Circle

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.

Given, 1st circle

$ =x^2+y^2-4 x-4 y+7=0 $

2nd circle $=x^2+y^2+4 x-4 y+6=0$

3rd circle $=x^2+y^2+4 x+4 y+5=0$

Now, find the center and radii of the given circles.

1st circle $x^2+y^2-4 x-4 y+7=0$

By complete the square

$ \begin{aligned} & (x-2)^2+(y-2)^2=2^2+2^2-7 \\ & (x-2)^2+(y-2)^2=1 \end{aligned} $

Center $C_1(2,2)$ Radius $R_1=1$

Similarly, 2nd Circle $(x+2)^2+(y-2)^2=2$

Center $C_2(-2,2)$ and Radius $R_2=\sqrt{2}$

3rd circle $(x+2)^2+(y+2)^2=3$

$ C_3(-2,-2), R_3=\sqrt{3} $

Now, let the Radius $R$ and center $(a, b)$ be the circle which cut each of given circles orthogonally, then the below condition must hold.

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

On subtract Eq. (i) from Eq. (ii), we get

$ a=\frac{1}{8} $

On subtracting Eq. (i) from Eq. (iii), we get

$ b=\frac{1}{8} $

On substitute $a$ and $b$ in Eq. (i)

$ \begin{aligned} &\left(\frac{1}{8}-2\right)^2+\left(\frac{1}{8}-2\right)^2=R^2+1 \\ &\left(\frac{-15}{8}\right)^2+\left(\frac{-15}{8}\right)^2=R^2+1 \\ & \frac{450}{64}=R^2+1 \\ & R^2=\frac{193}{32} \\ & R=\sqrt{\frac{193}{32}} \\ & R=\frac{\sqrt{193}}{4 \sqrt{2}} \end{aligned} $

Therefore, the radius of the circle that cuts all the given circle orthogonally is $\frac{\sqrt{193}}{4 \sqrt{2}}$ unit.

In $\triangle A P B$, we have

$\begin{aligned} & m_1=\text { Slope of } A P=\frac{y-3}{x-2} \\ & m_2=\text { Slope of } B P=\frac{y-1}{x+1}\end{aligned}$

Now, $\tan \theta=\left|\frac{m_1-m_2}{1+m_1 m_2}\right|$

$\begin{aligned} & \tan 90^{\circ}=\left|\frac{\frac{(y-3)}{(x-2)}-\frac{(y-1)}{(x+1)}}{1+\frac{(y-3)}{(x-2)} \frac{(y-1)}{(x+1)}}\right| \\ & \quad=\left|\frac{(y-3)(x+1)-(y-1)(x-2)}{(x-2)(x+1)+(y-3)(y-1)}\right| \\ & \quad \frac{1}{\infty}=0=\frac{(x-2)(x+1)+(y-3)(y-1)}{(y-3)(x+1)-(y-1)(x-2)} \\ & \Rightarrow x^2-x-2+y^2-4 y+3=0 \\ & \Rightarrow x^2-x+y^2-4 y+1=0\end{aligned}$

Given the equation of the circle:

$ x^2 + y^2 - 6x - 8y - 11 = 0 $

we can determine the center of the circle, $ C $, which is $ (3, 4) $, and the radius, $ r $, as follows:

$ r = \sqrt{9 + 16 + 11} = \sqrt{36} = 6 $

Now, let's consider the point $ P(15, 9) $.

The distance between the point $ P $ and the center $ C $ is calculated as:

$ CP = \sqrt{(15 - 3)^2 + (9 - 4)^2} = \sqrt{144 + 25} = \sqrt{169} = 13 $

To find the largest possible distance from the point $ P $ to any point on the circle, we add the radius to $ CP $:

$ \text{Greatest distance from } P \text{ to the circle} = CP + r = 13 + 6 = 19 $

Given the circle equation:

$ x^2 + y^2 - 8x - 12y + \alpha = 0 $

And the point $ P(6, 6) $ is an interior point of the circle. To determine the range of $\alpha$, we perform the following steps:

Point $ P(6, 6) $ lies inside the circle, thus:

$ 6^2 + 6^2 - 8 \cdot 6 - 12 \cdot 6 + \alpha < 0 $

$ 36 + 36 - 48 - 72 + \alpha < 0 $

Simplifying:

$ 72 - 48 - 72 + \alpha < 0 $

$ \alpha < 48 $

To ensure the circle does not touch the coordinate axes and remains in the first quadrant, the radius must be less than both the x-coordinate and y-coordinate of the center. The center of the circle $(h, k)$ is found by completing the square:

$ h = \frac{8}{2} = 4, \quad k = \frac{12}{2} = 6 $

Thus, the center is at $(4, 6)$ and the radius $ \sqrt{4^2 + 6^2 - \alpha} < 4 $ and also $ < 6 $:

$ \sqrt{52 - \alpha} < 4 \quad \text{and} \quad \sqrt{52 - \alpha} < 6 $

Solving these gives:

$ 52 - \alpha < 16 \quad \text{and} \quad 52 - \alpha < 36 $

Which simplifies to:

$ \alpha > 36 $

Combining the inequalities from steps 1 and 2, we have:

$ 36 < \alpha < 48 $

To find the equation of the circle whose diameter is the common chord of the circles given by:

$ x^2 + y^2 - 6x - 7 = 0 \quad (i) $

$ x^2 + y^2 - 10x + 16 = 0 \quad (ii) $

First, we find the common chord by subtracting equation (i) from equation (ii):

$ \begin{aligned} & \quad x^2 + y^2 - 10x + 16 - (x^2 + y^2 - 6x - 7) = 0, \\ & \Rightarrow -4x + 23 = 0, \\ & \Rightarrow x = \frac{23}{4}. \end{aligned} $

Substitute $ x = \frac{23}{4} $ back into either equation (let’s use equation (ii)) to find the $ y $-coordinates of the intersection points:

$ \begin{aligned} & x^2 + y^2 - 10x = -16, \\ & \left(\frac{23}{4}\right)^2 + y^2 - 10 \cdot \frac{23}{4} = -16, \\ & y = \frac{\pm 3 \sqrt{15}}{4}. \end{aligned} $

The points of intersection are $\left(\frac{23}{4}, \frac{-3 \sqrt{15}}{4}\right)$ and $\left(\frac{23}{4}, \frac{3 \sqrt{15}}{4}\right)$.

The midpoint of the chord (diameter) is:

$ \left(\frac{23}{4}, 0\right) $

The radius is the distance from this midpoint to either endpoint:

$ \begin{aligned} d &= \sqrt{\left(\frac{23}{4} - \frac{23}{4}\right)^2 + \left(0 - \frac{3 \sqrt{15}}{4}\right)^2} \\ &= \frac{3 \sqrt{15}}{4} \end{aligned} $

Thus, the equation of the circle is:

$ \begin{aligned} & \left(x - \frac{23}{4}\right)^2 + y^2 = \left(\frac{3 \sqrt{15}}{4}\right)^2, \\ & \Rightarrow \left(x - \frac{23}{4}\right)^2 + y^2 = \frac{135}{16}, \\ & \Rightarrow 16\left(x^2 - \frac{23}{2}x\right) + 16y^2 = 135, \\ & \Rightarrow 8x^2 + 8y^2 - 92x + 197 = 0. \end{aligned} $

This is the equation of the circle whose diameter is the common chord of the given circles.

Let mid-point be $P(h, k)$

and origin $O(0,0)$ is also the centre of the circle.

Since, chord making right angle at origin and $P$ is mid-point, $O P$ will bisect the right angle.

$O P=r \cos 45^{\circ}$, where $r$ is radius of given circle.

$ \begin{array}{ll} \Rightarrow & \sqrt{(h-0)^2+(k-0)^2}=5 \times \frac{1}{\sqrt{2}} \\ \Rightarrow & h^2+k^2=\left(\frac{5}{\sqrt{2}}\right)^2=\frac{25}{2} \\ \therefore & \quad O P=\frac{5}{\sqrt{2}} \end{array} $

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

$ O P=\sqrt{x_1^2+y_1^2}=\frac{5}{\sqrt{2}} $

Let $S_1: x^2+y^2+2 x+3 y+1=0$

$ \begin{aligned} & S_2: x^2+y^2+x-y+3=0 \\ & S_3: x^2+y^2-3 x+2 y+5=0 \end{aligned} $

The radical axis of the circles will be

$ \begin{array}{l} & S_1-S_2=0 \\ \Rightarrow & x+4 y-2=0 \quad \ldots \text { (i) } \\ \text { and } & S_2-S_3=0 \\ & 4 x-3 y-2=0 \quad \ldots \text { (ii) } \\ \text { and } & S_3-S_1=0 \\ \Rightarrow & -5 x-y+4=0 \quad \ldots \text { (iii) } \end{array} $

On multiply Eqs. (i) by 4 and then subtract from (ii), we get

$ y=\frac{6}{19} $

On substitute this value in Eq. (i), we get

$ x=\frac{14}{19} $

Since, $x$ and $y$ values also satisfies Eqs. (iii).

Thus, the radical centre of three circles is $\left(\frac{14}{19}, \frac{6}{19}\right)$.

In, figure $B A$ and $B C$ are tangents $y$ the circle. So, $O B$ is the angle bisecton of $\angle B$.

$ \text { Then, } \begin{aligned} \angle B & =60^{\circ} \\ \angle O B D & =\frac{60}{2}=30^{\circ} \end{aligned}$

$ \text { In } \triangle O B D, \angle O D B=90^{\circ} $

$ \begin{aligned} \tan 30^{\circ} & =\frac{O D}{B D} \\ \frac{1}{3} & =\frac{r}{\frac{a}{2}}=\frac{2 r}{a} \\ r & =\frac{a}{2 \sqrt{3}} \end{aligned} $

Now, diagonal of square $=$ diameter of circle

$ =2 r=2 \times \frac{a}{2 \sqrt{3}}=\frac{a}{\sqrt{3}} $

Now, area of square $=\frac{1}{2}(2 r)^2$

$ =\frac{1}{2} \times\left(\frac{a}{\sqrt{3}}\right)^2=\frac{a^2}{6} $

Slope of $A P=\frac{k}{h-1}$

Slope of $B P=\frac{k-1}{h}$

$ \begin{aligned} & \tan 45^{\circ}=\left|\frac{\frac{k}{h-1}-\frac{k-1}{h}}{1+\frac{k(k-1)}{h(h-1)}}\right| \\ & \pm 1=\frac{k h-(k-1)(h-1)}{h(h-1)+k(k-1)} \\ & \pm 1=\frac{k h-k h+h+k-1}{h^2-h+k^2-k} \\ & \Rightarrow \pm\left(h^2-h+k^2-k\right)=h+k-1 \end{aligned} $

If, $h+k-1=-\left(h^2-h+k^2-k\right)$,

$\Rightarrow h+k-1=-h^2+h-k^2+k$ $h^2+k^2-1$

If $h+k-1=h^2-h+k^2-12$

$ \Rightarrow h^2+k^2-2 h-2 k+1=0 $

Equation of locus of $P$ is $\left(x^2+y^2-1\right)=0$

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

Hence, equation of locus of $p$ is

$ \begin{aligned} & \left(x^2+y^2-1\right)\left(x^2+y^2-2 x-2 y+1\right)=0 \\ & x \neq 0,1 \end{aligned} $

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

Centre of circle lies on $2 x+y+3=0$

$ 2 h+k+3=0 $

The two tangents $3 x+4 y-18=0$ and $3 x+4 y+2=0$ are parallel as their slopes are equal

$ 2 r=\frac{|2+18|}{\sqrt{3^2+4^2}}=\frac{20}{5}=4 $

$ \therefore \quad r=2 \text { units } $

Now, $r=\frac{|3 h+4 k-18|}{\sqrt{3^2+4^2}}=2$

$ \begin{array}{ll} \Rightarrow & 3 h+4 k-18= \pm 10 \\ & 3 h+4 k-18+10=0 \\ \text { or } & 3 h+4 k-18-10=0 \\ \Rightarrow & 3 h+4 k-8=0 \\ \text { Or } & 3 h+4 k-28=0 \end{array} $

From Eqs. (i) and (ii), we get $h=-4$, $k=5$

From Eqs. (i) and (iii), we get $h=-8$, $k=13$

Neglecting $h=-8$ and $k=13$

because distance from centre to tangent $\neq 2$

$\therefore$ Centre of circle is $(-4,5)$

$\therefore$ Equation of circle is

$ \begin{array}{r} (x+4)^2+(y-5)^2=2^2 \\ x^2+y^2+8 x-10 y+16+25=4 \\ x^2+y^2+8 x-10 y+37=0 \end{array} $

The power of the point $(4, 2)$ with respect to the given circle is 9.

To find the circle's equation, use the power of the point:

$ 16 + 4 - 8\alpha + 12 + \alpha^2 - 16 = 9 $

Solving this equation gives:

$ \alpha^2 - 8\alpha + 7 = 0 $

which results in:

$ \alpha = 1 \quad \text{or} \quad \alpha = 7 $

Thus, the possible equations of the circles are:

$x^2 + y^2 - 2x + 6y - 15 = 0$

$x^2 + y^2 - 14x + 6y + 33 = 0$

For the equation $x^2 + y^2 - 2x + 6y - 15 = 0$:

Length of intercepts on the X-axis:

$ 2\sqrt{g^2 - c} = 2\sqrt{1 + 15} = 2\sqrt{16} = 8 $

Length of intercepts on the Y-axis:

$ 2\sqrt{f^2 - c} = 2\sqrt{9 + 15} = 2\sqrt{24} = 4\sqrt{6} $

For the equation $x^2 + y^2 - 14x + 6y + 33 = 0$:

Length of intercept on the X-axis:

$ 2\sqrt{g^2 - c} = 2\sqrt{49 - 33} = 2\sqrt{16} = 8 $

Length of intercept on the Y-axis is not possible (as the term under the square root is negative).

Therefore, the sum of the lengths of all possible intercepts is:

$ 8 + 8 + 4\sqrt{6} = 16 + 4\sqrt{6} $