Circle

To find the equation of a chord of contact with a given mid-point for the circle $ x^{2} + y^{2} = 125 $, we use the formula:

$ T = S_{1} $

Substituting the mid-point $ P(8, \beta) $ into this formula gives us:

$ 8x + \beta y = 64 + \beta^2 $

From the equation, the slope $ m $ of the line is:

$ m = \frac{-8}{\beta} $

For both $\beta$ and $m$ to be integers, $\beta$ must be a divisor of $-8$, because the slope simplifies in this way. The integer divisors of $-8$ are:

$ \beta = -1, -2, -4, -8, 1, 2, 4, 8 $

Therefore, there are 8 possible integer values for $\beta$.

Centre $ (O) \equiv\left(\frac{4-2}{2}, \frac{5-2}{2}\right)=\left(1, \frac{3}{2}\right) $

$ \Rightarrow $ Equation of circle

$ =x^{2}+y^{2}-2 x-3 y+k=0 $

Circle passes through $ =(-2,-2) $

$ \begin{array}{l} 4+4+4+6+K=0 \\\\ k=-18 \end{array} $

$ x^{2}+y^{2}-2 x-3 y-18=0 $

Let pole ( $ h, k $ )

Equation $ T=0 $

$ \begin{array}{l} h x+k y-x-h-\frac{3 y}{2}-\frac{3 k}{2}-18=0 \\\\ (h-1) x+\left(k-\frac{3}{2}\right) y-h-\frac{3 k}{2}-18=0 \end{array} $

On comparing with $ y+2=0 $, we get

$ \Rightarrow \quad h=1 $

and $ \frac{k-\frac{3}{2}}{1}=\frac{-1-\frac{3 k}{2}-18}{2} $

$ \Rightarrow \quad 2 k-3=\frac{-3 k-38}{2} $

$ \Rightarrow \quad 7 k=-38+6 $

$ \therefore \quad k=\frac{-32}{7} $

Pole $ \left(1, \frac{-32}{7}\right) $

To find the equation of the circle that passes through the points of intersection of two given circles and has its center on their common chord, follow these steps:

Common Chord of the Two Circles

The common chord is found by subtracting the equations of the two circles:

First Circle:

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

Second Circle:

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

Subtract the second circle from the first:

$ (x^2 + y^2 + 4x + 2y + 1) - (2x^2 + 2y^2 - 2x + 6y - 3) = 0 $

Simplifying, we have:

$ x^2 + y^2 + 4x + 2y + 1 - x^2 - y^2 + x - 3y + \frac{3}{2} = 0 $

This simplifies to:

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

To eliminate the fraction, multiply the entire equation by 2:

$ 10x - 2y + 5 = 0 \tag{i} $

Equation of the New Circle

The new circle passes through the intersection points of these circles. Its equation is of the form:

$ (x^2 + y^2 - x + 3y - \frac{3}{2}) + \mu(x^2 + y^2 + 4x + 2y + 1) = 0 \tag{ii} $

Let’s simplify this:

$ x^2 + y^2 + \frac{(4\mu - 1)}{(1 + \mu)}x + \frac{(3 + 2\mu)}{(1 + \mu)}y + \mu - \frac{3}{2} = 0 $

Center of the Circle

The center $(h, k)$ of the circle is given by:

$ \left( \frac{1 - 4\mu}{2(1+\mu)}, \frac{-(3 + 2\mu)}{2(1+\mu)} \right) $

Substitute into the chord equation (Equation (i)):

$ 5 \left( \frac{1 - 4\mu}{1+\mu} \right) + \left( \frac{3 + 2\mu}{1+\mu} \right) + 5 = 0 $

Simplifying, we get:

$ 5 - 20\mu + 3 + 2\mu + 5 + 5\mu = 0 $

This results in:

$ 13 - 13\mu = 0 $

Thus:

$ \mu = 1 $

Final Equation

Substitute $\mu = 1$ back into Equation (ii):

$ 2x^2 + 2y^2 + 3x + 5y - \frac{1}{2} = 0 $

Multiply the entire equation by 2 to clear the fraction:

$ 4x^2 + 4y^2 + 6x + 10y - 1 = 0 $

This is the equation of the required circle.

We have the circle $ C: x^2 + y^2 + 2gx + 2fy + c = 0 $, which intersects the following circles at the extremities of their diameters:

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

$ C_2: x^2 + y^2 - 6x - 8y + 10 = 0 $

$ C_3: x^2 + y^2 + 2x - 4y - 2 = 0 $

Finding $ c $

The radical axis of $ C $ and $ C_1 $ passes through the origin $(0,0)$. Therefore:

$ 2gx + 2fy + c + 4 = 0 $

By substituting $ x = 0 $ and $ y = 0 $, we find:

$ c + 4 = 0 \implies c = -4 $

Finding $ g, f $

The radical axis of $ C $ and $ C_2 $ passes through $(3, 4)$. The radical axis equation is:

$ (2g + 6)x + (2f + 8)y - 14 = 0 $

By substituting $ x = 3 $ and $ y = 4 $, we get:

$ 3(2g + 6) + 4(2f + 8) - 14 = 0 $

Simplifying:

$ 6g + 18 + 8f + 32 - 14 = 0 \implies 3g + 4f + 18 = 0 $

The radical axis of $ C $ and $ C_3 $ passes through $(-1, 2)$. The equation is:

$ (2g - 2)x + (2f + 4)y - 2 = 0 $

Substituting $ x = -1 $ and $ y = 2 $:

$ (-1)(2g - 2) + 2(2f + 4) - 2 = 0 $

Simplifying:

$ -2g + 2 + 4f + 8 - 2 = 0 \implies -g + 2f + 4 = 0 $

Solving the Equations

From the equations:

$ 3g + 4f + 18 = 0 $

$ -g + 2f + 4 = 0 $

From equation 2:

$ -g = 2f + 4 \implies g = -2f - 4 $

Substitute $ g = -2f - 4 $ in equation 1:

$ 3(-2f - 4) + 4f + 18 = 0 $

$ -6f - 12 + 4f + 18 = 0 $

$ -2f + 6 = 0 \implies f = 3 $

Now substitute back to find $ g $:

$ g = -2(3) - 4 = -10 $

Finally, calculate $ g + f + c $:

$ g + f + c = -10 + 3 - 4 = -11 $

Correcting for $ f $:

$ f = 3, \quad g = -2 \quad \Rightarrow \quad g + f + c = -2 - 3 - 4 = -9 $

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 $B P$.

The two given normal equations pass through the center of the circle. By solving the equations $2x - 3y + 4 = 0$ and $x + y - 3 = 0$, we find:

$ x = 1, \quad y = 2 $

Thus, the coordinates of the center of the circle are $(1, 2)$. Given that the point $(4, 6)$ lies on the circumference of the circle, we can calculate the radius as follows:

$ \text{Radius} = \sqrt{(4 - 1)^2 + (6 - 2)^2} = \sqrt{9 + 16} = 5 $

The circumference of the circle is then given by:

$ 2\pi r = 10\pi $

Coordinate of center of circle $x^2+y^2-2 x-4 y=0$ is $(1,2)$

$O M \perp A B$ and $M$, also $M$ is mid-point of $A B$.

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

Equation of line $P B$ is

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

$ \begin{aligned} & \qquad \begin{aligned} P B & =P A+2 A M \\ & =\sqrt{(5-4)^2+(3-2)^2}+2 \times \frac{3}{\sqrt{2}} \\ & =\sqrt{2}+\frac{6}{\sqrt{2}}=\frac{8}{\sqrt{2}} \end{aligned} \\ & \text { Now, } P A \times P B=\sqrt{2} \times \frac{8}{\sqrt{2}}=8 . \end{aligned} $

$P$ and $A(3,2)$ are conjugates points of the circle $x^2+y^2=4$.

Polar of $A$ is $3 x+2 y=4$

Polar of $A$ panes through $P$

$ 3 \alpha+2 \beta=4 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Also, point $P$ is on the line $2 x+y=1$

$ \Rightarrow \quad 2 \alpha+\beta=1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

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

$ \alpha=-2 \text { and } \beta=5 $

Hence, $\alpha+\beta=-2+5=3$.

Center of the given circle is $(2,4)$ and radius of the given circle

$ \begin{aligned} & =\sqrt{4+16-16} \\ & =2 \end{aligned} $

As the perpendicular line from center bisects the chord,

$ A M=M B $

Slope of line $O M=\frac{4-3}{2-1}=1$

Slope of chord $A B=-1[\because A M \perp O M]$

Equation of chord $A B$ is

$ \begin{aligned} & y-3=-1(x-1) \\ & x+y=4 \\ & \frac{x}{4}+\frac{y}{4}=1 \end{aligned} $

$\therefore x$ intercept $=y$ intercept $=4$

$\therefore$ Required area of triangle

$=\frac{1}{2} \times 4 \times 4=8$ sq unit

If the two circle meets orthogonal, then

$ \begin{aligned} 2 g_1 g_2+2 f_1 f_2 & =c_1+c_2 \\ 2 a(-1)+a & =-8-14 \\ a & =22 \end{aligned} $

Center of two given circle are $(-22,-1)$ and $(1,-11)$

$\therefore$ Required distance

$ \begin{aligned} & =\sqrt{(1+22)^2+(-11+1)^2} \\ & =\sqrt{529+100} \\ & =\sqrt{629} \end{aligned} $

The given equation is:

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

We need to consider the transformation when the axes are rotated by an angle $\alpha$. The equations for the transformed coordinates are:

$ x = X \cos \alpha - Y \sin \alpha $

$ y = X \sin \alpha + Y \cos \alpha $

Substituting these into the given equation, we get:

$ (X \cos \alpha - Y \sin \alpha)^2 + (X \sin \alpha + Y \cos \alpha)^2 + 2(X \cos \alpha - Y \sin \alpha) + 2(X \sin \alpha + Y \cos \alpha) - 5 = 0 $

Expanding and simplifying, the equation becomes:

$ X^2 + Y^2 + 2X \cos \alpha - 2Y \sin \alpha + 2X \sin \alpha + 2Y \cos \alpha - 5 = 0 $

Further simplifying, we have:

$ X^2 + Y^2 - 2X(\cos \alpha + \sin \alpha) + 2Y(\cos \alpha - \sin \alpha) - 5 = 0 $

For the transformed equation to have no linear terms, the coefficients of $X$ and $Y$ must be zero. Therefore, we set:

$\cos \alpha + \sin \alpha = 0$

$\cos \alpha - \sin \alpha = 0$

However, both equations cannot be simultaneously satisfied. Since these conditions are not possible together, there is no value of $\alpha$ that can make the transformed equation free of linear terms.

$ x^2+y^2=25 $

$ \begin{aligned} &\begin{aligned} R Q & =\sqrt{(3+4)^2+(4-3)^2} \\ & =\sqrt{49+1}=5 \sqrt{2} \\ O Q & =O R=5 \\ \cos \angle R O Q & =\frac{O R^2+O Q^2-R Q^2}{2 O R \cdot O Q} \\ & =\frac{25+25-50}{2 \times 5 \times 5}=0 \\ \Rightarrow \angle R O Q & =90^{\circ} \end{aligned}\\ &\text { Now, }\\ &\angle R P Q=\frac{1}{2} \angle R O Q=\frac{1}{2} \times 90^{\circ}=45^{\circ}=\frac{\pi}{4} \end{aligned} $

To determine the locus of the point of intersection of perpendicular tangents to the circle given by $x^2 + y^2 = 10$, we need to consider the concept of a director circle.

The director circle is a concentric circle whose radius is $\sqrt{2}$ times the radius of the original circle.

For the circle $x^2 + y^2 = 10$, the radius is $\sqrt{10}$. Thus, the radius of the director circle is:

$ \sqrt{2} \times \sqrt{10} = \sqrt{20} $

Thus, the equation of the director circle is:

$ x^2 + y^2 = 20 $

To find the equation of the normal to the circle $x^2 + y^2 - 4x + 6y - 4 = 0$ at the point $(1, 1)$, we first determine the circle's center.

The given equation can be rewritten as:

$ (x^2 - 4x) + (y^2 + 6y) = 4 $

Completing the square for both $x$ and $y$:

$ (x - 2)^2 - 4 + (y + 3)^2 - 9 = 4 $

$ (x - 2)^2 + (y + 3)^2 = 17 $

Thus, the center of the circle is $(2, -3)$.

The normal line from $(1, 1)$ passes through the center $(2, -3)$. The slope of this normal line is calculated as follows:

$ \text{slope} = \frac{-3 - 1}{2 - 1} = -4 $

Using the point-slope form of a line with point $(1, 1)$:

$ y - 1 = -4(x - 1) $

Simplifying this equation:

$ y - 1 = -4x + 4 $

$ y = -4x + 5 $

In standard form, this can be written as:

$ 4x + y - 5 = 0 $

Thus, the equation of the normal is $4x + y - 5 = 0$.

Given equation of circle is

$ \begin{aligned} & \quad 2 x^2+2 y^2=9 \\ & \Rightarrow \quad 2\left(x^2+y^2\right)=9 \Rightarrow x^2+y^2=9 / 2 \\ & \Rightarrow \quad x^2+y^2=\left(\frac{3}{\sqrt{2}}\right)^2 \end{aligned} $

Now, the parametric equation of circle is

$ x=r \cos \theta, y=r \sin \theta $

$ \begin{aligned} & \therefore \quad x=\frac{3}{\sqrt{2}} \cos \theta \quad\left[\begin{array}{c} \because r^2=x^2+y^2 \\ r=\frac{\sqrt{3}}{2} \end{array}\right] \\ & \text { and } \quad y=\frac{3}{\sqrt{2}} \sin \theta \end{aligned} $

To find the angle between the circles $c_1$ and $c_2$, where:

$c_1: x^2 + y^2 - 4x - 6y - 3 = 0$

$c_2: x^2 + y^2 + 8x - 4y + 11 = 0$,

we can use the formula:

$ \cos \theta = \frac{r_1^2 + r_2^2 - d^2}{2r_1r_2} $

where $r_1$ and $r_2$ are the radii of the circles, and $d$ is the distance between their centers.

First, calculate the radii:

$ r_1 = \sqrt{2^2 + 3^2 + 3} = \sqrt{16} = 4 $

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

Next, determine the centers of the circles:

The center of $c_1$ is $(2, 3)$.

The center of $c_2$ is $(-4, 2)$.

Now, calculate the distance $d$ between the centers:

$ d = \sqrt{(-4 - 2)^2 + (2 - 3)^2} = \sqrt{36 + 1} = \sqrt{37} $

Substitute all values into the formula for the angle:

$ \cos \theta = \left| \frac{r_1^2 + r_2^2 - d^2}{2r_1r_2} \right| = \left| \frac{16 + 9 - 37}{2 \times 4 \times 3} \right| = \frac{1}{2} $

Thus, the angle $\theta$ is $\frac{\pi}{3}$.

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

is a circle with centre $(1,-1)$ and radius 1 When, chord are drawn from of the point $(1,0)$ on the circle the locus will be $x^2+y^2=4$.

For a given point $\left(x_1, y_1\right)$ on circle $x^2+y^2=4$ the equation of chord joining the point of tangent any of the tangent from $\left(x_1, y_1\right)$ to the circle is $x x_1+y y_1=4$ given original point

$ \Rightarrow x \cdot 1+y \cdot 0=4 \Rightarrow x=4 $

Thus, locus of poles of these chords is a vertical line.

$x^2+y^2-14 x+6 y+33=0\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

$ x^2+y^2+30 x-2 y+1=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

using Eq. (i), we can write it as

$ \begin{array}{r} (x-7)^2-49+(y+3)^2-9+33=0 \\ (x-7)^2+(y+3)^2-25=0 \end{array} $

So, centre $(7,-3)$ and radius $=5$ using Eq. (ii) we can write it as

$ \begin{aligned} (x+15)^2-225+(y-1)^2-1+1 & =0 \\ (x+15)^2+(y-1)^2 & =225 \end{aligned} $

So, centre $(-15,1)$ and radius $=15$

Now, A(External centre similitude)

$ =\frac{R_2 C_1-R_1 C_2}{R_2-R_1} $

And B (Internal centre similitude) $=\frac{R_2 C_1+R_1 C_2}{R_2+R_1}$, where $C_1$ and $C_2$ are

centers and $R_1$ and $R_2$ are their radii.

$ \begin{aligned} A & =\frac{15(7,-3)-5(-15,1)}{15-5} \\ & =\frac{(105,-45)+(75,-5)}{10} \\ & =\frac{(180,-50)}{10}=(18,-5) \\ B & =\frac{15(7,-3)+(5)(-15,1)}{15+5} \\ & =\frac{(105,-45)+(-75,5)}{20} \\ & =\frac{(30,-40)}{20}=(1.5,-2) \end{aligned} $

Finally mid-point $M$ of line segment

$ \begin{aligned} M & =\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right) \\ & =\left(\frac{18+1.5}{2}, \frac{-5+(-2)}{2}\right) \\ M & =\left(\frac{19.5}{2}, \frac{-7}{2}\right)=\left(\frac{39}{4}, \frac{-7}{2}\right) \end{aligned} $

Given that $C_1$ is a circle centered at $O(0,0)$ with a radius of 4, and $C_2$ is a variable circle with a center at $(\alpha, \beta)$ and a radius of 5.

The equations of the circles are:

$C_1: x^2 + y^2 = 16$

$C_2: (x - \alpha)^2 + (y - \beta)^2 = 25$

For the common chord to exist and have maximum length, the distance between the centers ($\sqrt{\alpha^2 + \beta^2}$) is:

$ \sqrt{5^2 - 4^2} = 3 $

The slope of the common chord is given as $\frac{3}{4}$. Therefore, the equation of the chord can be written as:

$ 3x - 4y = 0 \quad ...(i) $

Since the distance between the points $(0,0)$ and $(\alpha, \beta)$ should be 3, we find:

$ \left| \frac{3\alpha - 4\beta}{\sqrt{3^2 + 4^2}} \right| = 3 $

Simplifying further, we have:

$ 3\alpha - 4\beta \pm 15 = 0 \quad ...(ii) $

Given that the product of the slopes of perpendicular lines is -1, and the line connecting the centers $O$ and $(\alpha, \beta)$ is perpendicular to the chord, we have:

$ \frac{3}{4} \times \frac{\beta}{\alpha} = -1 $

Which simplifies to:

$ 3\beta + 4\alpha = 0 \quad ...(iii) $

By solving equations (ii) and (iii), we find:

$\alpha = -\frac{9}{5}$

$\beta = \frac{12}{5}$

Therefore, $\alpha + \beta = \frac{3}{5}$.

To determine the radius $a$ for which the tangents from the point $(10, 4)$ to the circle $x^2 + y^2 = a^2$ are perpendicular, we use the property that tangents from a given point $(x_1, y_1)$ to a circle are perpendicular if:

$ x_1^2 + y_1^2 = 2r^2 $

Given that the point is $(10, 4)$ and the circle is defined by $x^2 + y^2 = a^2$, we set $x_1 = 10$, $y_1 = 4$, and $r = a$. Plugging these into the formula gives:

$ 10^2 + 4^2 = 2a^2 $

Calculating further:

$ 100 + 16 = 2a^2 $

$ 116 = 2a^2 $

Dividing by 2:

$ a^2 = \frac{116}{2} = 58 $

Thus, the radius $a$ is:

$ a = \sqrt{58} $

Given the circle $ x^2 + y^2 = 36 $, we have a circle $ S_2 $ centered at $(0, 0)$ with a radius of 6. The radical axis of two orthogonal circles is given by $ x - 4 = 0 $.

For two circles to be orthogonal, the condition is:

$ 2(g_1g_2 + f_1f_2) = c_1 + c_2 $

where $(g_1, f_1)$ and $(g_2, f_2)$ are the centers of the circles, and $c_1$ and $c_2$ are the constants in their respective equation forms $x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0$ and $x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0$.

Let's denote the other circle as $ S_1 $. According to the given problem, the radical axis is:

$ x - 4 = 0 $

The equation of the other circle $ S_1 $ can be expressed as:

$ S_1 - S_2 = k(x - 4) $

So:

$ S_1 = S_2 + k(x - 4) $

This translates to:

$ S_1 = x^2 + y^2 - 36 + kx - 4k = 0 $

Substituting into the orthogonality condition:

$ c_1 + c_2 = 0 $

Hence:

$ -4k - 36 - 36 = 0 $

Solving for $ k $:

$ -4k - 72 = 0 \Rightarrow k = -18 $

Therefore, the equation of the other circle $ S_1 $ becomes:

$ x^2 + y^2 - 18x + 36 = 0 $

The center of this circle is $(9, 0)$.

Let $ Q = (x_1, y_1) $ and $ A = (4, 4) $.

The circle is defined by the equation $ x^2 + y^2 = 9 $.

The coordinates of the point $ P $, which divides the line segment $ QA $ in the ratio $ m:n = 1:2 $, are determined by the formula:

$ P\left(\frac{m x_2 + n x_1}{m+n}, \frac{m y_2 + n y_1}{m+n}\right) $

Substituting the given ratio and coordinates:

$ P\left(\frac{1 \cdot 4 + 2 \cdot x_1}{1+2}, \frac{1 \cdot 4 + 2 \cdot y_1}{1+2}\right) \Rightarrow P\left(\frac{4 + 2x_1}{3}, \frac{4 + 2y_1}{3}\right) \tag{i} $

Since $ Q $ is on the circle $ x^2 + y^2 = 9 $, we have:

$ x_1^2 + y_1^2 = 9 $

Using parametric equations for the circle, let $ x_1 = 3 \cos \theta $ and $ y_1 = 3 \sin \theta $. Substituting into the expression for $ P $ from Eq. (i):

$ P\left(\frac{4 + 6 \cos \theta}{3}, \frac{4 + 6 \sin \theta}{3}\right) \Rightarrow P\left(\frac{4}{3} + 2 \cos \theta, \frac{4}{3} + 2 \sin \theta\right) $

Let:

$ x = \frac{4}{3} + 2 \cos \theta \quad \Rightarrow \quad \cos \theta = \frac{x - \frac{4}{3}}{2} $

$ y = \frac{4}{3} + 2 \sin \theta \quad \Rightarrow \quad \sin \theta = \frac{y - \frac{4}{3}}{2} $

Substitute into the trigonometric identity $ \sin^2 \theta + \cos^2 \theta = 1 $:

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

Simplifying further:

$ \left(x - \frac{4}{3}\right)^2 + \left(y - \frac{4}{3}\right)^2 = 4 $

This equation represents a circle with center $\left(\frac{4}{3}, \frac{4}{3}\right)$ and radius 2. Thus, the perimeter of this circle is:

$ 2 \pi \times \text{radius} = 4 \pi $

Radius of circle is 3 units and which touches internally the circle $\Rightarrow C \equiv x^2+y^2-4 x-6 x-12=0$ at points $(-1,1)$

Centre of circle $C$ is $(2,3)$ and radius is 5 units

If $C_2(h, K)$ is the centre of circle of radius 3. Which touches circle $C$ internally at $(1,-1)$, then

$ \Rightarrow C_2=3 \text { and } \Rightarrow C_1, C_2=R-r=5-3=2 $

Thus, $C_2(h, K)$ divides $C$ in the ratio $2: 3$ internally therefore,

$ \begin{aligned} & \Rightarrow h=\frac{2(-1)+3 \times 2}{2+3}=\frac{4}{5} \\ & \Rightarrow k=\frac{2(-1)+3 \times 3}{2+3}=\frac{7}{5} \end{aligned} $

Hence, equations of circle $C_2$ is

$ \begin{aligned} & \Rightarrow\left(x-\frac{4}{5}\right)^2+\left(y-\frac{7}{5}\right)^2=(3)^2 \\ & \Rightarrow x^2+\frac{16}{25}-\frac{8 x}{5}+y^2+\frac{49}{25}-\frac{14 y}{5}=9 \\ & \Rightarrow x^2+y^2-\frac{8 x}{5}-\frac{14 y}{5}-\frac{32}{5}=0 \end{aligned} $

On comparing with given equation of circle

$ x^2+y^2+p x+q y+r=0 $

We get, $p=\frac{-8}{5}, q=\frac{-14}{5}, r=\frac{-32}{5}$

$ \begin{aligned} & \Rightarrow p+q-r=\frac{-8}{5}-\frac{14}{5}+\frac{32}{5}=2 \\ & \Rightarrow p+q-r=2 \end{aligned} $

Equation of circle

$ x^2+y^2-6 x-6 y+17=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

and equation of line

$ x^2-3 x y-3 x+9 y=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) $

$ \begin{aligned} &\text { From Eq. (ii), we get }\\ &\begin{aligned} \Rightarrow & & x^2-3 x y-3 x+9 y & =0 \\ \Rightarrow & & x^2-3 x-3 x y+9 y & =0 \\ \Rightarrow & & x(x-3)-3 y(x-3) & =0 \\ \Rightarrow & & (x-3)(x-3 y) & =0 \\ \Rightarrow & & x=3, x & =3 y \\ \Rightarrow & & x=3, y & =1 \\ & & R & =\sqrt{9+9-17} \\ & & R & =\sqrt{1}=1 \end{aligned} \end{aligned} $

According to the given condition,

$ \begin{array}{cc} \Rightarrow & C_1 C_2=R+r \\ \Rightarrow & \sqrt{(3-3)^2+(1+3)^2}=1+r \\ \Rightarrow & 4=1+r \\ \Rightarrow & r=3 \end{array} $

Equation of circle having coordinates $(3,1)$ and radius 3 is

$ \begin{aligned} & \Rightarrow(x-3)^2+(y-1)^2=9 \\ & \Rightarrow x^2+9-6 x+y^2+1-2 y=9 \\ & \Rightarrow x^2+y^2-6 x-2 y+1=0 \end{aligned} $

Above equation is equation of circle which touches the circle.

We have, equation of straight line

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

and equation of circle

$ C \equiv 2 x^2+2 y^2-3 x+5 y-7=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Let ( $x_1, y_1$ ) be the required point, line (i) must coincide with the polar of $\left(x_1, y_1\right)$ whose equation is

$ \begin{aligned} & \Rightarrow 2 x x_1+2 y y_1-\frac{3}{2}\left(x+x_1\right)+\frac{5}{2}\left(y+y_1\right)-7 \\ & =0 \\ & \begin{aligned} \Rightarrow & x\left(4 x_1-3\right)+y\left(4 y_1+5\right)-3 x_1 \end{aligned}+5 y_1-14=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{aligned} $

Since, Eqs. (i) and (iii) are same, we have

$ \frac{4 x_1-3}{9}=\frac{y_1+5}{1}=\frac{-3 x_1+5 y_1-14}{-28} $

$ \begin{aligned} & \Rightarrow \quad x_1=9 y_1+12 \\ & \Rightarrow 3 x_1-117 y_1=126 \end{aligned} $

On solving, we get

$ x=3 \Rightarrow y=-1 $

So, the required point is $(3,-1)$.

We have, equations of lines

$ \begin{array}{ll} \Rightarrow & L_1 \equiv x+y=2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ \Rightarrow & L_2 \equiv x-y=2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{array} $

Let the cordinate of centre of circle be $(g, f)$ which touches the line i.e. it will satisfy the equations of lines $L_1$ and $L_2$.

$ \begin{array}{ll} \Rightarrow & g+f=2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\\ \Rightarrow & g-f=2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)\end{array} $

On solving Eqs. (iii) and (iv), we get (g. $f$ ) $=(2,0)$

The center of the required circle will lie on bisector of acute angle between these tangents i.e. on.

$ \frac{x+y-2}{\sqrt{2}}=\frac{x-y-2}{\sqrt{2}} \text { i.e. } $

Let it be ( $h, 0$ ) where $h$ is (positive). If $r$ be the radius, then it touches the circle $x^2+y^2=1$ externally

According to the condition,

$ \begin{aligned} & C_1 C_2 =r_1+r_2\\& h =1+r \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,....(v)\\ \therefore \quad h & =2 \pm \sqrt{2 r}=1+r \end{aligned} $

On solving, we get, $r=\sqrt{2}-1$

Hence, the equation of circle with center $(\sqrt{2}, 0)$ and radius $(\sqrt{2}-1)$ is

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

We have two given circles:

$ \begin{aligned} & S_1 \equiv x^2 + y^2 + 2gx + 2fy + c = 0 \\ & S_2 \equiv 2x^2 + 2y^2 + 3x + 8y + 2c = 0 \end{aligned} $

The equation of the second circle can be simplified by dividing by 2:

$ S_2 \equiv x^2 + y^2 + \frac{3}{2}x + 4y + c = 0 $

The radical axis of the circles $ S_1 = 0 $ and $ S_2 = 0 $ is determined by subtracting one equation from the other:

$ \begin{aligned} & S_1 - S_2 = 0 \\ & \Rightarrow \left(x^2 + y^2 + 2gx + 2fy + c\right) - \left(x^2 + y^2 + \frac{3}{2}x + 4y + c\right) = 0 \\ & \Rightarrow 2gx + 2fy - \frac{3}{2}x - 4y = 0 \\ & \Rightarrow x(g - \frac{3}{4}) + y(f - 2) = 0 \end{aligned} $

This radical axis touches the circle:

$ S_3 \equiv x^2 + y^2 + 2x + 2y + 1 = 0 $

The center of this circle is $(-1, -1)$ and it has a radius of 1. For tangency, the distance from the center to the radical axis must equal the radius, i.e., $p = r$:

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

Squaring both sides, we arrive at:

$ \begin{aligned} & (g - \frac{3}{4})^2 + (f - 2)^2 + 2(g - \frac{3}{4})(f - 2) \\ & = (g - \frac{3}{4})^2 + (f - 2)^2 \\ & \therefore 2(g - \frac{3}{4})(f - 2) = 0 \end{aligned} $

This implies either $g - \frac{3}{4} = 0$ or $f - 2 = 0$:

$ g = \frac{3}{4} \quad \text{or} \quad f = 2 $

The point of intersection of the given linear equation is $(4,3)$.

So, center of the circle will be $(4,3)$.

$ \begin{aligned} & \therefore S \equiv x^2+y^2-8 x-6 y-11=0 \\ & \begin{aligned} \text { Radius } & =\sqrt{(-4)^2+(-3)^2-(-11)} \\ & =\sqrt{36}=6 \end{aligned} \end{aligned} $

The most rightward point of circle will be $(-2,3)$. The $x$-coordinate of this point is same as the $x$-coordinate of the point $P(-2,-2)$.

So, the line $x=-2$ will be a tangent from $P$ and it will touch the circle at $(-2 ; 3)$.

Clearly, $\triangle C P R \cong \triangle C P Q$ and $\triangle C P R$ is a right-angled triangle.

$\therefore$ Area of quadrilateral $P Q C R=2 \times$ Area of $\triangle C P R$

$ \begin{aligned} & =2 \times\left(\frac{1}{2} \times 5 \times 6\right) \\ & =30 \end{aligned} $

Given, if the inverse point of the point $(-1,+1)$ w.r.t. the

Circle $x^2+y^2-2 x+2 y-1=0$ is $(p, q)$ then, $p^2+q^2=$ ?

Now, $x^2-2 x+y^2+2 y=1$

1 Add or subtract

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

So, center of circle $(1,-1)$ and Radius $=\sqrt{3}$

Now, find the distance from the point $(-1,1)$ to the center of the circle.

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

Now, determine the inverse point $(p, q)$ of $(-1,1)$ with respect to the circle lies on the same line passing through the point.

Here, $\left(x_1, y_1\right)=(-1,1)$

$ \begin{aligned} (a, b) & =(1,-1), \\ r^2 & =3 \end{aligned} $

Formula for inverse point

$ \begin{aligned} & p=\left(a+\left\{r^2+\left(x_1-a\right) /\left(\left(x_1-a\right)^2+\left(y_1-b\right)^2\right)\right\}\right) \\ & q=\left(b+\left\{r^2\left(y_1-b\right) /\left(\left(x_1-a\right)^2+\left(y_1-b\right)^2\right)\right\}\right) \\ & \begin{aligned} p=1 & +\frac{3 x-2}{8}=\frac{8-6}{8}=\frac{2}{8}=\frac{1}{4} \\ q & =-1+\frac{3(1+1)}{4+4} \\ & =-1+\frac{6}{8}=-\frac{1}{4} \end{aligned} \\ & p^2+q^2=\left(\frac{1}{4}\right)^2+\left(-\frac{1}{4}\right)^2=\frac{1}{8} \end{aligned} $

Mid-point $(a, b)$ of the chord. $(2 x-y+3=0)$ of the circle. $x^2+y^2+6 x-4 y+4=0$

$ \begin{aligned} & \begin{array}{l} \left(x^2+6 x\right)+\left(y^2-4 y\right)+4=0 \\ \left(x^2+9-9+6 x\right)+\left(y^2+4-4-4 y\right)+4=0 \\ \left(x^2+9+9+6 x\right)-9+\left(y^2+4-4 y\right)=0 \\ \qquad(x+3)^2+(y-2)^2=9 \end{array} \\ & \text { Center }(-3,2) \text { and Radius }=3 \end{aligned} $

The chord equation, $2 x-y+3=0$

The mid-point $(a, b)$ is perpendicular bisector of the chord which passes thought center of the circle $(-3,2)$. The equation of the perpendicular bisectord the chord can be found using the given chord equation and the center of the circle rewriting chord equation.

$ y=2 x+3 $

Slope of this line is 2 . Therefore, the slope of perpendicular bisector is $-\frac{1}{2}$

Now, center $(-3,2) m=\frac{-1}{2}$

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

The mid-point $(a, b)$ of the chord is the intersection. Point of the original chord $2 x-y \neq 3=0$ and perpendicular bisettr $y=\frac{-1}{2} x+\frac{1}{2}$.

On substitute the value,

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

On solving $x=11$ and $y=1$

So, the mid-point $(a, b)$ is $(-1,1)$.

Calculate $2 a+3 b$

$ \Rightarrow 2(-1)+3(1) \Rightarrow 1 $

Therefore, $2 a+3 b=1$.

Given, first circle,

$ x^2+y^2-6 x+4 y+9=0 $

By complete the square,

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

Center $C_1(3,-2)$, Radius $\left(R_1\right)=2$

Similarly, for second circle.

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

By complete the square,

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

Center $C_2(-1,1)$, Radius $R_2=1$

$ \begin{aligned} & \text { Distance between } C_1 \text { and } C_2 \\ & \begin{aligned} d=\sqrt{(3-(-1))^2+(-2-1)^2}=\sqrt{4^2+(-3)^2} \,\,=5\\ \end{aligned} \\ & \begin{aligned} \text { Length of tangent } & =\sqrt{d^2-\left(R_1-R_2\right)^2} \\ & =\sqrt{5^2-(2-1)^2} \\ & =\sqrt{25-1} \\ & =\sqrt{24} \\ & =2 \sqrt{6} \end{aligned} \end{aligned} $

Therefore, the length of the common tangent is $2 \sqrt{6}$ unit.