Circle

Equation parallel to $X$-axis is given as

$y=a$

or $y-a=0$ ....... (i)

If Eq. (i) touches the circle

$x^2+y^2-6 x-4 y-12=0$ ..... (ii)

Then, Eq. (i) becomes tangent to circle.

Centre of Eq. (ii) is $(3,2)$

Radius of Eq. (ii) is $\sqrt{9+4+12}=5$

Then, perpendicular distance from centre to line (i) is radius i.e.

Perpendicular distance between $(3,2)$ and $y-a=0$ is equal to 5

$\begin{aligned} & \text { i.e. } \quad\left|\frac{2-a}{l}\right|=5 \\ & \Rightarrow \quad 2-a= \pm 5 \\ & \Rightarrow \quad a=7,-3 \\ \end{aligned}$

$\therefore$ Equation of pair of lines $=(y+3)(y-7)=0$

$\begin{aligned} & \text { i.e. }(y+3)(y-7)=0 \\ & \Rightarrow y^2-4 y-21=0 \end{aligned}$

Given circle,

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

It cut the $X$-axis, when $y=0$

Put $y=0$ in Eq. (i), we obtain

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

Thus, circle (i) touches $X$-axis at $(2,0),(1,0)$.

$\begin{array}{lrl} & x^2+y^2+8 x+10 y-8=0 \\ \Rightarrow & (x+4)^2+(y+5)^2-16-25-8=0 \\ \Rightarrow & (x+4)^2+(y+5)^2=49 \\ \Rightarrow & (x+4)^2+(y+5)^2=(7)^2 \end{array}$

which is standard form of circle.

Radius = 7 and Centre = $(-4,-5)$

We know an important result:

If you draw tangents to the circle $x^2 + y^2 = a^2$, and then find their poles with respect to the circle $(x + a)^2 + y^2 = 2a^2$, all those poles will be on the curve $y^2 + 4a x = 0$.

In our question, $a = 2$. So, we are looking at the circles $x^2 + y^2 = 4$ and $(x + 2)^2 + y^2 = 8$.

Using the result, the poles of the tangents to $x^2 + y^2 = 4$ with respect to $(x+2)^2 + y^2 = 8$ will be on:

$y^2 + 4 \times 2\, x = 0$

which simplifies to:

$y^2 + 8x = 0$

Given equation of circle is

$S: x^2+y^2+4 x-6 y-a=0$ ...... (i)

Power of point w.r.t. Eq. (i) is $-16$ , i.e.

$S(1,6) =-16$

or $(1)^2+(6)^2+4-36-a =-16$

$\Rightarrow 5-a =-16$

$\Rightarrow a =21$

$\begin{aligned} & S_1: x^2+y^2+4 x+6 y+7=0 \\ & S_2: x^2+y^2+2 x+3 y-\frac{9}{4}=0 \end{aligned}$

Equation of radical axis

$\begin{array}{left} S_1-S_2=0 \\ \left(x^2+y^2+4 x+6 y+7\right) -\left(\left(x^2+y^2+2 x+3 y-\frac{9}{4}\right)=0\right. \\ \Rightarrow 2 x+3 y+\frac{37}{4}=0 \\ \text { or } \quad 8 x+12 y+37=0 \end{array}$

${S_1}:{x^2} + {y^2} - 4x + 6y - 10 = 0$ ...... (i)

${S_2}:{x^2} + {y^2} + 2x - 6y + 2 = 0$ ...... (ii)

Equation of radical axis

${S_1} - {S_2} = 0$

$\begin{array}{ccc} \Rightarrow & \left(x^2+y^2-4 x+6 y-10\right) \\ & \quad-\left(x^2+y^2+2 x-6 y+2\right)=0 \\ \Rightarrow & 6 x-12 y+12=0 \\ \Rightarrow & x-2 y+2=0 \quad \text{......(iii)} \end{array}$

Given that Eq. (iii) cut the circle (i),

Using Eq. (iii) in Eq. (i) as, $x=2 y-2$

$\begin{aligned} & (2 y-2)^2+y^2-4(2 y-2)+6 y-10=0 \\ & \Rightarrow 4 y^2+4-8 y+y^2-8 y+8+6 y-10=0 \\ & \Rightarrow \quad 5 y^2-10 y+2=0 \\ & \Rightarrow \quad y=\frac{10 \pm \sqrt{100-40}}{10}=\frac{5 \pm \sqrt{15}}{5} \\ & \Rightarrow \quad x=2\left(\frac{5 \pm \sqrt{15}}{5}\right) \\ \end{aligned}$

Thus, Eq. (iii) cuts $S_1$ at two real and distinct points.

Let the point be $(h, k)$.

So, we have

$\begin{aligned} (h-3)^2+(k+2)^2 & =4^2 \\ \Rightarrow h^2+9-6 h+k^2+4+4 k & =16 \\ \Rightarrow \quad h^2+k^2-6 h+4 k-3 & =0 \end{aligned}$

Replace $h \rightarrow x$ and $k \rightarrow y$

$x^2+y^2-6 x+4 y-3=0$

$\begin{aligned} 2 \sqrt{g^2-c} & =2 ; \\ 2 \sqrt{f^2-c} & =3 \text { and } c=0 \\ g & = \pm 1, f= \pm \frac{3}{2} \\ x^2+y^2 \pm 2 x & \pm 3 y=0 \end{aligned}$

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

So, centre must be in second quadrant or fourth guadrant.

The $x$-coordinate of centre is negative.

$\therefore \quad x^2+y^2+2 x-3 y=0$

${(x + 2)^2} + {(y + 2)^2} = {3^2}$

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

$\sin \theta = {3 \over {3\sqrt 2 }}$

$ \Rightarrow \theta = 45\Upsilon $

$\therefore$ $2\theta = 90\Upsilon $

$x^2+y^2-4 x-8 y-5=0$

$\begin{aligned} & \text { Centre }=(2,4) \\ & \text { Radius }=2 \sqrt{4+16+5}=10 \end{aligned}$

$\text { Line : } 3 x-4 y-m=0$

Distance of $L$ from Centre < Radius

$\begin{aligned} & \frac{|3 \cdot 2-4 \cdot 4-m|}{\sqrt{3^2+4^2}}<10 \\ & \Rightarrow \quad \frac{|6-16-m|}{5} < 10 \\ & \Rightarrow \quad|-10-m| < 50 \\ & \Rightarrow \quad-50 < -10-m < 50 \\ & \Rightarrow \quad-40 < -m < 60 \\ & \Rightarrow \quad-60 < m < 40 \\ \end{aligned}$

Only (d) satisfies the above inequality.

Let $A = (3\cos \theta ,3\sin \theta )$

Then, $B = \left( {3\cos \left( {\theta + {{2\pi } \over 3}} \right),3\sin \left( {\theta + {{2\pi } \over 3}} \right)} \right)$

Let (h, k) be the required point

$2h = 3\cos \theta + 3\cos \left( {\theta + {{2\pi } \over 3}} \right)$

$ = 3\left[ {\cos \theta + \cos \theta \left( { - {1 \over 2}} \right) - \sin \theta \left( {{{\sqrt 3 } \over 2}} \right)} \right]$

$ = {3 \over 2}(\cos \theta - \sqrt 3 \sin \theta )$

$2k = 3\sin \theta + 3\sin \left( {\theta + {{2\pi } \over 3}} \right)$

$ = 3\left[ {\sin \theta + \sin \theta \left( {{{ - 1} \over 3}} \right) + \cos \theta \left( {{{\sqrt 3 } \over 2}} \right)} \right]$

$ = {3 \over 2}(\sin \theta + \sqrt 3 \cos \theta )$

$\cos \theta - \sqrt 3 \sin \theta = {{4h} \over 3} \Rightarrow \sqrt 3 \cos \theta + \sin \theta = {{4k} \over 3}$

$\sqrt 3 \cos \theta - 3\sin \theta = {{4h} \over {\sqrt 3 }}$

$4\sin \theta = {4 \over 3}(k - \sqrt 3 h)$

$\sin \theta = \left( {{{k - \sqrt 3 h} \over 3}} \right)$

$\cos \theta = \left( {{{3k + \sqrt 3 h} \over {3\sqrt 3 }}} \right) = \left( {{{\sqrt 3 k + h} \over 3}} \right)$

$ \Rightarrow {\left( {{{k - \sqrt 3 h} \over 3}} \right)^2} + {\left( {{{\sqrt 3 k + h} \over 3}} \right)^2} = 1$

$ \Rightarrow {k^2} + 3{h^2} - 2\sqrt 3 hk + 3{k^2} + {h^2} + 2\sqrt 3 kh = 9$

$ \Rightarrow 4({k^2} + {h^2}) = 9 \Rightarrow {x^2} + {y^2} = {9 \over 4}$

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

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

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

$\angle A P B=90 \Upsilon$

$\Rightarrow$ 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 $\left(8, \frac{-15}{2}\right)$.

Since, the circle touches X-axis

(x $-$ h)² + (y $-$ k)² = k² ..... (i)

Also, it passes through the points (1, $-$2) and (3, $-$4).

(1 $-$ h)² + ($-$2 $-$ k)² = k² ...... (ii)

and (3 $-$ h)² + ($-$4 $-$ k)² = k² ...... (iii)

Subtracting Eq. (iii) from Eq. (ii), we get

h = k + 5

On solving these equations, we get

By putting the values of (h, k) = ($-$5, $-$10) or (3, $-$2) in Eq. (i), we get

x² + y² + 10x + 20y + 25 = 0 or x² + y² $-$ 6x + 4y + 9 = 0

$ \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 $(+3,-2)$.