Circle

$\begin{gathered} x^2+y^2+3 x+5 y+4=0 \\ x^2+y^2+5 x+3 y+4=0 g P-2 x+2 y=0 \\ \Rightarrow \quad x=y \Rightarrow 2 x^2+8 x+4=0 \\ \Rightarrow \quad x^2+4 x+2=0 \Rightarrow(x+2)^2=2\end{gathered}$

$\begin{aligned} & \Rightarrow \quad x+2= \pm \sqrt{2} \Rightarrow x=-2 \pm \sqrt{2} \\ & \text { Length }=\sqrt{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2}=\sqrt{2}\left(x_1-x_2\right) \\ & =\sqrt{2}(-2+\sqrt{2})-(-2-\sqrt{2})=4\end{aligned}$

$\begin{aligned} & \left(x^2+y^2-8 x-6 y+21\right)+\lambda\left(x^2+y^2-2 x-15\right)=0 \\ & \Rightarrow \quad(1+4-8-12+21)+\lambda(1+4-2-15)=0 \\ & \Rightarrow \quad 6+(-12 \lambda)=0 \Rightarrow \lambda=1 / 2 \\ & \Rightarrow \quad 3\left(x^2+y^2\right)-18 x-12 y+27=0\end{aligned}$

$\angle A P B=90 \Upsilon$

$\Rightarrow$ Locus of $P$ is a circle with $A$ and $B$ as extremities of diameter.

$\therefore$ Locus of $P$ is given by

$\begin{array}{r} (x+2)(x-3)+(y-1)(y-0)=0 \\ x^2-x-6+y^2-y=0 \\ \text{or}\quad x^2+y^2-x-y-6=0 \end{array}$

Let $S=x^2+y^2-4=0$, then $T=x x_1+y y_1-4$

Equation of tangent from $(4,0)$ is $x_1=4, y_1=0$ and $S_1=x_1^2+y_1^2-4$

$S S_1=T^2$

$\begin{array}{rlrl}\Rightarrow & \left(x^2+y^2-4\right)(16-4) =(4 x-4)^2 \\ \Rightarrow 3\left(x^2+y^2-4\right) & =4(x-1)^2 \\ \Rightarrow & 3 x^2+3 y^2-12 & =4\left(x^2-2 x+1\right) \\ \Rightarrow & 3 y^2 =(x-4)^2 \\ \Rightarrow & y = \pm \frac{1}{\sqrt{3}}(x-4)\end{array}$

Let $P(-9,-1)$

On the circle

$\begin{aligned} & \quad x^2+y^2+4 x+8 y-38=0 \\ & \because 2 g=4 \Rightarrow g=2 \\ & 2 f=8 \Rightarrow f=4 \\ & \text { Centre }=(-g,-f) \end{aligned}$

Whose centre is $C(-2,-4)$.

Let $P B$ be a diameter. Then, $C$ in mid-point of $P B$ and Let $B(x, y)$.

Then, $(-2,-4)=\left(\frac{-9+x}{2}, \frac{-1+y}{2}\right)$

$\begin{aligned} & \Rightarrow \quad \frac{-9+x}{2}=-2 \text { and } \frac{-1+y}{2}=-4 \\ & \Rightarrow \quad x=5 \text { and } y=-7 \\ & \therefore \quad B(5,-7) \end{aligned}$

Tangent at $(5,-7)$

$\begin{aligned} 5 x-7 y+2(x+5)+4(y-7)-38 & =0 \\ 5 x-7 y+2 x+10+4 y-28-38 & =0 \\ \Rightarrow \quad7 x-3 y & =56 \end{aligned}$

Given, radius = 5

and points on X-axis at 4 units distance from origin be ($-$4, 0) and (4, 0).

Let equation of circle be

$x^2+y^2+2gx+2fy+c=0$ ..... (i)

Equation (i) passes through ($-$4, 0) and (4, 0)

$\therefore \quad 16-8g+c=0$

and $\quad 16+8g+c=0$

On solving, we have

$c=-16,g=0$

$\therefore$ radius = 5

$\Rightarrow \sqrt{g^2+f^2-c}=5$

or $f^2+16=25$

or $f=\pm3$

$\therefore$ Required equation of circle

$x^2+y^2\pm6y-16=0$

Equation of normal passing through

(4, 3) and (2, 1)

$y-3=\frac{1-3}{2-4}(x-4)$

$y-3=x-4$

$x-y=1$ ..... (i)

Equation of diameter

$2x-y=2$ ...... (ii)

Solving Eqs. (i) and (ii), we have coordinate of centre of circle

$\therefore \quad x=1, y=0$

Centre of Circle $(1,0)$

Radius of circle $=$ distance between $(1,0)$ and $(2,1)$

$\begin{aligned} & =\sqrt{(2-1)^2+(1-0)^2}=\sqrt{2} \\ \text { Radius } & =\sqrt{2} \end{aligned}$

$\therefore$ Equation of circle

$\begin{array}{rlrl} & (x-1)^2+(y-0)^2 =(\sqrt{2})^2 \\ \Rightarrow & x^2+y^2-2 x-1 =0 \end{array}$

Let $P(x, y)$ be any point on the circle, therefore it will satisfy the circle

$\left(x_1-3\right)^2+\left(y_1+2\right)^2=5 r^2$ ..... (i)

The length of tangent drawn from point $P\left(x_1, y_1\right)$ to the circle $(x-3)^2+(y+2)^2=r^2$ is

$\begin{aligned} \sqrt{\left(x_1-3\right)^2+\left(y_1+2\right)^2-r^2} & =\sqrt{5 r^2-r^2} \text { [from Eq. (i)] } \\ & =\sqrt{4 r^2}=2 r \end{aligned}$

According to the question,

$\begin{aligned} 16= & 2 r \\ \Rightarrow \quad r & =8 \\ \therefore\text { Area between two circles } & =5 \pi r^2-\pi r^2 \\ & =4 \pi r^2=4 \pi \cdot 8^2 \\ & =256 \pi \text { squnits } \end{aligned}$

Let equation of circle be

${x^2} + {y^2} + 2gx + 2fy + c = 0$ ..... (i)

This circle is orthogonal to

${x^2} + {y^2} - 2x + 3y - 7 = 0$ .... (ii)

${x^2} + {y^2} + 5x - 5y + 9 = 0$ .... (iii)

${x^2} + {y^2} + 7x - 9y + 29 = 0$ ... (iv)

By condition of orthogonality, we have

$ - 2g + 3f = c - 7$ .... (vi)

$5g - 5f = c + 9$ .... (vii)

$7g - 9f = c + 29$ .... (viii)

from Eq. (vii) $c = 5g - 5f - 9$

From Eqs. (vi) and (vii), we get

$ - 7g + 8f = - 16$

and $g - 2f = 10$ {on subtracting Eq. (vii) from Eq. (viii)}

On solving, we have $g = - 8,f = - 9,c = - 4$

Required equation of circle,

$\therefore$ ${x^2} + {y^2} - 16x - 18y - 4 = 0$

Given circle $x^2+y^2=50$ ..... (i)

and a line $x+7=0$ ..... (ii)

Point of intersection of Eqs. (i) and (ii) is $x=-7, y= \pm 1$

$\therefore(-7,1)$ and $(-7,-1)$ are points of contact of Eqs. (i) and (ii),

for circle $x^2+y^2=r^2$

Equation of tangent at $\left(x_1, y_1\right)$ is

$x x_1+y y_1=r^2$

$\therefore$ Equation of tangent of Eq. (i) at $(-7,1)$ and $(-7,-1)$ are

$\text { } \quad \begin{array}{ll} -7 x+y=50 \text { or } 7 x-y+50=0 \\ \text { and } -7 x-y=50 \text { or } 7 x+y+50=0 \end{array}$

(c) Let $\left(x_1, y_1\right)$ be any point on the circle

$x^2+y^2=r_1^2$

Then, $x_1^2+y_1^2=r_1^2$ ..... (i)

Equation of chord of contact of tangents from $\left(x_1, y_1\right)$ to the circle $x^2+y^2=r_2^2$ is

$x x_1+y y_1=r_2^2$ .... (ii)

$\because$ Eq. (ii) touches the circle

$x^2+y^2=r_3^2$

The length of perpendicular from centre $(0,0)$ on Eq. (ii) is radius $=r_3$

$\begin{aligned} & \Rightarrow \quad \frac{r_2^2}{\sqrt{x_1^2+y_1^2}}=r_3 \Rightarrow \frac{r_2^2}{\sqrt{r_1^2}}=r_3 \\ & \Rightarrow \quad r_2^2=r_1 r_3 \\ & \Rightarrow \quad r_1, r_2, r_3 \text { are in GP. } \end{aligned}$

Let $(h, k)$ be centre and $r$ be the radius

Then, $h=3$

Equation of circle

$(x-3)^2+(y-k)^2=r^2$

It passes through $(3,0)$ and $(1,-2)$

$\begin{aligned} \Rightarrow k^2 & =r^2 \\ \text { and } 4+k^2+4 k+4 & =r^2 \end{aligned}$

$\begin{array}{rlrl} \Rightarrow & & 8+4 k & =0 \\ \Rightarrow & k & =-2 \end{array}$

$\therefore$ Required equation of circle

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

Let $L_1: y=m x$ (equation of line passes through origin)

Intercept made by $L_1$ and $L_2$ are equal $\Rightarrow L_1$ and $L_2$ are at the same distance from centre of circle $x^2+y^2-x+3 y=0$

Now, centre $\left(\frac{1}{2}, \frac{-3}{2}\right)$

$[\because \text { centre }=(-g,-f)]$

Distance of $L_1$ from $\left(\frac{1}{2},-\frac{3}{2}\right)=\frac{\left|m \cdot \frac{1}{2}+\frac{3}{2}\right|}{\sqrt{m^2+1}}$

Distance of $L_2$ from $\left(\frac{1}{2},-\frac{3}{2}\right)=\frac{\left|\frac{1}{2}-\frac{3}{2}-1\right|}{\sqrt{1+1}}$

$\because L_2: x+y-1=0$

$[\because y=m x \Rightarrow m x-y=0]$

$\Rightarrow \quad \frac{\left|\frac{m+3}{2}\right|}{\sqrt{m^2+1}}=\frac{2}{\sqrt{2}}$

Squaring both sides, we get

$\frac{(m+3)^2}{4\left(m^2+1\right)}=\frac{4}{2} $

$\begin{aligned} & \Rightarrow \quad m^2+6 m+9=8\left(m^2+1\right) \\ & \Rightarrow \quad 7 m^2-6 m-1=0 \\ & \Rightarrow \quad m=\frac{6 \pm \sqrt{36+28}}{14}=\frac{6 \pm 8}{14} \\ & =1, \frac{-1}{7} \\ & \therefore \quad L_1 \text { is } y=x \\ & \text { and } \\ & y=\frac{-1}{7} x \\ & \text { or } \\ & x-y=0 \\ & \text { or } \\ & x+7 y=0 \\ & \end{aligned}$

Given, equation of circle

$x^2+y^2+4 x+16 y-30=0$ ..... (i)

Comparing with $x^2+y^2+2 g_1 x+2 f_1 y+c_1=0$

We have, $g_1=2, f_1=8, c_1=-30$

Let equation of circle whose centre at $(1,2)$ is

$x^2+y^2-2 x-4 y+c=0$ ...... (ii)

Here, $g_2=-1, f_2=-2, c_2=c$

Circles Eqs. (i) and (ii) are orthogonal, then

$\begin{aligned} \Rightarrow & & 2\left(g_1 g_2+f_1 f_2\right) & =c_1+c_2 \\ \text { or } & & 2(-2-16) & =-30+c \\ & & c & =-6 \end{aligned}$

From Eqs. (ii), we get

$\begin{gathered} x^2+y^2-2 x-4 y-6=0 \\ \text { Radius }=\sqrt{(-1)^2+(-2)^2-(-6)}=\sqrt{11} \end{gathered}$

$\begin{array}{r} \text { Let } S_1: x^2+y^2-8 x+40=0 \\ S_2: x^2+y^2-5 x+16=0 \\ S_3: x^2+y^2-8 x+16 y+160=0 \end{array}$

The point which has same power is obtained by solving

$S_2-S_1=0$

and $S_3-S_1=0$

$\Rightarrow \quad 3 x-24=0$

and $\quad 16 y+120=0$

$\Rightarrow \quad x=8 \text { and } y=\frac{-15}{2} \Rightarrow 2 a t=2$

$\therefore$ Required point is $\left(8, \frac{-15}{2}\right)$.

$ \begin{aligned} & \text { Circle } z \bar{z}+a z+\bar{a} \bar{z}-4=0 \\ & {\left[\because z \bar{z}=|z|^2\right]} \\ & \Rightarrow|z|^2+a z+(\overline{a z})-4=0 \\ & {\left[\because \bar{z}_1 \bar{z}_2=\overline{z_1 z_2}\right]} \\ & \Rightarrow|z|^2+2 \cdot \operatorname{Re}(a z)-4=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & {[z+\bar{z}=2 \operatorname{Re}(z)]} \end{aligned} $

$ \begin{aligned} & \text { Now, given } z=x+i y \Rightarrow|z|^2=x^2+y \\ & \text { and } \quad a=1+i \text {, then } a z=(1+i)(x+i y \\ & \therefore \quad a z=(x-y)+i(x+y) \\ & \Rightarrow \operatorname{Re}(a z)=x-y \\ & \therefore \text { From Eq. (i), put } \operatorname{Re}(a z)=x-y \\ & \text { and } \,\,\,\,\,\,\,\quad|z|^2=x^2+y^2 \\ & \Rightarrow \quad x^2+y^2+2(x-y)-4=0 \\ & \quad x^2+y^2+2 x-2 y-4=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

is a circle and given line

$ \begin{array}{ll} & L: z+\bar{z}-i(z-\bar{z})+2=0 \\ \Rightarrow & L: 2 \operatorname{Re}(z)-i 2 \operatorname{Im}(z)+2=0 \\ \Rightarrow & L: 2 x-i(2 y i)+2=0 \\ \Rightarrow & L: x+y+1=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{array} $

Now, equation of circle passing through point of intersection of circle $S$ and line $L$ is

$ \begin{align*} S+\lambda L=0\left(x^2\right. & \left.+y^2+2 x-2 y-4\right) \\ & +\lambda(x+y+1)=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv) \end{align*} $

Pass through $(0,0)$

∴ Put in Eq. (iv), we get

$ -4+\lambda=0 \Rightarrow \lambda=4 $

Now, required circle is

$ \begin{array}{cc} \Rightarrow & x^2+y^2+2 x-2 y-4 \\ & +4(x+y+1)=0 \\ \Rightarrow & x^2+y^2+6 x+2 y=0 \end{array} $

Let point $P$ is $(h, k)$. Given $Q(0,2)$ and $R(-1,0)$ then given $P Q=\frac{1}{\sqrt{2}} P R$

Squaring on both sides,

$ \begin{gathered} 2(P Q)^2=(P R)^2 \\ 2\left(h^2+(k-2)^2\right)=(h+1)^2+k^2 \\ 2 h^2+2 k^2-8 k+8=h^2+1+2 h+k^2 \\ \therefore \quad h^2+k^2-2 h-8 k+7=0 \end{gathered} $

∴ Locus of point $P$

$ \begin{equation*} x^2+y^2-2 x-8 y+7=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{equation*} $

∴ Eq. (i) is circle.

∴ Centre $(1,4)$ and radius $=\sqrt{1+16-7} =\sqrt{10}$

Hence, locus at point $P$ is circle with centre $(1,4)$ and radius $\sqrt{10}$

Equation of circle $x^2+y^2-a^2+\lambda (x \cos \alpha+y \sin \alpha-p)=0$ is the smallest circle, then centre $\left(\frac{-\lambda \cos \alpha}{2}, \frac{-\lambda \sin \alpha}{2}\right)$ lies on the line

$ \begin{equation*} x \cos \alpha+y \sin \alpha=p \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{equation*} $

∴ Put centre in the line (i), then we getting

$ \begin{aligned} & \frac{-\lambda \cos ^2 \alpha}{2}-\frac{\lambda \sin ^2 \alpha}{2}=p \\ \Rightarrow \quad & \frac{-\lambda}{2}=p \\ \Rightarrow \quad & \lambda=-2 p \end{aligned} $

Given, from point $P(1,1)$, tangent $P A$ and $P B$ is drawn

$ \begin{aligned} & \therefore \quad P A=\sqrt{S_1} \\ & \Rightarrow \quad P A=\sqrt{1+1+g+g-2}=\sqrt{2 g} \\ & \text { Radius of circle }=\sqrt{\frac{g^2}{4}+\frac{g^2}{4}+2} \\ & \Rightarrow \quad A C=\frac{1}{\sqrt{2}} \sqrt{g^2+4} \end{aligned} $

$ \begin{aligned} & \therefore \text { Area of } \triangle P A C =\frac{1}{2} P A \times A C \\ & =\frac{1}{2} \times \sqrt{2 g} \times \frac{1}{\sqrt{2}} \sqrt{g^2+4} \\ & \therefore \text { Area of } P A C B =2 \times \text { area of } \triangle P A C \\ & =\frac{1}{\sqrt{2}} \sqrt{2 g} \sqrt{g^2+4} \\ & =\sqrt{g^3+4 g} \end{aligned} $

Given circle,

$ S_1: x^2+y^2-8 x-6 y+21=0 $

and $S_2: x^2+y^2-2 y-15=0$

Circle $S_1$ : Centre $C_1(4,3)$, radius

$ r_1=\sqrt{16+9-21}=2 $

$S_2$ : Centre $C_2(0,1)$, radius

$ r_2=\sqrt{1+15}=4 $

Now $C_1 C_2=\sqrt{4^2+2^2}=\sqrt{20}$

$ \therefore C_1 C_2

⇒ circle $C_1$ and $C_2$ put with other.

Let point of intersection of tangent on the circle $S_1$ and $S_2$ is $P$.

Now, point $P$ divide line joining $C_2 C_1$ in the ratio of their radii externally

$ \therefore \quad \frac{C_2 P}{C_1 P}=\frac{r_2}{r_1} \Rightarrow \frac{C_2 P}{C_1 P}=\frac{4}{2}=\frac{2}{1} $

∴ By using external division formula

$ \begin{aligned} & =\left(\frac{m_1 n_2-m_2 n_1}{m_1-m_2}, \frac{m_1 y_2-m_2 y_1}{m_1-m_2}\right) \\ & =\left(\frac{2 \times 4-1 \times 0}{2-1}, \frac{2 \times 3-1 \times 1}{2-1}\right) \\ & =(8,5) \end{aligned} $

As radical axis of the two circles bisects the all common tangents of the circles.

Now, the mid-point of $A(1,1)$ and $B(0,1)$ is $M\left(\frac{1}{2}, 1\right)$ and mid-point of $C\left(\frac{3}{5}, \frac{4}{5}\right)$ and $D\left(-\frac{1}{5}, \frac{7}{5}\right)$ is $N\left(\frac{1}{5}, \frac{11}{5}\right)$

∴ Equation of required radical axis $M N$ is

$ \begin{aligned} y-1 & =\frac{\frac{11}{10}-1}{\frac{1}{5}-\frac{1}{2}}\left(x-\frac{1}{2}\right) \\ \Rightarrow \quad 3(y-1) & =-x+\frac{1}{2} \\ \Rightarrow \quad 2 x+6 y & =7 \end{aligned} $

Let the equation of the required circle be

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

and centre is $(-g,-f)$

Given circles,

$ C_1: x^2+y^2-2 x-4 y-4=0 $

Here, $g_1=-1, f_1=-2, c_1=-4$

$ C_2: x^2+y^2-10 x+12 y+52=0 $

Here, $g_2=-5, f_2=6, c_2=52$

Condition of two circles cuts orthogonally

$ 2 g_1 g_2+2 f_1 f_2=c_1+c_2 $

For $C_1: 2(g)(-1)+2(f)(-2)=C-4$

or $2 g+4 f=-c+4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

For $C_2: 2(g)(-5)+2(f)(6)=c+52$

$ \begin{equation*} 10 g-12 f=-c-52\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{equation*} $

Subtracting on Eqs. (i) and (ii), we get

$ -8 g+16 f=56 $

or     $ g-2 f=-7 $

or     $ \begin{equation*} g=2 f-7\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii) \end{equation*} $

Multiplying Eq. (i) by 5 and then subtracting

Eq. (ii), we get

or $ 8 f=-c+18 \text { or } c=-8 f+18 $

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

$ \begin{aligned} & =\sqrt{(2 f-7)^2+f^2(-8 f+18)} \\ & =\sqrt{5 f^2-20 f+31} \\ & =\sqrt{5\left(f^2-4 f\right)+31}=\sqrt{5(f-2)^2+11} \end{aligned} $

For minimum, value $f-2=0$

$ \Rightarrow \quad f=2 $

Putting the value of ' $f$ ' in Eq. (iii), we get

$ g=2 \times 2-7 \Rightarrow g=-3 $

Therefore, coordinate of centre is $(+3,-2)$.

A circle is passing through $(3,4),(3$, $2)$ and ( 1,4 ).

Clearly, $\triangle A B C$ is a right angle triangle - centre of circle is mid-point of $A C$ i.e., $O(2,3)$

$ \begin{aligned} r & =A O=\sqrt{(3-2)^2+(2-3)^2} \\ & =\sqrt{1+1}=\sqrt{2} \end{aligned} $

Equation of circle in parameter form is $x=2+\sqrt{2} \cos \theta \Rightarrow y=3+\sqrt{2} \sin \theta$

$ \quad \begin{aligned} \therefore\,\,\,\,\,\, a & =2, b=3, r=\sqrt{2} \\ b^a \cdot r^a & =(3)^2(\sqrt{2})^2=9 \times 2=18 \end{aligned} $

Given equation of circle,

$ \begin{gathered} x^2+y^2-4 x-6 y+9=0 \\ R=P C=\sqrt{4+9-9}=2 \\ x^2+y^2-4 x-6 y+12=0 \\ r=Q C=\sqrt{4+9-12}=1 \end{gathered} $

In $\triangle P Q C$,

$ \cos \theta=\frac{Q C}{P C}=\frac{1}{2} \Rightarrow \theta=\frac{\pi}{3} $

∴ Area of $\triangle Q R C$,

$ \begin{aligned} & =\frac{1}{2}(C Q)^2 \times \sin 2 \theta=\frac{1}{2} \times 1 \times \sin \frac{2 \pi}{3} \\ & =\frac{1}{2} \times 1 \times \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{4} \end{aligned} $

Equation of tangent drawn from origin, ' $O$ ' of the circle is

$ \begin{aligned} & \quad x^2+y^2+2 g x+2 f y+g^2=0 \\ & \text { } \\ &or \,\,\,\,\,\,\,\,\,\,\,\,\, \quad T^2=S S_1 \\ & \therefore\left(0+0+2 g\left(\frac{x+0}{2}\right)+2 f \frac{(y+0)}{2}+g^2\right)^2 \\ & \quad=\left(x^2+y^2+2 g x+2 f y+g^2\right)\left(g^2\right) \end{aligned} $

$ \begin{aligned} & \Rightarrow\left(g x+f y+g^2\right)^2 \\ & \quad=g^2\left(x^2+y^2\right)+2 g^3 x+2 f g^2 y+g^4 \\ & \begin{aligned} \Rightarrow & g^2 x^2+f^2 y^2+g^4+2 g^3 x +2 g^2 f y+2 g f x y \\ & =g^2 x^2+g^2 y^2+2 g^3 x+2 f g^2 y+g^4 \\ \Rightarrow & f^2 y^2+2 g^2 f y+2 g f x y=g^2 y^2+2 f g^2 y \\ \Rightarrow & y^2\left(g^2-f^2\right)-2 g f y x=0 \\ \Rightarrow & y\left[\left(g^2-f^2\right) y-2 g f x\right]=0 \\ \Rightarrow & y=0 \\ & \left(g^2-f^2\right) y-2 g f x=0 \end{aligned} \end{aligned} $

$2 x+y=0$ is equation of chord of circle,

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

$2 x+y=0$ is also diameter of the required circle

∵ Centre of circle $\left(\lambda-1, \frac{\lambda-6}{2}\right)$ lie on chord

$ \begin{aligned} \therefore & & 2(\lambda-1)+\frac{\lambda-6}{2} & =0 \\ \Rightarrow & & \lambda & =2 \end{aligned} $

∵ Equation of circle

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

$(-2,1)$ is lie on circle

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

Radical axis of circle

$ x^2+y^2+2 \alpha x+2 \beta y+c=0 $

and $\quad x^2+y^2+\frac{3}{2} x+4 y+c=0$ is

$ \begin{aligned} & \left(2 \alpha-\frac{3}{2}\right) x+(2 \beta-4) y=0 \\ & (4 \alpha-3) x+4(\beta-2) y=0 \end{aligned} $

$(4 \alpha-3) x+(\beta-8) y=0$ touches the circle

$ \therefore \quad 1=\left|\frac{(4 \alpha-3)(-1)+(3-8)(-1)}{\sqrt{(4 \alpha-3)^2+(4 \beta-8)^2}}\right| $

$ \begin{aligned} & \Rightarrow(4 \alpha-3)^2+(4 b-8)^2 \\ & \quad=(4 \alpha+4 \beta-11)^2 \\ & \Rightarrow 16 \alpha^2-24 \alpha+9+16 \beta^2-64 \beta+64 \\ & \quad=16 \alpha^2+16 \beta^2+121+32 \alpha \beta-88 \alpha-88 \beta \end{aligned} $

$ \begin{aligned} \Rightarrow 32 \alpha \beta-88 \alpha+24 \alpha- & 88 \beta+64 \beta \\ & =9+64-121 \\ 32 \alpha \beta-64 \alpha-24 \beta & =-48 \\ \Rightarrow 4 \alpha \beta-8 \alpha-3 \beta & =-6 \\ \Rightarrow 4 \alpha \beta-8 \alpha-3 \beta+10 & =-6+10 \\ & =4 \end{aligned} $

Given circle

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

Centre $(2,1)$

Equation of diameter of circle passes through origin.

$ x-2 y=0 $

Equation of circle through the end point of diameter of circle and line $x-2 y=0$ is

$ x^2+y^2-4 x-2 y-4+\lambda(x-2 y)=0 $

Since, this is passes through $(1,2)$

$ \quad \begin{aligned} \because \quad 1+4-4-4-4+\lambda(-3) & =0 \\ \lambda & =\frac{-7}{3} \end{aligned} $

Required equation of circle is

$ \begin{array}{r} x^2+y^2-4 x-2 y-4-\frac{7}{3}(x-2 y)=0 \\ 3\left(x^2+y^2\right)-19 x+8 y-12=0 \\ 3 x^2+3 y^2-19 x+8 y-12=0 \end{array} $

We have, ( $2, \alpha$ ) is mid-point of chord of circle

$ x^2+y^2-4 x+8 y+6=0 $

$\because(2, \alpha)$ inside the circle

$ \begin{aligned} & \therefore \quad 4+\alpha^2-8+8 \alpha+6 \leq 0 \\ & \alpha^2+8 \alpha+2 \leq 0 \\ & \alpha=\frac{-8 \pm \sqrt{64-8}}{2} \\ & \alpha=\frac{-8 \pm 2 \sqrt{14}}{2}=-4 \pm \sqrt{14} \\ & \alpha \in(-4-\sqrt{14},-4+\sqrt{14}) \end{aligned} $

$C_1$ and $C_2$ are external and internal similitude of the circle

$ x^2+y^2-2 x+4 y+1=0 $

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

Centre of circle

$ \begin{aligned} x^2+y^2-2 x+4 y+1 & =0 \text { is } \\ (1,-2), r_1 & =2 \end{aligned} $

Centre of circle

$ \begin{array}{r} x^2+y^2+4 x-6 y+4=0 \\ (-2,3), r_2=1 \end{array} $

Coordinate of $C_1=\frac{-5}{3}$

$ \left(\frac{2(-2)-1}{2-1}, \frac{6+2}{2-1}\right) \equiv(-5,8) $

Coordinate of

$ \begin{aligned} C_2= & \left(\frac{2(-2)+1}{2+1}, \frac{6-2}{2+1}\right) \equiv\left(-1, \frac{4}{3}\right) \\ & C_1 C_2=\sqrt{(-5+1)^2+\left(8-\frac{4}{3}\right)^2} \\ & C_1 C_2=\sqrt{16+\frac{400}{9}} \\ \Rightarrow & C_1 C_2=\sqrt{\frac{544}{9}}=\frac{4 \sqrt{34}}{3} \end{aligned} $

$ \begin{aligned} &\text { Here, } C_1 C_2=2 r\\ &\begin{aligned} \therefore \quad \frac{4}{3} \sqrt{34} & =2 r \\ \frac{9}{2} r & =3 \sqrt{34} \end{aligned} \end{aligned} $

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

orthogonally the circle

$ \begin{aligned} & x^2+y^2-4 x-6 y+11=0 \text { and } \\ & \quad x^2+y^2-10 x-4 y+21=0 \end{aligned} $

$ \begin{array}{lll} \therefore & & -4 g-6 f=c+11\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \\ \text { and } & & -10 g-4 f=c+21 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \text { } \end{array} $

Centre of circle $x^2+y^2+2 g x+2 f y+c=0$ is lie in bisector of angle between the positive coordinate axis

$ \therefore \quad g=f \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$

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

$ \begin{array}{r} -4 f-6 f=c+11 \\ -10 f=c+11 ...(iv)\end{array} $

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

$ -14 f=c+21 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(v)$

From Eqs. (iv) and (v), we get

$ \begin{array}{rlrl} & g=f=\frac{-5}{2}, c=14 \\ \therefore & 2 g+2 f+c-5-5+14= & 4 \end{array} $

Given $S_1=x^2+y^2=16$

Centre $(0,0) r=4$

Clearly, the diameter of $S_1$ will be common chord. Let $P Q$ be common chord and centre of $S_2$ be $(h, k)$

We have $A P=5, P B=4$,

$ \therefore \quad A B=3 $

Where $B=(0,0)$

Slope of $P Q=3 / 4$

∴ Slope of $A B=\frac{-4}{3}$

By parameter equation

$ \begin{aligned} \frac{h-0}{\frac{-3}{5}} & =\frac{k-0}{\frac{4}{5}}= \pm 3 \\ x=\mp \frac{9}{5}, y & =\frac{ \pm 12}{5} \end{aligned} $

∴ Centre $\left(\frac{-9}{5}, \frac{12}{5}\right)$ and $\left(\frac{9}{5}, \frac{-12}{5}\right)$

Equation of circle passing through the points $(1,1),(2,-1)$ and $(3,2)$ is

$ \begin{aligned} & \left|\begin{array}{cccc} x^2+y^2 & x & y & 1 \\ 1+1 & 1 & 1 & 1 \\ 4+1 & 2 & -1 & 1 \\ 9+4 & 3 & 2 & 1 \end{array}\right|=0 \\ & \left|\begin{array}{cccc} x^2+y^2 & x & y & 1 \\ 2 & 1 & 1 & 1 \\ 5 & 2 & -1 & 1 \\ 13 & 3 & 2 & 1 \end{array}\right|=0 \end{aligned} $

$ \begin{aligned} & \left|\begin{array}{cccc} x^2+y^2-2 & x-1 & y-1 & 0 \\ -3 & -1 & 2 & 0 \\ -8 & -1 & -3 & 0 \\ 13 & 3 & 2 & 1 \end{array}\right|=0 \\ & \left|\begin{array}{ccc} x^2+y^2-2 & x-1 & y-1 \\ -3 & -1 & 2 \\ -8 & -1 & -3 \end{array}\right|=0 \\ & \left(x^2+y^2-2\right)(3+2)-(x-1)(9+16)+(y-1)(3-8)=0 \\ & \quad \begin{array}{l} 5\left(x^2+y^2-2\right)-25(x-1)-5(y-1)=0 \\ x^2+y^2-2-5(x-1)-(y-1)=0 \\ x^2+y^2-2-5 x+5-y+1=0 \\ x^2+y^2-5 x-y+4=0 \end{array} \end{aligned} $

Now, by putting all the points, we see $\left(\frac{5}{2}+\sqrt{\frac{5}{2}}, \frac{1}{2}\right),\left(\frac{5+\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right)$ passes through above circle.

Let $P$ be $\left(x_1, y_1\right)$

Now, equation of circle of radius $r$ which touches the coordinate axes and lies in the first quadrant is

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

Now, polar of $P$ with respect to the circle $x^2+y^2-2 x r-2 y r-r^2=0$ is

$ \begin{aligned} & x x_1+y y_1-\left(x+x_1\right) r-\left(y+y_1\right) r-r^2=0 \\ & \Rightarrow \quad\left(x_1-r\right) x+\left(y_1-r\right) y=\left(x_1+y_1-r\right) r \end{aligned} $

But it is given that polar of $P$ with respect to above circle is $x+2 y=4 r$

$ \begin{aligned} & \therefore \quad \frac{x_1-r}{1}=\frac{y_1-r}{2}=\frac{x_1+y_1-r}{4} \\ & \Rightarrow \quad x_1-r=\frac{x_1+y_1-r}{4} \\ & \text { and } x_1-r=\frac{y_1-r}{2} \\ & \Rightarrow \quad 4 x_1-4 r=x_1+y_1-r \text { and } \\ & 2 x_1-2 r=y_1-r \\ & \Rightarrow \quad 3 x_1-3 r=y_1 \text { and } 2 x_1-r=y_1 \\ & \therefore \quad 3 x_1-3 r=2 x_1-r \\ & \Rightarrow \quad x_1=2 r \text { and } y_1=2 x_1-r \\ & \quad 4 r-r=3 r \\ & \quad \quad P \text { is }(2 r, 3 r) \end{aligned} $

The centres of the given circles are $C_1(1,3+\sqrt{7})$ and $C_2(4,3)$ and corresponding radii are

$ \begin{aligned} r_1 & =\sqrt{1^2+(3+\sqrt{7})^2-(8+6 \sqrt{7})}=3 \\ \text { and } r_2 & =\sqrt{4^2+3^2-k^2}=\sqrt{25-k^2} \end{aligned} $

Now, $C_1 C_2=\sqrt{(4-1)^2+(3-3-\sqrt{7})^2}=4$

Clearly, $C_1 C_2

$ \begin{aligned} & \Rightarrow \quad 4<3+\sqrt{25-k^2} \Rightarrow 1<\sqrt{25-k^2} \\ & \Rightarrow \quad 1<25-k^2 \Rightarrow k^2<24 \end{aligned} $

$ \therefore k=0, \pm 1, \pm 2, \pm 3, \pm 4 \quad[\because k \in \mathbf{Z}] $

Let the equation of circle $S$ is

$ \begin{aligned} & S=x^2+y^2+2 g x+2 f y+c=0 \\ & C_1=x^2+y^2-8 x-2 y+16=0 \\ & C_2=x^2+y^2-4 x-4 y-1=0 \end{aligned} $

$S$ is orthognoally to $C_1$ and $C_2$

$ \begin{array}{lrl} \therefore & -8 g-2 f & =C+16 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & -4 g-4 f & =C-1 \quad \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \end{array} $

From Eqs. (i) and (ii),

$ -4 g+2 f=17 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$

Common chord of $S$ and $C_1=$

$ (2 g+8) x+(2 f+2) y+c-16=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)$

Given common chord is

$ 2 x+13 y-15=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(v)$

Eqs. (iv) and (v) are coinside

$ \begin{array}{ll} \therefore & \frac{2 g+8}{2}=\frac{2 f+2}{13}=\frac{c-16}{-15} \\ \Rightarrow & 13 g-2 f=-50 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(vi)\end{array} $

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

$ g=\frac{11}{3} f=\frac{-7}{6} $

∴ Centre of circle, $S=\left(\frac{-11}{3}, \frac{7}{6}\right)$

We have,

$ \begin{aligned} & S_1: x^2+y^2+k x-4 y-1=0 \\ & S_2: 3 x^2+3 y^2-14 x+23 y-15=0 \\ & \quad=x^2+y^2-\frac{14}{3} x+\frac{23}{3} y-5=0 \end{aligned} $

Since, $S_1$ and $S_2$ are orthogonal

$ \begin{array}{ll} \therefore & 2\left[\left(\frac{-k}{2}\right)\left(\frac{7}{3}\right)+(2)\left(\frac{-23}{6}\right)\right]=-1-5 \\ \Rightarrow & 2\left[\frac{-7 k}{6}-\frac{46}{6}\right]=-6 \\ \Rightarrow & \frac{-7 k}{6}-\frac{46}{6}=-3 \\ \Rightarrow & \frac{7 k}{6}=3-\frac{46}{6} \\ \Rightarrow & \frac{7 k}{6}=\frac{18-46}{6} \\ \Rightarrow & 7 k=-28 \\ \Rightarrow & k=-4 \\ \therefore & S_1: x^2+y^2-4 x-4 y-1=0 \\ & S_2: x^2+y^2-\frac{14}{3} x+\frac{23}{3} y-5=0 \end{array} $

Equation of circle passing through intersection of $S_1$ and $S_2$ is

$ \begin{aligned} & x^2+y^2-4 x-4 y-1 \\ & +\lambda\left(x^2+y^2-\frac{14 x}{3}+\frac{23}{3} y-5\right)=0 \end{aligned} $

Since above circle passes through $(-1,-1)$, so

$ 1+1+4+4-1+\lambda\left(1+1+\frac{14}{3}-\frac{23}{3}-5\right)=0 $

$ \begin{aligned} &9-6 \lambda=0 \Rightarrow \lambda=\frac{3}{2}\\ &\text { ∴ Equation of required circle is }\\ &\begin{aligned} & x^2+y^2-4 x-4 y-1+\frac{3}{2} \\ & \left(x^2+y^2-\frac{14 x}{3}+\frac{23 y}{3}-5\right)=0 \\ & \Rightarrow \quad 2 x^2+2 y^2-8 x-8 y-2+3 x^2+3 y^2 -14 x+23 y-15=0 \\ & \Rightarrow \quad 5 x^2+5 y^2-22 x+15 y-17=0 \end{aligned} \end{aligned} $