r^2 \\ \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\,\,\,\, r^2 & =25 \\ \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, r & =5 \end{array} $

To find the equation of the circle whose diameter is the common chord of two given circles, we start by considering the equations of the circles:

$ S_1 = x^2 + y^2 + 2x + 3y + 1 = 0 \quad \text{(Equation 1)} $

$ S_2 = x^2 + y^2 + 4x + 3y + 2 = 0 \quad \text{(Equation 2)} $

The common chord of these two circles is also known as the radical axis, which can be found by subtracting the second equation from the first:

$ S_1 - S_2 = 0 $

This results in:

$ (x^2 + y^2 + 2x + 3y + 1) - (x^2 + y^2 + 4x + 3y + 2) = 0 $

Simplifying this, we obtain:

$ 2x + 1 = 0 $

$ \Rightarrow x = -\frac{1}{2} $

Now, considering this line as $ S_3 = 2x + 1 = 0 $.

To find the circle that passes through the points of intersection of $ S_1 $ and $ S_3 $, its general equation is:

$ S_1 + \lambda S_3 = 0 $

Substituting the values, we have:

$ (x^2 + y^2 + 2x + 3y + 1) + \lambda(2x + 1) = 0 $

Simplifying further:

$ x^2 + y^2 + (2 + 2\lambda)x + 3y + (1 + \lambda) = 0 $

The center of this circle is $(-1 - \lambda, \frac{3}{2})$.

The line $ 2x + 1 = 0 $ will be a diameter if the center of our circle lies on this line. So, we solve:

$ 2(-1 - \lambda) + 1 = 0 $

$ -2 - 2\lambda + 1 = 0 $

$ -2\lambda - 1 = 0 $

$ \lambda = -\frac{1}{2} $

Substituting $\lambda = -\frac{1}{2}$ back into our circle equation gives:

$ x^2 + y^2 + (2 - 1)x + 3y + \frac{1}{2} = 0 $

$ \Rightarrow 2(x^2 + y^2) + 2x + 6y + 1 = 0 $

Therefore, the equation of the circle whose diameter is the common chord is:

$ 2x^2 + 2y^2 + 2x + 6y + 1 = 0 $

The equation of circle are

$ x^2+y^2-2 x-6 y+9=0 $

and $x^2+y^2+6 x-2 y+1=0$

Centre and radii of the circles are $c_1(1,3)$ and $r_1=1$

and $c_2=(-3,1)$ and $r_2=3$

Clearly,

$ \begin{aligned} & c_1 c_2=\sqrt{(1+3)^2+(3-1)^2}=\sqrt{16+4}=\sqrt{20} \\ & r_1+r_2=4=\sqrt{16} \end{aligned} $

Thus, $c_1 c_2>r_1+r_2$

Hence, the circles are non-intersecting. Thus, there will be four common tangents.

Transverse common tangents are tangents drawn from the point $P$ which divides $c_1, c_2$ internally in the ratio of radii 1: 3 coordinate of $P$ are

$ \left(\frac{1 \cdot(-3)+3 \cdot 1}{1+3}, \frac{1 \cdot 1+3 \cdot 3}{1+3}\right)=\left(0, \frac{5}{2}\right) $

Equation of tangent through the point $P\left(0, \frac{5}{2}\right)$.

Any line through $P\left(0, \frac{5}{2}\right)$ is

$ y-\frac{5}{2}=m(x-0) \Rightarrow m x-y+\frac{5}{2}=0 $

Now, $\frac{m \cdot 1-3+\frac{5}{2}}{\sqrt{m^2+1}}=1$

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

$\Rightarrow m^2+\frac{1}{4}-m=m^2+1 \Rightarrow m=\frac{-3}{4}$ and $^{\prime} \infty$ '

Thus, $x=0$ is a tangent and $y-\frac{5}{2}=-\frac{3 x}{4}$

$ \Rightarrow 3 x+4 y-10=0 $

is another tangent.

Direct common tangents are tangents drawn from the point $Q$ which divides $c_1 c_2$ externally in the ratio $1: 3$.

Direct tangent, are tangent drawn from the point $Q(3,4)$.

Now, proceeding as for transverse tangents their equations are $y=4$,

$ 4 x-3 y=0 $

Thus, the number of common tangent are ' 4 '.

The given straight line,

$ 9 x+y-28=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

and the given circle is

$ \begin{aligned} & \quad 2 x^2+2 y^2-3 x+5 y-7=0 \\ & \text { i.e. } \quad x^2+y^2-\frac{3}{2} x+\frac{5}{2} y-\frac{7}{2}=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

Let, $(h, k)$ be the pole of the given line Eq. (i) w.r.t. the circle Eq. (ii).

The equation of the polar

$ \begin{aligned} & h x+k y-\frac{3}{4}(x+h)+\frac{5}{4}(y+k)-\frac{7}{2}=0 \\ & \Rightarrow x\left(h-\frac{3}{4}\right)+y\left(k+\frac{5}{4}\right)-\frac{3}{4} h+\frac{5}{4} k-\frac{7}{2}=0 \\ & \Rightarrow x(4 h-3)+y(4 k+5)-3 h+5 k-14=0 \end{aligned} $

This equation and $9 x+y-28=0$ represent same line so,

$ \begin{aligned} \frac{4 h-3}{9} & =\frac{4 k+5}{1}=\frac{-3 h+5 k-14}{-28}=\lambda(\text { say }) \\ \Rightarrow \quad h & =\frac{9 \lambda+3}{4}, k=\frac{\lambda-5}{4},-3 h+5 k-14 \\ & =-28 \lambda \end{aligned} $

Now,

$ \begin{aligned} & -3\left(\frac{9 \lambda+3}{4}\right)+5\left(\frac{\lambda-5}{4}\right)-14=-28 \lambda \\ & \Rightarrow \quad-9-27 \lambda+5 \lambda-25-56=-112 \lambda \\ & \Rightarrow-22 \lambda-90=-112 \lambda \Rightarrow 90 \lambda=90 \\ & \Rightarrow \quad \lambda=1 \end{aligned} $

Thus, $h=\frac{3+9.1}{4}=\frac{12}{4}=3$

$ \begin{aligned} k & =\frac{1-5}{4}=\frac{-4}{4}=-1 \\ \therefore(h, k) & =(3,-1) \end{aligned} $