Circle

Curve : $x^2+y^2=1$

Let $\left(\frac{1}{a}, \frac{1}{b}\right)=\left(x_1, y_1\right)$ touches the circles, then equation of tangent at $\left(x_1, y_1\right)$ is

$\begin{aligned} x x_1+y y_1 & =1 \\ \Rightarrow \quad x \cdot\left(\frac{1}{a}\right)+y \cdot\left(\frac{1}{b}\right) & =1 \Rightarrow \frac{x}{a}+\frac{y}{b}=1 \end{aligned}$

$\begin{aligned} & 4 x-3 y-10=0 \\ & \Rightarrow \quad x=\frac{3 y+10}{4} \\ & \text { and } x^2+y^2-2 x+4 y-20=0 \\ & \Rightarrow \quad\left(\frac{3 y+10}{4}\right)^2+y^2-2\left(\frac{3 y+10}{4}\right)+4 y-20=0 \\ & \Rightarrow \quad 9 y^2+100+60 y+16 y^2-24 y-80 +64y-320=0\\ & \Rightarrow \quad 25 y^2+100 y-300=0 \Rightarrow y^2+4 y-12=0 \\ & \Rightarrow \quad y^2+6 y-2 y-12=0 \Rightarrow y(y+6)-2(y+6)=0 \\ & \Rightarrow \quad(y+6)(y-2)=0 \\ & y=-6,2 \\ & y=-6 \Rightarrow x=\frac{-18+10}{4}=-2 \Rightarrow(-2,-6) \\ & y=2 \Rightarrow x=\frac{6+10}{4}=4 \Rightarrow(4,2) \end{aligned}$

Required length

$\begin{aligned} & =\sqrt{(4+2)^2+(2+6)^2} \\ & =\sqrt{36+64}=\sqrt{100}=10 \end{aligned}$

Let the pole be $\left(x_1, y_1\right)$, then the equation of the polar with respect to the circle $x^2+y^2=c^2$ is $x x_1+y y_1=c^2$

$\Rightarrow \quad \frac{x x_1}{c^2}+\frac{y y_1}{c^2}=1 \quad \text{... (i)}$

Eq. (i) and $\frac{x}{a}+\frac{y}{b}=1$ represents the same line.

$\begin{aligned} \therefore \quad \frac{x_1}{c^2} & =\frac{1}{a} \Rightarrow x_1=\frac{c^2}{a} \\ \frac{y_1}{c^2} & =\frac{1}{b} \Rightarrow y_1=\frac{c^2}{b} \end{aligned}$

Pole : $\left(x_1, y_1\right)=\left(\frac{c^2}{a}, \frac{c^2}{b}\right)$

Line: $5 x-2 y+6=0$

On $Y$-axis, $x=0 \Rightarrow y=\frac{6}{2}=3$

i.e. $Q \equiv(0,3)$

$\begin{aligned} & \text { Circle : } x^2+y^2+6 x+6 y-2=0 \\ & \text { Centre }(-g,-f) \equiv(-3,-3) \\ & \text { Radius }=\sqrt{g^2+f^2-c} \\ & =\sqrt{9+9+2}=\sqrt{20} \\ & =2 \sqrt{5} \\ & C Q=\sqrt{(0+3)^2+(3+3)^2}=\sqrt{9+36}=\sqrt{45} \\ & \text { In } \triangle P Q C \text {, } \\ & (C Q)^2=(C P)^2+(Q P)^2 \\ & \Rightarrow \quad 45=20+(Q P)^2 \\ & \Rightarrow \quad 25=(Q P)^2 \Rightarrow Q P=5 \\ & \end{aligned}$

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}$