Circle

Let the equation of the desired circle be:

$ S \equiv x^2 + y^2 + 2gx + 2fy + c = 0 $

The circle $ S $ is orthogonal to three given circles. Using the property of orthogonal circles, we find the equations involving $ g, f, $ and $ c $.

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

Since $ S $ and this circle cut orthogonally, we have:

$ 2 \cdot g \cdot (-1) + 2 \cdot f \cdot 3 = c + 0 $

Simplifying:

$ -2g + 6f = c \quad \text{...(i)} $

For the Circle $ x^2 + y^2 - 4x - 2y + 6 = 0 $:

Orthogonality gives us:

$ 2g \cdot (-2) + 2f \cdot (-1) = c + 6 $

Simplifying:

$ -4g - 2f = c + 6 $

For the Circle $ x^2 + y^2 - 12x + 2y + 3 = 0 $:

Orthogonality yields:

$ 2g \cdot (-6) + 2f \cdot 1 = c + 3 $

Simplifying:

$ -12g + 2f = c + 3 $

From equation (i), substituting $ c $ from the orthogonality conditions, we proceed to simplify:

Solving the System of Equations:

Using equations:

$ -4g - 2f = c + 6 $

$ -12g + 2f = c + 3 $

Rearrange equation (i):

$ c = -2g + 6f \quad \text{(equation i)} $

From $-4g - 2f = -2g + 6f + 6$:

$ -2g - 8f = 6 \quad \Rightarrow g + 4f = -3 \quad \text{...(ii)} $

From $-12g + 2f = -2g + 6f + 3$:

$ -10g - 4f = 3 \quad \Rightarrow 10g + 4f = -3 $

Solving equations (ii) and the above, we find:

$ g = 0, \quad f = -\frac{3}{4} $

Substituting into equation (i) for $ c $:

$ c = 6 \times \left(-\frac{3}{4}\right) = -\frac{9}{2} $

Equation of the Circle $ S $:

$ x^2 + y^2 - \frac{3}{2}y - \frac{9}{2} = 0 $

Finding the Equation of the Tangent at $ (0, 3) $:

The equation of the tangent to this circle at the point $(x_1, y_1)$ is given by:

$ x x_1 + y y_1 - \frac{3}{2} \left(\frac{y + y_1}{2}\right) - \frac{9}{2} = 0 $

Substituting point $ (0, 3) $:

$ 3y - \frac{3}{2} \left(\frac{y + 3}{2}\right) - \frac{9}{2} = 0 $

Simplifying:

$ 3y - \frac{3}{4}y - \frac{9}{4} - \frac{9}{2} = 0 $

Convert all terms to common denominator:

$ \frac{12y}{4} - \frac{3y}{4} - \frac{9}{4} - \frac{18}{4} = 0 $

Re-arrange the equation:

$ \frac{9y}{4} = \frac{27}{4} $

Solve for $ y $:

$ y = 3 $

Thus, the equation of the tangent is:

$ y = 3 $

We have,

Equation, of circle is

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

Centre of circle $=(\lambda-2)$

radius $=\sqrt{9+4-12}=1$

$ \begin{aligned} O P & =\sqrt{(3-2)^2+(-2-3)^2} \\ & =\sqrt{1+25}=\sqrt{26} \\ A P & =\sqrt{O P^2-A O^2} \\ & =\sqrt{26-1}=5 \end{aligned} $

$\ln \triangle M O P, \tan \frac{\theta}{2}=\frac{A O}{A P}=\frac{1}{5}$

$ \tan \theta=\frac{2 \tan \frac{\theta}{2}}{1-\tan ^2 \frac{\theta}{2}}=\frac{2 \times \frac{1}{5}}{1-\left(\frac{1}{5}\right)^2}=\frac{\frac{2}{5}}{\frac{24}{25}}=\frac{5}{12} $

Hence, $\theta=\tan ^{-1} \frac{5}{12}$

Let ( $a, b$ be the point on the line

$ x+2 y+K=0 $

We have, equation of circle is

$ x^2+y^2+8 x-6 y-24=0 $

Equation of polar for the given circle is obtained as

$ \begin{aligned} & a x+b y+4(x+d)-3(y+b)-24=0 \\ & (a+4 x+(b-3) y+4 a-3-24 \end{aligned} $

Now, comparing this line with

$2 x-3 y+3=0$ we get

$ \begin{aligned} & \frac{a+4}{2}=\frac{b-3}{-3}=\frac{4 a-y-24}{3} \\ & -3 a-12=2 b-6 \Rightarrow 3 a+2 b=-6 \\ & \text { and } 3 a+12=8 a-6 b-4 s \\ & \Rightarrow 5 a-6 b=60 \ldots \text { (ii) } \end{aligned} $

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

$ a=3 \text { and } b=\frac{-15}{2} $

Now. $\left(3, \frac{15}{2}\right)$ lies on line $x+2 y+k=0$

$ 3-15+K=0 \Rightarrow K=12 $

Required length of tangent from a point

$ \begin{aligned} &\left(\frac{12}{4}, \frac{12}{3}\right) \text { i.e. }(3,4) \\ &=\sqrt{3^3+4^2+24-24-24} \\ &=\sqrt{9+16-24}=1 \end{aligned} $

The equation of pole with respect ts the point $\beta, 2$ to the given circle is

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

Inverse of point $P(1,2)$ is the foot $(h, k)$ of the perpendicalar from the point $(1,2)$

$ \text { to the line } x-2 y+1=0 $

$ \begin{array}{ll} \Rightarrow & \frac{h-1}{1}=\frac{k-2}{-2}=-\left[\frac{1-4+1}{1^2+\left(-2^3\right.}\right] \\ \Rightarrow & h-1=\frac{K-2}{-2}=\frac{2}{5} \\ \Rightarrow & h-1=\frac{2}{5} \text { and } \frac{K-2}{-2}=\frac{2}{5} \\ \Rightarrow & h=\frac{7}{5} \text { and } K=\frac{-4}{5}+2=\frac{6}{5} \end{array} $

Hence, $2 h+K=\frac{14}{5}+\frac{6}{5}=\frac{20}{5}=4$

We have, equation of two circlet

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

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

Centre and radius of both circles are

$ \begin{aligned} & c_1=\left(-2,0, r_1=\sqrt{4+5}=3\right. \\ & c_2=(0,3), r_2=\sqrt{9-8}=1 \end{aligned} $

Let $P$ and $Q$ be the internal and external centre of similitude, it divides $C_1 C_2$ is ine ratio 3 : 1.

$ P=\frac{x_2+C_1}{3+1}=\left(\frac{-2}{4}, \frac{9}{4}\right)=\left(\frac{-1}{2} ; \frac{9}{4}\right) $

Here, $a=\frac{-1}{2}$ and $b=\frac{9}{4}$

$ Q=\frac{3 C_2-C_1}{C_1-C_2}=\left(\frac{2}{2}, \frac{9}{2}\right)=\left(1, \frac{9}{2}\right) $

Here, $C=1$ and $d=\frac{9}{2}$

$ \begin{aligned} \text { Hence, }(a+\Delta)(b+c) & =\left(\frac{-1}{2}+\frac{9}{2}\right)\left(\frac{9}{4}+1\right) \\ & =\frac{4 \times 13}{4}=13 \end{aligned} $

The equation of a circle that passes through the intersection points of the given circles is determined as follows:

First, start with the general form that includes both circles and a parameter $\lambda$:

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

Simplifying this equation leads to:

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

To rewrite in standard circle form, factor out the coefficients:

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

The center of the circle $(h, k)$ from this equation is:

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

Since this center lies on the line $x - y + 6 = 0$, we substitute the center into the line's equation:

$ 1 - \lambda - \left(\lambda - 1\right) + 6(1 + \lambda) = 0 $

Solving:

$ 1 - \lambda - \lambda + 1 + 6\lambda + 6 = 0 \quad \Rightarrow \quad 4\lambda + 8 = 0 \quad \Rightarrow \quad \lambda = -2 $

Substitute $\lambda = -2$ back into the center, we get:

$ (h, k) = \left(-3, 3\right) $

Finally, calculate the radius of the circle:

$ \text{Radius} = \sqrt{(-3)^2 + 3^2 - \frac{2 - (-2)}{1 + (-2)}} = \sqrt{(-3)^2 + 3^2 - \frac{4}{-1}} $

Solving further:

$ \text{Radius} = \sqrt{9 + 9 + 4} = \sqrt{14} $

Let the line cut the axes in $A$ and $B$ and if $(h, k)$ be the mid-point of $A B$, then

$2 h=\frac{1}{\cos \theta}, 2 k=\frac{1}{\sin \theta}$

In order to find the locus, eliminate the variable $\theta$ by $\cos ^2 \theta+\sin ^2 \theta=1$

$\frac{1}{4 h^2}+\frac{1}{4 k^2}=1 \Rightarrow \frac{1}{x^2}+\frac{1}{y^2}=4$

Equation of the given tangents are

$\begin{aligned} E_1=3 x-4 y+4 & =0 \\ E_2=6 x-8 y-7 & =0 \\ \Rightarrow \quad 3 x-4 y-\frac{7}{2} & =0 \end{aligned}$

Here, $a=3, b=-4 c_1=4, c_2=\frac{-7}{2}$

Since, slopes of the given tangents are equal, i.e. $\frac{3}{4}$.

$\therefore$ The given tangents are parallel,

$\begin{aligned} & d=\left|\frac{c_2-c_1}{\sqrt{a^2+b^2}}\right|=\left|\frac{4-\left(\frac{-7}{2}\right)}{\sqrt{3^2+(-4)^2}}\right| \\ & d=\frac{\frac{15}{2}}{\sqrt{9+16}}=\frac{15}{2 \times 5}=\frac{3}{2} \end{aligned}$

As we know that distance between two parallel tangents of a circle is equal to the diameter of that circle.

$\begin{aligned} \therefore \text { Diameter of given circle } & =\frac{3}{2} \\ \therefore \text { Radius of the given circle } & =\frac{\text { Diameter }}{2} \\ & =\left(\frac{3 / 2}{2}\right)=\frac{3}{4} \end{aligned}$

Since, the line $(x-2) \cos \theta+(y-2) \sin \theta=1$ touches a circle. So it is a tangent equation to a circle.

Equation of tangent to a circle at $(x_1, y_1)$ is $(x-h) x_1+(y-k) y_1=a^2$ to a circle

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

Then, after comparing

$\begin{aligned} & x-h=x-2 y-k=y-2 \text { and } a^2=1 \\ & x_1=\cos \theta, y_1=\sin \theta \end{aligned}$

$\therefore$ Required equation of circle

$\begin{gathered} (x-2)^2+(y-2)^2=1 \\ \Rightarrow \quad x^2+y^2-4 x-4 y+7=0 \end{gathered}$

Given equation $\left(x^2+y^2-4 x-2 y-20=0\right)$

So, $C=(2,1)$ and radius $=\sqrt{(g)^2+(f)^2-c}$

$=\sqrt{(2)^2+(1)^2+20}$

Radius $=5$

When substituted $x=10$ and $y=7$ in equation, then value becomes

$(10)^2+(7)^2-4(10)-2(7)-20=75$

which is greater than zero.

Thus, the point $(10,7)$ lies outside the circle.

Its distance from the centre of the circle $(2,1)$ is

$\sqrt{(10-2)^2+(7-1)^2}=10 \text { units }$

So, the minimum distance from the circle therefore becomes $10-5=5$ units

Let $x_1$ and $x_2$ are roots of equation $x^2+2 x-a^2=0$ and $y_1$ and $y_2$ are roots of equation $y^2+4 b-b^2=\left(x-x_1\right)\left(x-x_2\right)=0$ and $y^2+4 b-b^2 \rightarrow y_1, y_2=\left(y-y_1\right)\left(y-y_2\right)$ (roots of equation)

Let $A\left(x_1, y_1\right)$ and $B\left(x_2, y_2\right)$.

Now, the equation of circle with diameter $A B$ is given by $\Rightarrow\left(x-x_1\right)\left(x-x_2\right)+\left(y-y_1\right)\left(y-y_2\right)=0$

$\begin{aligned} & \Rightarrow x^2+2 x-a^2+y^2+4 y-b^2=0 \\ & \Rightarrow x^2+y^2+2 x+4 y=a^2+b^2 \\ & \Rightarrow x^2+y^2+2 x+4 y-a^2-b^2=0 \end{aligned}$

Let $S_1: x^2+y^2-1=0\quad \text{.... (i)}$

$S_2: x^2+y^2-8 x+15=0\quad \text{.... (ii)}$

and $S_3: x^2+y^2+10 y+24=0\quad \text{.... (iii)}$

From Eqs. (i) and (ii),

$\begin{aligned} & 8 x-15=1 \\ \Rightarrow \quad & x=2 \end{aligned}$

Now, from Eqs. (i) and (iii),

$\begin{aligned} & -10 y-24=1 \\ & \Rightarrow \quad-10 y=25 \Rightarrow y=\frac{-5}{2} \end{aligned}$

Hence, the radical centre is $\left(2, \frac{-5}{2}\right)$.

$\begin{aligned} \text { Let } x & =\frac{8 t}{1+t^2} \quad \text{.... (i)}\\ y & =\frac{4\left(1-t^2\right)}{1+t^2} \quad \text{.... (ii)} \end{aligned}$

Squaring and adding Eqs. (i) and (ii), we get

$\begin{aligned} x^2+y^2 & =\frac{64 t^2}{\left(1+t^2\right)^2}+\frac{16\left(1-2 t^2+t^4\right)}{\left(1+t^2\right)^2} \\ & =\frac{16\left(1+2 t^2+t^4\right)}{\left(1+t^2\right)^2}=\frac{16\left(1+t^2\right)^2}{\left(1+t^2\right)^2}=16 \\ \Rightarrow x^2+y^2 & =(4)^2 \end{aligned}$

which is circle of radius 4 units.

Centre of circle passing through $(1,2),(5,2)$ and $(5,-2)$ is mid-point of $C(1,2)$ and $A(5,-2)$. We know that diameter always form a right angle on a circle.

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

$\begin{aligned} \text { and Radius } & =\frac{\text { Diameter }}{2} \\ & =\frac{\sqrt{(5-1)^2+(-2-2)^2}}{2}=2 \sqrt{2} \end{aligned}$

Thus, area is $\pi(2 \sqrt{2})^2=8 \pi$ (units)$^2$

To find the shortest and longest distances from the point $(2,-7)$ to the circle

$x^2+y^2-14 x-10 y-151=0$

So, to find the solution we first draw the diagram,

So, we have to find the coordinates of the centre 0 and the radius of the circle, i.e $O M=O N=$ radius Thus, the shortest distance $P M=O M-O P$ and the longest distance $P N=M N-P M$ First we have to determine the point $P(2,-7)$ lies in which side of circle. So, we put the value $x=2,y=-7$ in the left hand side of the equation, we get $x^2+y^2-14 x-10 y-151 =(2)^2$

$=(2)^2+(-7)^2-14\times2-10(-7)-151=-56 < 0$

Thus, the point lies inside of the circle. Thus, we know any equation of the circle is in the from $x^2+y^2+2 g x+2 f y+c=0$.

Thus, $g=-7$, and $f=-5, c=-151$

So, centre is $(-g,-f)=(7,5)$

$\begin{aligned} \text { and radius } & =r=O M=O N \\ & =\sqrt{7^2+5^2-(-151)} \\ & =\sqrt{49+25+151} \\ & =\sqrt{225}=15 \end{aligned}$

Now, distance between centre $O(7,5)$ and $P(2,-7)$ is $O P=\sqrt{(7-2)^2+(5+7)^2}$

$=\sqrt{25+144}=\sqrt{169}=13$

Thus, the shortest distance from the point $P(2,-7)$ to the circle is $P M=O M-O P=15-13=2$ units and the longest distance is

$\begin{aligned} P N=M N-P M & =(O M+O N)-P M \\ & =(15+15)-2=28 \text { units } \end{aligned}$

Thus, the required ratio is $28: 2=14: 1$

Let the centre of circle be $(h, k)$. Equation of circle will be

$x^2+y^2-2 h x-2 k y+c=0 \quad \text{... (i)}$

Circle passes through $(2,3)$

$\begin{array}{llrl} & & 2^2+3^2-4 h-6 k+c & =0 \\ \Rightarrow & & 4 h+6 k-c=13 \\ \text { or } & & c=4 h+6 k-13 \end{array}$

Equation of circle be

$x^2+y^2-2 h x-2 k y+4 h+6 k-13$

This circle makes intercept of length 3 on the line $x=2$

$\left|y_1-y_2\right|=3$, when $x=2$ is substituted in the equation of circle, where $y_1$ and $y_2$ are the two roots

$\begin{aligned} 4+ & y^2-4 h-2 k y+4 h+6 k-13=0 \\ \Rightarrow & y^2-2 k y+6 k-9=0 \\ & y_1+y_2=2 k \text { and } y_1 y_2=6 k-9 \\ & \left|y_1-y_2\right|=3 \\ & \sqrt{\left(y_1+y_2\right)^2-4 y_1 y_2}=3 \\ \Rightarrow & 4 k^2-4(6 k-9)=9 \\ \Rightarrow & 4 k^2-24 k+36-9=0 \Rightarrow 4 k^2-24 k+27=0 \\ \Rightarrow & k=\frac{9}{2}, k \neq \frac{3}{2} \end{aligned}$

[$\because$ when substitute $k=\frac{3}{2}$ in $y_1 y_2$, we get $y_1 y_2$ as 0$]$

Now, similarly substitute $y=3$

$\begin{aligned} & x^2+9-2 h x-6 k+4 h+6 k-13=0 \\ \Rightarrow & x^2-2 h x+4 h-4=0 \\ \Rightarrow & x_1+x_2=2 h_1, x_1 x_2=4 h-4 \\ \because \quad & \left|x_1-x_2\right|=4 \\ & \sqrt{\left(x_1+x_2\right)^2-4 x_1 x_2}=4 \\ \Rightarrow \quad & 4 h^2-16 h+16=16 \Rightarrow 4 h(h-4)=0 \\ & h=4 \text { and } h \neq 0 \end{aligned}$

$\therefore$ Equation of the required circle is

$x^2+y^2-8 x-9 y+30=0$

The given equation of circles

$\begin{aligned} & S_1: x^2+y^2+8 x+2 y+10=0 \\ & S_2: x^2+y^2+7 x+3 y+10=0 \end{aligned}$

As we know the equation of radical axis of two circles is given by

$\begin{aligned} S_1- & S_2=0 \\ \Rightarrow & \left(x^2+y^2+8 x+2 y+10\right) \\ & -\left(x^2+y^2+7 x+3 y+10\right)=0 \\ \Rightarrow & (8 x-7 x)+(2 y-3 y)=0 \\ \Rightarrow & x-y=0 \end{aligned}$

Thus, equation of radical axis of the given circle is

$x-y=0$

Hence, the image of the point $(3,4)$ with respect to $y=x$ line is $(4,3)$.

Common tangents which are parallel are

$\begin{array}{ll} & 3 x-4 y+4=0 \\ \text { and } & 6 x-8 y-7=0 \\ \Rightarrow & 3 x-4 y+4=0 \\ \text { and } & 3 x-4 y-\frac{7}{2}=0 \end{array}$

The diameter is equal to the distance between these parallel lines.

So, the radius $=\frac{1}{2} \times\left|\frac{4-\left(-\frac{7}{2}\right)}{\sqrt{3^2+(-4)^2}}\right|$

$=\frac{1}{2} \times \frac{15}{2} \times \frac{1}{5}=\frac{3}{4}$

The centre of the circle lies on the line parallel to the given lines at a distance of $\frac{3}{4}$ from both the lines.

So, let the equation be $3 x-4 y+k=0\quad \text{....(i)}$

Then, distance between $3 x-4 y+k=0$ and $3 x-4 y+4$ is same as distance between $3 x-4 y+k=0$ and $3 x-4 y-\frac{7}{2}=0$

$\begin{aligned} & \left|\frac{k-4}{\sqrt{3^2+(-4)^2}}\right|=\left|\frac{k-\left(-\frac{7}{2}\right)}{\sqrt{(3)^2+(-4)^2}}\right| \\ & \Rightarrow \quad|k-4|=-\left|k+\frac{7}{2}\right| \\ & \text { and }-(k-4)=k+\frac{7}{2} \\ & \therefore \quad 2 k=\frac{7}{2}-4=-\frac{1}{2} \Rightarrow k=-\frac{1}{4} \\ & \text { and } 2 k=4-\frac{7}{2}=\frac{1}{2} \Rightarrow k=\frac{1}{4} \end{aligned}$

$\therefore$ For $k=\frac{1}{4}$ distance of Eq. (i) from other line is $\frac{3}{4}$.

$\therefore$ Locus of centers of the circles, is

$3 x-4 y+\frac{1}{4}=0$

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}$

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

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

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

$\angle A P B=90 \Upsilon$

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