Circle

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

Tangent at $(9,3)$

$ \begin{aligned} & T=0 \Rightarrow 9 x+3 y-5(x+9)=0 \\ & -4 x+3 y=45 \end{aligned} $

$ \begin{aligned} &\text { Normal at }(9,3)\\ &\begin{aligned} & 3 x-4 y=\lambda \Rightarrow 3 \times 9-4 \times 3=\lambda \\ & \Rightarrow \lambda=15 \Rightarrow 3 x-4 y=15 \\ & \text { Area of triangle }=\frac{1}{2} \times\left(\frac{45}{4}-5\right) \times 3=\frac{75}{8} \end{aligned} \end{aligned} $

Given, circles

$ \begin{aligned} & x^2+y^2-4=0 \text { and } \\ & (x-3)^2+(y-4)^2=49 \\ & c_1\left(0,0, r_1=2 \text { and } c_2(3,4), r_2=7\right. \\ & \text { Now, } c_1 c_2=\sqrt{9+16}=5 \\ & \left|r_1-r_2\right|=5 \\ & c_1 c_2=\left|r_1-r_2\right| \end{aligned} $

⇒ Circles touch internally.

Given, circle

$ x^2+y^2+2 x+8 y-23=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Equation of new circle touching

Eq. (i) at point $(1,2)$ is

$ \begin{gathered} x^2+y^2+2 x+8 y-23+\lambda(x-1)^2 \left.+(y-2)^2\right]=0 \\ \Rightarrow(1+\lambda) x^2+(1+\lambda) y^2+(2-2 \lambda) x +(8-4 \lambda) y-23+5 \lambda=0 \\ \Rightarrow x^2+y^2-2 x\left(\frac{\lambda-1}{\lambda+1}\right)-4 y\left(\frac{\lambda-2}{\lambda+1}\right) +\frac{5 \lambda-23}{\lambda+1}=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \text { } \end{gathered} $

$ \begin{aligned} &\text { Given that } r=\sqrt{10}\\ &\begin{aligned} & \Rightarrow\left(\frac{\lambda-1}{\lambda+1}\right)^2+4\left(\frac{\lambda-2}{\lambda+1}\right)^2-\frac{5 \lambda-23}{\lambda+1}=10 \\ & \Rightarrow \lambda^2-2 \lambda+1+4 \lambda^2-16 \lambda+16 -\left(5 \lambda^2-18 \lambda-23\right)=10(\lambda+1)^2 \\ & \Rightarrow \quad 10 \lambda^2+20 \lambda-30=0 \\ & \Rightarrow \quad \lambda^2+2 \lambda-3=0 \end{aligned} \end{aligned} $

$ \Rightarrow \quad \lambda=-3,1 $

At $\lambda=1 \Rightarrow$ Eq. (ii)becomes

$ \begin{array}{lr} \Rightarrow & x^2+y^2+2 y-9=0 \\ \Rightarrow & |a+b+c|=7 \end{array} $

At $\lambda=-3 \Rightarrow$ Eq. (ii) becomes

$ \begin{aligned} & \Rightarrow x^2+y^2-4 x-10 y+19=0 \\ & \Rightarrow \quad|a+b+c|=5 \end{aligned} $

$x^2+y^2-4 x-6 y+10=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Let the circle passing through origin centred on line $x-y=0$

$ x^2+y^2+2 g x+2 f y=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

∵ Line (i) and circle (ii) cuts orthogonally

$ \begin{aligned} & \left\{\begin{array}{l} g=f \\ \text { and } c=0 \end{array}\right. \\ & -4 g-6 g=0+10 \Rightarrow g=-1 \end{aligned} $

Now, radius of (ii) circle

$ r=\sqrt{g^2+f^2-c}=\sqrt{2} $

Diameter $=2 r=2 \sqrt{2}$

$ \begin{aligned} &\text { Equation of required circle is }\\ &\begin{aligned} & \begin{array}{l} x^2+y^2+6 x+4 y-12+\lambda \\ \quad\left(x^2+y^2-4 x-6 y-12\right)=0 \end{array} \\ & \Rightarrow x^2+y^2+\frac{(6-4 \lambda) x}{\lambda+1}+\frac{(4-6 \lambda) y}{\lambda+1} -\frac{12(\lambda+1)}{\lambda+1}=0 \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { Radius }=\sqrt{13} \\ & \begin{array}{l} \Rightarrow \sqrt{\frac{(2 \lambda-3)^2}{(\lambda+1)^2}+\frac{(3 \lambda-2)^2}{(\lambda+1)^2}+12}=\sqrt{13} \\ \Rightarrow \quad(\lambda+1)^2=13 \lambda^2+13-24 \lambda \\ \Rightarrow 12 \lambda^2-26 \lambda+12=0 \\ \Rightarrow \quad 6 \lambda^2-13 \lambda+6=0 \\ \quad \lambda=\frac{3}{2}, \frac{2}{3} \end{array} \\ & \text { At } \lambda=\frac{3}{2} \Rightarrow x^2+y^2-2 y-12=0 \\ & \text { At } \lambda=\frac{2}{3} \Rightarrow x^2+y^2+2 x-12=0 \end{aligned} $

Let, coordinates of $P=(x, y)$

So, distance of $P$ from

$ (0,2)=\sqrt{(x-0)^2+(y-2)^2} $

and distance of $P$ from

$ (-1,0)=\sqrt{(x+1)^2+(y-0)^2} $

According to the question,

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

On squaring both sides, we get

$ \begin{aligned} & 2\left(x^2+y^2-4 y+4\right)=x^2+y^2+2 x+1 \\ & \Rightarrow x^2+y^2-2 x-8 y+7=0 \end{aligned} $

which is the equation of a circle.

$ \begin{aligned} \text { Centre of circle } & =\left(\frac{-(-2)}{2}, \frac{-(-8)}{2}\right) \\ & =(1,4) \end{aligned} $

and radius $(r)=\sqrt{(1)^2+(4)^2-7}=\sqrt{10}$

Hence, a circle with centre $(1,4)$ and radius $\sqrt{10}$ units.

$(3,4),(3,2)$ and $(1,4)$

$ \begin{gathered} x=a+r \cos \theta, y=b+r \sin \theta, \\ \cos \theta=\frac{x-a}{r}, \sin \theta=\frac{y-b}{r} \\ \because \quad \sin ^2 \theta+\cos ^2 \theta=1 \\ \left(\frac{x-a}{r}\right)^2+\left(\frac{y-b}{r}\right)^2=1 \end{gathered} $

Equation of circle $(x-a)^2+(y-b)^2=r^2$

$ x^2+y^2-2 a x-2 b y+a^2+b^2=r^2 $

Putting the point $(3,4),(3,2),(1,4)$ respectively.

$ \begin{aligned} & (3,4) \rightarrow 9+16-6 a-8 b+a^2+b^2=r^2 \\ & a^2+b^2-6 a-8 b+25=r^2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & (3,2) \rightarrow 9+4-6 a-4 b+a^2+b^2=r^2 \\ & a^2+b^2-6 a-4 b+13=r^2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ & (1,4) \rightarrow 1+16-2 a-8 b+a^2+b^2=r^2 \\ & a^2+b^2-2 a-8 b+17=r^2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{aligned} $

Subtracting Eq. (ii) from Eq. (i),

$ \begin{array}{ll} & -4 b+12=0 \\ \Rightarrow & b=3 \end{array} $

Subtracting Eq. (iii) from Eq. (i),

$ \begin{array}{r} -4 a+8=0 \\\Rightarrow \,\,\,\,\,\,\,\,\, a=2 \end{array} $

Putting values of $a$ and $b$ in Eq. (i),

$ \begin{array}{lc} & 4+9-12-24+25=r^2 \\ \Rightarrow & r^2=2 \Rightarrow r=\sqrt{2} \\ \therefore & b^a r^a=3^2(\sqrt{2})^2=9 \times 2=18 \end{array} $

Given, equation of circle is

$ x^2+y^2=4 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

On differentiating Eq. (i) w.r.t. $x$ on both sides, we get

$ \begin{aligned} & 2 x+2 y \frac{d y}{d x}=0 \\ \Rightarrow \quad & 2 y \frac{d y}{d x}=-2 x \Rightarrow \frac{d y}{d x}=\frac{-x}{y} \end{aligned} $

∴ Slope of tangent to $x^2+y^2=4$ at

$P(\sqrt{3}, 1)$ is given by $\left(\frac{d y}{d x}\right)_{(\sqrt{3}, 1)}$.

So, $\left(\frac{d y}{d x}\right)_{(\sqrt{3}, 1)}=\frac{-\sqrt{3}}{1}=-\sqrt{3}$

So, slope of line $L=\frac{-1}{-\sqrt{3}}=\frac{1}{\sqrt{3}}$

$ \left[\because \text { slope of line }=\frac{-1}{d y / d x}\right] $

Let equation of line $L$ be $x-\sqrt{3} y+c=0$

As, $L$ is tangent to $(x-3)^2+y^2=1$.

So, perpendicular distance of $L$ from its centre should be equal to its radius i.e. 1. Centre $(3,0)$

$ \therefore \quad \frac{|3+c|}{\sqrt{4}}=1 \Rightarrow c=-1 \text { or }-5 $

So, equation of line $L$ is $x-\sqrt{3} y=1$ or $x-\sqrt{3} y=5$

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

$ \begin{aligned} & P(6,-5) \\ & (x-1)^2+(y+2)^2=2 \end{aligned} $

Center $(1,-2)$

Radius $=\sqrt{2}$

Equation of tangent to given circle

$ \begin{aligned} y+2 & =m(x-1) \pm \sqrt{2} \sqrt{1+m^2} \\ (\because y-k & \left.=m(x-h) \pm a \sqrt{1+m^2}\right) \end{aligned} $

$m$ is slope of the tangent.

Tangent is passing through $(6,-5)$

$ \begin{aligned} -5+2 & =m(6-1) \pm \sqrt{2} \sqrt{1+m^2} \\ -3 & =5 m \pm \sqrt{2} \sqrt{1+m^2} \\ -3-5 m & = \pm \sqrt{2} \sqrt{1+m^2} \end{aligned} $

Squaring on both sides,

$ \begin{array}{rr} & 9+25 m^2+30 m=2\left(1+m^2\right) \\ \Rightarrow & 23 m^2+30 m+7=0 \\ \Rightarrow & 23 m^2+23 m+7 m+7=0 \\ \Rightarrow & (23 m+7)(m+1)=0 \\ \Rightarrow & m_1=-1, m_2=-\frac{7}{23} \end{array} $

$ \begin{array}{ll} & \therefore \quad \tan \theta=\left|\frac{m_1-m_2}{1+m_1 m_2}\right|=\left|\frac{-1+\frac{7}{23}}{1+\frac{7}{23}}\right| \\ \Rightarrow & \tan \theta=\left|\frac{-16}{30}\right| \Rightarrow \tan \theta=\frac{16}{30}=\frac{8}{15} \\ \Rightarrow & \tan \theta=\frac{8}{15} \Rightarrow \cot \theta=\frac{15}{8} \end{array} $

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

Centre $\left(c_1\right)=(2,3)$

and $r_1=\sqrt{(2)^2+(3)^2-k}=\sqrt{13-k}$

and $x^2+y^2+8 x-4 y+11=0$

Centre $\left(c_2\right)=(-4,2)$

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

Now,

$ d=c_1 c_2=\sqrt{(2+4)^2+(3-2)^2}=\sqrt{37} $

As, we know

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

$ \begin{aligned} & \Rightarrow \cos \pi / 2=\frac{13-k+9-37}{2(\sqrt{13-k}) \times 3} \\ & \Rightarrow \quad 0=22-37-k \Rightarrow k=-15 \end{aligned} $

$(x \pm \lambda)^2+(y \pm \lambda)^2=\lambda^2$

Centres of four circles are $\left(\lambda_{,}{ }^A \lambda\right),\left(-\lambda_{,}^B, \lambda\right),\left(-\lambda_{,}{ }^C-\lambda\right)$ and $(\lambda,-\lambda)$. Radius of circles are $\lambda$.

Radius of possible circles are, $O P$ or $O Q$

$ \begin{aligned} O Q & =O A+A Q=O A+\lambda \\ \text { and } O A & =\sqrt{(\lambda-0)^2+(\lambda-0)^2} \\ & =\lambda \sqrt{2} \end{aligned} $

So, $O Q=\lambda(\sqrt{2}+1)$

and $O P=\lambda \sqrt{2}-\lambda=\lambda(\sqrt{2}-1)$

From the given options

Then, radius $=(\sqrt{2}-1) \lambda$

$ \begin{aligned} & \text { } S_1: x^2+y^2-1=0 \\ & S_2: x^2+y^2-2 x-3=0 \\ & S_3: x^2+y^2-2 y-3=0 \end{aligned} $

For radical circle,

$ \begin{aligned} & S_1-S_2=0 \\ & \Rightarrow-1+2 x+3=0 \Rightarrow x=-1 \\ & \text { and } S_2-S_3=0 \Rightarrow-2 x-3+2 y+3=0 \\ & \Rightarrow x=y=-1 \end{aligned} $

So, centre of radical circle $=C(\alpha, \beta)$

$ =(-1,-1) $

and $r_1=1, r_2=\sqrt{(1)^2-(-3)}=2$

and $r_3=\sqrt{(+1)^2-(-3)}=2$

So, $r=r_1+r_2+r_3=5$

Therefore, equation of circle

$ (x+1)^2+(y+1)^2=25 $

Equation of $H F$, which is mid-parallel to $A D$ and $B C$

$ x+y-4=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Equation of $G E$, which is mid-parallel to $A B$ and $C D$

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

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

$ x=(1 / 2) \text { and } y=(7 / 2) $

∴ Centre $=(1 / 2,7 / 2)$

$ \text { and radius }=\frac{1}{2} \times \frac{|-6-(-2)|}{\sqrt{1^2+1^2}}=\frac{2 \sqrt{2}}{2}=\sqrt{2} $

∴ Equation of circle

$ \begin{aligned} \left(x-\frac{1}{2}\right)^2+\left(y-\frac{7}{2}\right)^2=(\sqrt{2})^2 \\ \Rightarrow \quad 2 x^2+2 y^2-2 x-14 y+21 & =0 \end{aligned} $

We have,

$ S=x^2+y^2+2 g x+2 f y+c=0 $

which touch the positive $X$-axis and positive $Y$-axis.

$ \text { So, }-f=-g=r=\sqrt{g^2+f^2-c} $

and equation of circles

$ \Rightarrow \quad(x-r)^2+(y-r)^2=r^2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

$\because P(2,4)$ point lies on the circle,

$ \begin{aligned} & (2-r)^2+(4-r)^2=r^2 \\ & \Rightarrow 4+r^2-4 r+16+r^2-8 r=r^2 \\ & \Rightarrow \quad r^2-12 r+20=0 \\ & \Rightarrow \quad r=10,2 \end{aligned} $

Hence, the difference of their areas

$ \begin{aligned} & =\pi\left(r_1^2-r_2^2\right)=\pi\left(10^2-2^2\right) \\ & =\pi(100-4)=96 \pi \end{aligned} $

We have,

$ \begin{array}{r} 2 x-3 y+3=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ 2 x+y+1=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ 6 x+4 y+1=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{array} $

represent the sides of a triangle. On solving Eqs. (i) and (ii), we get Point $A\left(-\frac{3}{4}, \frac{1}{2}\right)$.

Similarly, on solving Eq. (ii) with Eq. (iii) and Eq. (i) with Eq. (iii), we get point $B\left(-\frac{3}{2}, 2\right)$ and point $C\left(\frac{8}{13}, \frac{-15}{26}\right)$ respectively.

Let $E$ and $F$ is the mid-point of $A C$ and $A B$.

$ \therefore E \equiv\left(-\frac{7}{104}, \frac{-1}{26}\right) \text { and } F \equiv\left(\frac{-9}{8}, \frac{5}{4}\right) $

Now, equation of line passing through $E$ and perpendicular to $A C$ is

$ 312 x+208 y=-29\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv) $

and equation of line passing through $F$ and perpendicular to $A B$ is

$ 8 x-16 y=-29\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(v) $

$\because A B$ and $A C$ represent the chord of the circle and we know that perpendicular line from two chords of a circle intersect at the centre of circle.

On solving Eqs. (iv) and (v), we get

$ x=\frac{-9}{8}, y=\frac{10}{8} $

So, centre of circle is $O\left(\frac{-9}{8}, \frac{10}{8}\right)$.

$ \begin{aligned} \text { Distance }(A O) & =\sqrt{\left(\frac{-9}{8}+\frac{3}{4}\right)^2+\left(\frac{10}{8}-\frac{1}{2}\right)^2} \\ & =\text { radius }(r) \\ \Rightarrow \quad r & =\frac{3}{8} \sqrt{5} \end{aligned} $

Hence, required equation of circle with centre $\left(\frac{-9}{8}, \frac{10}{8}\right)$ and radius $\frac{3}{8} \sqrt{5}$ is

$ \begin{array}{r} \quad\left(x+\frac{9}{8}\right)^2+\left(y-\frac{10}{8}\right)^2=\left(\frac{3}{8} \sqrt{5}\right)^2 \\ \Rightarrow \quad 8 x^2+8 y^2+18 x-20 y+17=0 \end{array} $

Wrong information

As, $S^{\prime}=x^2+y^2+4 x+4=0$

represent the circle whose centre is ( $-2,0$ ) and radius is zero.

52. (a) We know that two circles

$ x^2+y^2+2 g x+2 f y+c=0 $

and $x^2+y^2+2 g^{\prime} x+2 f^{\prime} y+c^{\prime}=0$ cuts orthogonally, if $2 g g^{\prime}+2 f f^{\prime}=c+c^{\prime}$

So, $2 g(-2)+2 f(-2)=4+(-4)$

$ \Rightarrow \quad g+f=0 $

We also know that if angle of intersection of two circles is $\theta$, then

$ \cos \theta=\left|\frac{r_1^2+r_2^2-d^2}{2 r_1 r_2}\right| \text {, where } d $

represents the distance between centre of these circles.

For the circle $x^2+y^2+2 g x+2 f y+4=0$, the centre is ( $-g,-f$ ) and radius $r_1$ is

$ \sqrt{g^2+f^2-4}=\sqrt{2 f^2-4} $

$ \,\,\,\, \,\,\,\, \,\,\,\, \,\,\,\, \,\,\,\, \,\,\,\, \,\,\,\, \,\,\,\, \,\,\,\, \,\,\,\, \,\,\,\, \,\,\,\, \,\,\,\, \,\,\,\, \,\,\,\, \,\,\,\, \,\,\,\, [\because g+f=0] $

For the circle $x^2+y^2+4 x+4 y+4=0$, the centre is $(-2,-2)$ and radius $r_2$ is $\sqrt{(-2)^2+(-2)^2-4}=2$

Now, $d=\sqrt{(-g+2)^2+(-f+2)^2}$

$ \begin{array}{r} \Rightarrow d^2=(g-2)^2+(f-2)^2 \\ \Rightarrow d^2=g^2+4-4 g+f^2+4-4 f=2 f^2+8 \\ \quad[\because g+f=0] \end{array} $

$ \begin{aligned} & \therefore \quad \cos 60^{\circ}=\left|\frac{\left(2 f^2-4\right)+4-\left(2 f^2+8\right)}{2 \sqrt{2 f^2-4}(2)}\right| \\ & \Rightarrow \quad \frac{1}{2}=\left|\frac{-8}{4 \sqrt{2 f^2-4}}\right| \\ & \Rightarrow \quad 2 f^2-4=16 \Rightarrow 2 f^2=20 \end{aligned} $

$ \begin{aligned}Thus,\,\, r_1 & =\sqrt{2 f^2-4} \\ & =\sqrt{20-4}=\sqrt{16}=4 \end{aligned} $

We know that two circles

$ x^2+y^2+2 g x+2 f y+c=0 $

and $x^2+y^2+2 g^{\prime} x+2 f^{\prime} y+c^{\prime}=0$ cuts orthogonally, if $2 g g^{\prime}+2 f f^{\prime}=c+c^{\prime}$

$ \begin{aligned} & \text { So, } 2 g(2)+2 f(2)=c+7 \\ & \Rightarrow \quad 4 g+4 f=c+7 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & \quad 2 g(-2)+2 f(2)=c+7 \\ & \Rightarrow \quad-4 g+4 f=c+7 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ & 2 g(-2)+2 f(-2)=c+7 \\ & \Rightarrow \quad 4 g-4 f=c+7 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{aligned} $

on solving the Eqs. (i), (ii) and (iii), we get

$ g=0, f=0 \text { and } c=-7 $

Now, the equation of circle $s=0$ becomes

$ \begin{aligned} x^2+y^2-7 & =0 \\ \Rightarrow \quad 2 x+2 y \frac{d y}{d x}-0 & =0 \Rightarrow \frac{d y}{d x}=-\frac{x}{y} \end{aligned} $

$ \begin{aligned} &\Rightarrow \quad\left(\frac{d y}{d x}\right)_{(\sqrt{3}, 2)}=-\frac{\sqrt{3}}{2}\\ &\text { ∴ The required equation be }\\ &\begin{array}{ll} & y-2=-\frac{\sqrt{3}}{2}(x-\sqrt{3}) \\ \Rightarrow & \sqrt{3} x+2 y-7=0 \end{array} \end{aligned} $

$C(2,3)$ is centre

$x+y+1=0$ is chord

As, $\angle A C B=60^{\circ}$

$\therefore \triangle A B C$ is equilateral

$ \Rightarrow A C=A B=B C=\text { radius of circle }=r(\text { let }) $

Draw $C M \perp A B$

$ \Rightarrow \quad A M=M B $

Now, $C M=\left|\frac{2+3+1}{\sqrt{1^2+1^2}}\right|=\frac{6}{\sqrt{2}}=3 \sqrt{2}$

$\triangle A C M$ is right angled triangle, then

$ \begin{aligned} A C^2 & =C M^2+A M^2 \\ r^2 & =18+\left(\frac{r}{2}\right)^2 \Rightarrow \frac{3}{4} r^2=18 \\ r^2 & =24 \end{aligned} $

Equation of circle $(x-2)^2+(y-3)^2=24$

$ \Rightarrow x^2+y^2-4 x-6 y-11=0 $

$X$-intercept $=6$

$ \begin{aligned} \Rightarrow & & 2 \sqrt{g^2-c} & =6 \\ \Rightarrow & & c & =g^2-9 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

$Y$-intercept $=8$

$ \begin{aligned} \Rightarrow & & 2 \sqrt{f^2-c} & =8 \\ \Rightarrow & & c & =f^2-16\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \end{aligned} $

From Eqs. (i) and (ii), we get

$ \begin{array}{r} f^2-g^2=7 \\& (f+g)(f-g)=7 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{array} $

$\because g x+f y=1$ passes through $(1,-1)$

Then,

$ f-g=-1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)$

From Eq. (iii), we get

$ f+g=-7 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(v)$

On solving Eqs. (iv) and (v), we get

$ f=-4, g=-3 $

From Eq. (i), we get

$ \begin{aligned} c & =0 \\ \text { Radius } & =\sqrt{g^2+f^2-c} \\ & =\sqrt{\left(-3^2\right)+(-4)^2-0}=5 \end{aligned} $

Let equation of circle be

$ x^2+y^2+2 g x+2 f y+c=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Circle passes through $(3,1)$ and $(-2,4)$

$ \begin{array}{rr} \therefore \quad 6 g+2 f+c+10=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ -4 g+8 f+c+20=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{array} $

Also, centre $(-g,-f)$ lies on the line

$ \begin{array}{ll} & x-y+1=0 \\ \therefore & -g+f+1=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)\end{array} $

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

$ \begin{array}{r} 10 g-6 f-10=0 \\&or\,\, g-\frac{3}{5} f-1=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(v)\end{array} $

or On solving Eqs. (iv) and (v), we get

$ f=0 \text { and } g=1 $

Using Eq. (ii), we get

$ c=-16 $

Equation of circle is

$ \begin{aligned} x^2+y^2+2 x-16 & =0 \\ (x+1)^2+y^2 & =(\sqrt{17})^2 \end{aligned} $

Parametric form

$ \begin{gathered} x+1=\sqrt{17} \cos \theta, y=\sqrt{17} \sin \theta \\or\,\, x=-1+\sqrt{17} \cos \theta \\ y=\sqrt{17} \sin \theta \end{gathered} $

$S \equiv x^2+y^2-8 x+10 y+5=0$

$ P(1,1) \text { and } Q(1,-1) $

Equation of polar of $P(1,1)$

$ \begin{array}{} S_1=0 \\ 3 x-6 y+6=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{array} $

Equation of chord with $Q$ as mid-point is

$ \begin{aligned} S_1 & =S_{11} \\ -3 x+4 y+7 & =0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

Solving Eqs. (i) and (ii), we have

$ x=11, y=\frac{13}{2} $

∴ Point of intersection $\left(11, \frac{13}{2}\right)$

Given, circles

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

and  $ x^2+y^2+2 x-2 y+k=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

On comparing with

$ x^2+y^2+2 g x+2 f y+c=0 $

$ \begin{aligned}and\,\, & x^2+y^2+2 g_1 x+2 f_1 y+c_1=0 \\ & g=-1, f=1, c=1 \end{aligned} $

and  $ g_1=1, f_1=-1, c_1=k $

Angle between circles (i) and (ii) is $\frac{\pi}{3}$.

$ \begin{array}{rlr} \therefore \cos \frac{\pi}{3} & =\frac{\left(c+c_1\right)-2\left(g g_1+f f_1\right)}{2 \sqrt{g^2+f^2-c} \sqrt{g_1^2+f_1^2-c_1}} \\ \Rightarrow & \frac{1}{2} =\frac{1+k-2(-1-1)}{2 \sqrt{1+1-1} \sqrt{1+1-k}} \\ \Rightarrow & \sqrt{2-k}=k+5 \\ \Rightarrow & 2-k=k^2+10 k+25 \\ \Rightarrow & k^2+11 k+23=0 \\ \Rightarrow & k =\frac{-11 \pm \sqrt{121-92}}{2} \\ & =\frac{-11 \pm \sqrt{29}}{2} \end{array} $

$\therefore k$ is an irrational number.

Given, line

$ x-y+1=0 $

and circle $x^2+y^2+2 x+2 y+1=0$

For intersection substituting $y=x+1$ in Eq. (ii), we have

$ \begin{aligned} &y=x+1 \text { in Eq. (ii), we have }\\ &\begin{array}{lr} x^2+(x+1)^2+2 x+2(x+1)+1=0 \\ x^2+x^2+2 x+1+2 x+2 x+2+1=0 \\ \Rightarrow 2 x^2+6 x+4=0 \\ \Rightarrow x^2+3 x+2=0 \\ \Rightarrow (x+1)(x+2)=0 \\ \Rightarrow x=-1, x=-2 \end{array} \end{aligned} $

∴ For $x=-1, y=0$

and For $x=-2, y=-1$

$A(-1,0)$ and $B(-2,-1)$ are diameter of a circle.

Equation of circle is

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

On comparing with

$ \begin{aligned} & x^2+y^2+2 g x+2 f y+c=0 \\ & g=\frac{3}{2}, f=\frac{1}{2}, c=1 \\ & g+f=\frac{3}{2}+\frac{1}{2}=2=2 c \end{aligned} $

The given two diameters are,

$ \begin{aligned} & x-y=2 \\ & 2 x+3 y=14 \end{aligned} $

Now, solving these equations, we get

$ x=4 \text { and } y=2 $

So, centre of circle is $(4,2)$.

Now, equation of circle is

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

where, ' $r$ ' is radius.

Since, it also passes through $(1,-2)$, then

$ \begin{array}{rlrl} (1-4)^2+(-2-2)^2 & =r^2 \\ \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, 9+16 & =r^2 \\ \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\,\,\,\, r^2 & =25 \\ \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, r & =5 \end{array} $

To find the equation of the circle whose diameter is the common chord of two given circles, we start by considering the equations of the circles:

$ S_1 = x^2 + y^2 + 2x + 3y + 1 = 0 \quad \text{(Equation 1)} $

$ S_2 = x^2 + y^2 + 4x + 3y + 2 = 0 \quad \text{(Equation 2)} $

The common chord of these two circles is also known as the radical axis, which can be found by subtracting the second equation from the first:

$ S_1 - S_2 = 0 $

This results in:

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

Simplifying this, we obtain:

$ 2x + 1 = 0 $

$ \Rightarrow x = -\frac{1}{2} $

Now, considering this line as $ S_3 = 2x + 1 = 0 $.

To find the circle that passes through the points of intersection of $ S_1 $ and $ S_3 $, its general equation is:

$ S_1 + \lambda S_3 = 0 $

Substituting the values, we have:

$ (x^2 + y^2 + 2x + 3y + 1) + \lambda(2x + 1) = 0 $

Simplifying further:

$ x^2 + y^2 + (2 + 2\lambda)x + 3y + (1 + \lambda) = 0 $

The center of this circle is $(-1 - \lambda, \frac{3}{2})$.

The line $ 2x + 1 = 0 $ will be a diameter if the center of our circle lies on this line. So, we solve:

$ 2(-1 - \lambda) + 1 = 0 $

$ -2 - 2\lambda + 1 = 0 $

$ -2\lambda - 1 = 0 $

$ \lambda = -\frac{1}{2} $

Substituting $\lambda = -\frac{1}{2}$ back into our circle equation gives:

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

$ \Rightarrow 2(x^2 + y^2) + 2x + 6y + 1 = 0 $

Therefore, the equation of the circle whose diameter is the common chord is:

$ 2x^2 + 2y^2 + 2x + 6y + 1 = 0 $

The equation of circle are

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

and $x^2+y^2+6 x-2 y+1=0$

Centre and radii of the circles are $c_1(1,3)$ and $r_1=1$

and $c_2=(-3,1)$ and $r_2=3$

Clearly,

$ \begin{aligned} & c_1 c_2=\sqrt{(1+3)^2+(3-1)^2}=\sqrt{16+4}=\sqrt{20} \\ & r_1+r_2=4=\sqrt{16} \end{aligned} $

Thus, $c_1 c_2>r_1+r_2$

Hence, the circles are non-intersecting. Thus, there will be four common tangents.

Transverse common tangents are tangents drawn from the point $P$ which divides $c_1, c_2$ internally in the ratio of radii 1: 3 coordinate of $P$ are

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

Equation of tangent through the point $P\left(0, \frac{5}{2}\right)$.

Any line through $P\left(0, \frac{5}{2}\right)$ is

$ y-\frac{5}{2}=m(x-0) \Rightarrow m x-y+\frac{5}{2}=0 $

Now, $\frac{m \cdot 1-3+\frac{5}{2}}{\sqrt{m^2+1}}=1$

$ \Rightarrow \quad\left(m-\frac{1}{2}\right)=\sqrt{m^2+1} $

$\Rightarrow m^2+\frac{1}{4}-m=m^2+1 \Rightarrow m=\frac{-3}{4}$ and $^{\prime} \infty$ '

Thus, $x=0$ is a tangent and $y-\frac{5}{2}=-\frac{3 x}{4}$

$ \Rightarrow 3 x+4 y-10=0 $

is another tangent.

Direct common tangents are tangents drawn from the point $Q$ which divides $c_1 c_2$ externally in the ratio $1: 3$.

Direct tangent, are tangent drawn from the point $Q(3,4)$.

Now, proceeding as for transverse tangents their equations are $y=4$,

$ 4 x-3 y=0 $

Thus, the number of common tangent are ' 4 '.

The given straight line,

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

and the given circle is

$ \begin{aligned} & \quad 2 x^2+2 y^2-3 x+5 y-7=0 \\ & \text { i.e. } \quad x^2+y^2-\frac{3}{2} x+\frac{5}{2} y-\frac{7}{2}=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

Let, $(h, k)$ be the pole of the given line Eq. (i) w.r.t. the circle Eq. (ii).

The equation of the polar

$ \begin{aligned} & h x+k y-\frac{3}{4}(x+h)+\frac{5}{4}(y+k)-\frac{7}{2}=0 \\ & \Rightarrow x\left(h-\frac{3}{4}\right)+y\left(k+\frac{5}{4}\right)-\frac{3}{4} h+\frac{5}{4} k-\frac{7}{2}=0 \\ & \Rightarrow x(4 h-3)+y(4 k+5)-3 h+5 k-14=0 \end{aligned} $

This equation and $9 x+y-28=0$ represent same line so,

$ \begin{aligned} \frac{4 h-3}{9} & =\frac{4 k+5}{1}=\frac{-3 h+5 k-14}{-28}=\lambda(\text { say }) \\ \Rightarrow \quad h & =\frac{9 \lambda+3}{4}, k=\frac{\lambda-5}{4},-3 h+5 k-14 \\ & =-28 \lambda \end{aligned} $

Now,

$ \begin{aligned} & -3\left(\frac{9 \lambda+3}{4}\right)+5\left(\frac{\lambda-5}{4}\right)-14=-28 \lambda \\ & \Rightarrow \quad-9-27 \lambda+5 \lambda-25-56=-112 \lambda \\ & \Rightarrow-22 \lambda-90=-112 \lambda \Rightarrow 90 \lambda=90 \\ & \Rightarrow \quad \lambda=1 \end{aligned} $

Thus, $h=\frac{3+9.1}{4}=\frac{12}{4}=3$

$ \begin{aligned} k & =\frac{1-5}{4}=\frac{-4}{4}=-1 \\ \therefore(h, k) & =(3,-1) \end{aligned} $

The given equation of circles are

$ S_1=x^2+y^2-5 x+6 y+12=0 $

and $S_2=x^2+y^2+6 x-4 y-14=0$

Now, equation of radical axis of two circle is $S_1-S_2=0$

$ \begin{array}{ll} \Rightarrow & \left(x^2+y^2-5 x+6 y+12\right) \\ & -\left(x^2+y^2+6 x-4 y-14\right)=0 \\ \Rightarrow & -11 x+10 y+26=0 \end{array} $

$ \Rightarrow \quad 11 x-10 y-26=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

So, slope of line (i) is $\frac{11}{10}\left(\right.$ say $\left.m_1\right)$.

Let slope of perpendicular line be $m_2$. Thus, $m_1 m_2=-1 \Rightarrow m_2=-\frac{10}{11}$ Now, equation of line is $y=-\frac{10}{11} x+c$ which passes through $(1,1)$

$ \Rightarrow \quad c=\frac{21}{11} $

Thus, the required equation of line is

$ \begin{aligned} y & =-\frac{10}{11} x+\frac{21}{11} \\ \Rightarrow 11 y+10 x-21 & =0 \\ \Rightarrow 10 x+11 y-21 & =0 \end{aligned} $

The given equations of circles are,

$ x^2+y^2-2 x-4 y+c=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

and $x^2+y^2-4 x-2 y+4=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

From Eq. (i), center $c_1=(1,2)$ and radius $=\sqrt{5-c}=r_1$ (say)

From Eq. (ii), centre $c_2=(2,1)$ and radius $=1=r_2$ (say)

Now, $d=c_1 c_2=\sqrt{(1-2)^2+(2-1)^2}=\sqrt{2}$

$ \therefore \quad d^2=2 $

Now, $\theta=60^{\circ}(given)$

$ \begin{aligned} & \cos \theta=\frac{r_1^2+r_2^2-d^2}{2 r_1 r_2} \\ \Rightarrow & \cos 60^{\circ}=\frac{(5-c)+1-2}{2 \sqrt{(5-c)} \times 1} \Rightarrow \frac{1}{2}=\frac{4-c}{2 \sqrt{5-c}} \\ \Rightarrow & \quad 5-c=(4-c)^2 \\ \Rightarrow & \quad 5-c=16-8 c+c^2 \\ \Rightarrow & \quad c^2-7 c+11=0 \\ \because & \quad c=\frac{7 \pm \sqrt{49-44}}{2} \Rightarrow c=\frac{7 \pm \sqrt{5}}{2} \end{aligned} $

Given, equation of circle

$ \begin{aligned} & x^2+y^2-4 x+6 y-12=0 \\ & C \equiv(2,-3) \\ & \quad l\left(C C^{\prime}\right)=\sqrt{(-3-2)^2+(2-(-3))^2} \end{aligned} $

$\begin{aligned} & =\sqrt{25+25}=5 \sqrt{2} \\ R^2 & =(5 \sqrt{2})^2+5^2=50+25=75=25 \times 3 \\ R & =5 \sqrt{3}\end{aligned}$

$\begin{aligned} \therefore \quad \frac{k+1}{2} & =\frac{1}{2} \sqrt{(h-1)^2+(k-1)^2} \\ (k+1)^2 & =(h-1)^2+(k-1)^2 \\ k^2+2 k+1 & =(h-1)^2+k^2-2 k+1 \\ (h-1)^2 & =4 k \\ \therefore \quad(x-1)^2 & =4 y\end{aligned}$

$\because \alpha<0 \Rightarrow$ centre of circle lies on 2nd quadrant

circle is touching $Y$-axis

$ \begin{array}{ll} \Rightarrow & r=|\alpha|=-\alpha \\ \text { and } & \beta=3 \end{array} $

Also, $A B=2$

$ \begin{aligned} & \text { In } \triangle P Q A \text {, } \\ & P Q=3, P A=r, A Q=1 \\ & P Q^2+A Q^2=P A^2 \\ & 9+1=r^2 \Rightarrow r=\sqrt{10} \end{aligned} $

(given)

From Eq. (i), we get $\alpha=-\sqrt{10}$

$ \text { Centre } \equiv(-\sqrt{10}, 3) $

Given equation of circle

$ x^2+y^2=4 $

Let the slope of tangent be $m$

$ \begin{aligned} y & =m(x-4) \\ \Rightarrow y-m x+4 m & =0 \end{aligned} $

Distance of $L$ from origin will be 2 .

$ \begin{aligned} & \frac{|4 m|}{\sqrt{1+m^2}} \doteq 2 \\ & \Rightarrow \quad 16 m^2=4 m^2+4 \\ & \Rightarrow \quad 12 m^2=4 \\ & \Rightarrow \quad m^2=\frac{1}{3} \\ & \Rightarrow \quad m=\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}} \\ & \sqrt{3} y= \pm(x-4) \end{aligned} $

To find image of circle in the line find the image of its centre and radius will be equal to radius of given circle.

$ \begin{aligned} \frac{x-x_1}{a} & =\frac{y-y_1}{b} \\ & =-\frac{2\left(a x_1+b y_1\right)+c}{a^2+b^2} \\ \frac{x}{1} & =\frac{y}{1}=-2 \frac{(-1)}{1+1} \\ x & =1, y=1 \end{aligned} $

Centre of image circle is $(1,1)$ and radius $=1$

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

Given that the point $ Q(1,3) $ is the inverse of $ P(-3,5) $ with respect to a circle, we need to determine the polar of point $ P $ with respect to the circle.

Since $ Q $ is the inverse of $ P $, the line $ PQ $ is related to the polar we are looking for. To find the polar line, we first determine the slope of the line segment joining the points $ P(-3,5) $ and $ Q(1,3) $.

The slope $ m_1 $ of line $ PQ $ is calculated as:

$ m_1 = \frac{5 - 3}{-3 - 1} = -\frac{1}{2} $

The polar of $ P $ will be perpendicular to this line, so its slope $ m_2 $ must satisfy the condition:

$ m_1 \cdot m_2 = -1 \quad \Rightarrow \quad -\frac{1}{2} \cdot m_2 = -1 \quad \Rightarrow \quad m_2 = 2 $

Using the slope $ m_2 = 2 $, we can write the equation of the line passing through $ Q(1,3) $ with this slope:

$ y - 3 = 2(x - 1) $

Rearrange this equation:

$ y - 3 = 2x - 2 \quad \Rightarrow \quad 2x - y + 1 = 0 $

Thus, the equation of the polar of the point $ P $ with respect to the circle is:

$ 2x - y + 1 = 0 $

Length of tangent from an external point is $\sqrt{S_1}$

$ \begin{aligned} \Rightarrow \quad P Q & =\sqrt{0+9+0+18-2} \\ & =\sqrt{25}=5 \end{aligned} $

Solving the given equations,

$\begin{aligned} x^2+(m x+c)^2 & =a^2 \\ \Rightarrow \quad x^2+m^2 x^2+c^2+2 m x c & =a^2 \\ \Rightarrow \quad\left(1+m^2\right) x^2+2 m c x+c^2-a^2 & =0 \end{aligned}$

Since, the straight line touches the circle at a point.

So, the discriminant $=0$

$\begin{aligned} & \Rightarrow(2 m c)^2-4\left(1+m^2\right)\left(c^2-a^2\right)=0 \\ & \Rightarrow 4 m^2 c^2-4 c^2-4 m^2 c^2+4 a^2+4 a^2 m^2=0 \\ & \Rightarrow-4 c^2=-4 a^2\left(1+m^2\right) \Rightarrow c^2=a^2\left(1+m^2\right) \end{aligned}$

The normal line will makes $45^{\circ}$ angle with positive $\mathrm{X}$-axis because it is inclined at $45^{\circ}$ with the straight line $y=6$ which is parallel to the $\mathrm{X}$-axis.

$\therefore$ Slope of normal line $=m=\tan 45^{\circ}=1$

The centre of the circle, $x^2+y^2=25$ is $(0,0)$.

So, the normal line will passes through $(0,0)$, because normal line at any point $P$ on the circle passes through its centre.

Thus, the equation of normal line be $y=x$ and the point $P$ lies on the straight line $y=x$.

The equation of tangent to the given circle at $(\cos \theta, \sin \theta)$ be, $x \cos \theta+y \sin \theta=1$

It meets the $X$-axis at $(\sec \theta, 0)$ and $Y$-axis at $(0, \operatorname{cosec} \theta)$ Let $R(h, k)$ be the mid-point of $P Q$.

$\begin{gathered} \therefore \quad \frac{\sec \theta+0}{2}=h \text { and } \frac{0+\operatorname{cosec} \theta}{2}=k \\ \Rightarrow \quad \frac{1}{2 h}=\cos \theta \quad \text{... (i)} \end{gathered}$

and $\frac{1}{2 k}=\sin \theta \quad \text{.... (ii)}$

On squaring and adding Eqs. (i) and (ii), we get

$ \frac{1}{4 h^2}+\frac{1}{4 k^2}=\cos ^2 \theta+\sin ^2 \theta $

$ \begin{aligned} & \Rightarrow \quad \frac{1}{4}\left(\frac{k^2+h^2}{h^2 k^2}\right)=1 \\ & \Rightarrow \quad h^2+k^2=4 h^2 k^2 \\ & \Rightarrow \quad h^2+k^2-4 h^2 k^2=0 \\ \end{aligned} $

$ \text { Hence, the required locus be } x^2+y^2-4 x^2 y^2=0 $

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

$ \Rightarrow {(x - 2)^2} + {y^2} = 1$

$AO = \sqrt {{{(OC)}^2} - {{(AC)}^2}} $

$ = \sqrt {4 - 1} = \sqrt 3 $

$\sin \theta = {1 \over 2} \Rightarrow \theta = {\pi \over 6}$

Also, $AO = BO$

Area of $\Delta OAB = {1 \over 2}\,.\,OA\,.\,OB\sin 60^\circ $

$ = {1 \over 2} \times \sqrt 3 \,.\,\sqrt 3 \,.\,{{\sqrt 3 } \over 2} = {{3\sqrt 3 } \over 4}$

Let $P(h,k)$ be the orthocentre of $\Delta$ABC

Then

$h = {{\sin t + \cos t + a} \over 3},\,k = {{ - \cos t + \sin t + b} \over 3}$

(Orthocentre coincide with centroid)

$\therefore$ ${(3h - a)^2} + {(3k - b)^2} = 2$

$\therefore$ ${\left( {h - {a \over 3}} \right)^2} + {\left( {k - {b \over 3}} \right)^2} = {2 \over 9}$

$\because$ Orthocentre lies on circle with centre $\left( {1,{1 \over 3}} \right)$

$\therefore$ $a = 3,\,b = 1$

$\therefore$ ${a^2} - {b^2} = 8$

$QR = 2r = 4$

$P = \left( {{1 \over 2} + 2\cos {\pi \over 4}, - 1 + 2\sin {\pi \over 4}} \right)$

$ = \left( {{1 \over 2} + \sqrt 2 , - 1 + \sqrt 2 } \right)$

Area of $\Delta PQR = {1 \over 2} \times 4 \times \sqrt 2 $

$ = 2\sqrt 2 $ sq. units

Equation of C₁

${x^2} + {y^2} - 4x = 0$

Intersection with

$y = 2x$

${x^2} + 4{x^2} - 4x = 0$

$5{x^2} - 4x = 0$

$ \Rightarrow x = 0,{4 \over 5}$

$y = 0,{8 \over 5}$

$A:\left( {{4 \over 5},{8 \over 5}} \right)$

Tangent of C₂ at $A\left( {{4 \over 5},{8 \over 5}} \right)$

$x + 2y = 4 \Rightarrow P:(4,0),\,Q:(0,2)$

$QA:AP = 1:4$

Circle : ${x^2} + {y^2} - 2gx + 6y - 19c = 0$

It passes through $h(6,1)$

$ \Rightarrow 36 + 1 - 12g + 6 - 19c = 0$

$ = 12g + 19c = 43$ ..... (1)

Line $x - 2cy = 8$ passes through centre

$ \Rightarrow g + 6c = 8$ ...... (2)

From (1) & (2)

$g = 2,\,c = 1$

$C:{x^2} + {y^2} - 4x + 6y - 19 = 0$

x intercept $= 2\sqrt {{g^2} - C} $

$ = 2\sqrt {4 + 19} $

$ = 2\sqrt {23} $

Abscissae of PQ are roots of ${x^2} - 4x - 6 = 0$

Ordinates of PQ are roots of ${y^2} + 2y - 7 = 0$

and PQ is diameter

$\Rightarrow$ Equation of circle is

${x^2} + {y^2} - 4x + 2y - 13 = 0$

But, given ${x^2} + {y^2} + 2ax + 2by + c = 0$

By comparison $a = - 2,b = 1,c = - 13$

$ \Rightarrow a + b - c = - 2 + 1 + 13 = 12$

${c_1}:{x^2} + {y^2} = {r^2}$ ; center = (0, 0) and radius = r

${c_2}:{(x - 1)^2} + {(y - 1)^2} = {r^2}$ ; center = (1, 1) and radius = r

${c_3}:{(x - 2)^2} + {(y - 1)^2} = {r^2}$ ; center = (2, 1) and radius = r

Distance of $y = mx + c$ line from center (0, 0) is,

$\left| {{{0 + 0 + c} \over {\sqrt {{m^2} + 1} }}} \right| = r$ ..... (1)

Distance of $y = mx + c$ line from center (1, 1) is,

$\left| {{{m - 1 + c} \over {\sqrt {{m^2} + 1} }}} \right| = r$ ..... (2)

Distance of $y = mx + c$ line from center (2, 1) is,

$\left| {{{2m - 1 + c} \over {\sqrt {{m^2} + 1} }}} \right| = r$ .... (3)

From (1) and (2), we get

$\left| {{c \over {\sqrt {1 + {m^2}} }}} \right| = \left| {{{m - 1 + c} \over {\sqrt {1 + {m^2}} }}} \right|$

$ \Rightarrow m - 1 + c = \pm \,c$ ..... (4)

taking positive sign,

$m - 1 + c = c$

$ \Rightarrow m - 1 = 0$

$ \Rightarrow m = 1$

From (2) and (3), we get

$\left| {{{m - 1 + c} \over {\sqrt {1 + {m^2}} }}} \right| = \left| {{{2m - 1 + c} \over {\sqrt {{m^2} + 1} }}} \right|$

$ \Rightarrow (m - 1 + c) = \, \pm \,(2m - 1 + c)$ ...... (5)

taking positive sign,

$m - 1 + c = 2m - 1 + c$

$ \Rightarrow m = 0$

By taking positive sign we get two different value of m so it is not acceptable.

From equation (4), taking negative sign,

$m - 1 + c = - c$

$ \Rightarrow m + 2c - 1 = 0$ ..... (6)

From equation (5), taking negative sign

$m - 1 + c = - (2m - 1 + c)$

$ \Rightarrow 3m + 2c - 2 = 0$ ..... (7)

Solving equation (6) and (7), we get

$3m + 1 - m - 2 = 0$

$ \Rightarrow 2m = 1$

$ \Rightarrow m = {1 \over 2}$

$\therefore$ $2c = 1 - {1 \over 2}$

$ \Rightarrow c = {1 \over 4}$

Putting value of $m = {1 \over 2}$ and $c = {1 \over 4}$ in equation (1), we get

$r = \left| {{{{1 \over 4}} \over {\sqrt {1 + {1 \over 4}} }}} \right|$

$ = \left| {{1 \over 4} \times {2 \over {\sqrt 5 }}} \right|$

$ = {1 \over {2\sqrt 5 }}$

$\therefore$ $20({r^2} + c)$

$ = 20\left( {{1 \over {4 \times 5}} + {1 \over 4}} \right)$

$ = 20\left( {{{1 + 5} \over {20}}} \right)$

$ = 6$

Note:

For equation of circle ${x^2} + {y^2} + 2gx + 2fy + c = 0$, center is $( - g,\, - f)$ and radius $r = \sqrt {{g^2} + {f^2} - c} $

Given,

equation of circle is

${x^2} - \sqrt 2 (x + y) + {y^2} = 0$

$ \Rightarrow {x^2} + {y^2} - \sqrt 2 x - \sqrt 2 y = 0$

$ \Rightarrow {x^2} + {y^2} + 2\left( { - {1 \over {\sqrt 2 }}} \right)x + 2\left( { - {1 \over {\sqrt 2 }}} \right) = 0$

$\therefore$ $g = - {1 \over {\sqrt 2 }}$ and $f = - {1 \over {\sqrt 2 }}$

$\therefore$ Center $ = ( - g,\, - f) = \left( {{1 \over {\sqrt 2 }},{1 \over {\sqrt 2 }}} \right)$

And Radius $ = r = \sqrt {{{\left( { - {1 \over {\sqrt 2 }}} \right)}^2} + {{\left( { - {1 \over {\sqrt 2 }}} \right)}^2} - 0} $

$ = \sqrt {{1 \over 2} + {1 \over 2}} = \sqrt 1 = 1$

As AB and AC makes an angle 90$^\circ$ then line BC passes through the center of circle and BC is the diameter of the circle.

$\therefore$ Length of BC = 2r = 2 $\times$ 1 = 2

$\therefore$ AC² = BC² $-$ AB²

= 2² $-$ ($\sqrt2$)²

= 2

$\Rightarrow$ AC = $\sqrt2$

$\therefore$ Area of right angle triangle ABC

= ${1 \over 2}$ $\times$ AC $\times$ AB

= ${1 \over 2}$ $\times$ $\sqrt2$ $\sqrt2$

= 1 square unit.

Equation of tangent at point M is

$T = 0$

$ \Rightarrow x{x_1} + y{y_1} = 2$

$ \Rightarrow - x + y = 2$

$ \Rightarrow y = x + 2$

Putting this value to equation of circle C₂,

${(x - 3)^2} + {(y - 2)^2} = 5$

$ \Rightarrow {(x - 3)^2} + {x^2} = 5$

$ \Rightarrow {x^2} - 6x + 9 + {x^2} = 5$

$ \Rightarrow 2{x^2} - 6x + 4 = 0$

$ \Rightarrow {x^2} - 3x + 2 = 0$

$ \Rightarrow (x - 2)(x - 1) = 0$

$ \Rightarrow x = 1,2$

when $x = 1,$ $y = 3$

and when $x = 2,$ $y = 4$

$\therefore$ Point $A(1,3)$ and $B(2,4)$

Now, equation of tangent at $A(1,3)$ on circle ${(x - 3)^2} + {(y - 2)^2} = 5$ or ${x^2} + {y^2} - 6x - 4y + 8 = 0$ is

$T = 0$

$x{x_1} + y{y_1} + g(x + {x_1}) + f(y + {y_1}) + C = 0$

$ \Rightarrow x + 3y - 3(x + 1) - 2(y + 3) + 8 = 0$

$ \Rightarrow x + 3y - 3x - 3 - 2y - 6 + 8 = 0$

$ \Rightarrow - 2x + y - 1 = 0$

$ \Rightarrow 2x - y + 1 = 0$ ....... (1)

Similarly tangent at $B(2,4)$ is

$2x + 4y - 3(x + 2) - 2(y + 4) + 8 = 0$

$ \Rightarrow 2x + 4y - 3x - 6 - 2y - 8 + 8 = 0$

$ \Rightarrow - x + 2y - 6 = 0$

$ \Rightarrow x - 2y + 6 = 0$ ...... (2)

Solving equation (1) and (2), we get

$x - 2(2x + 1) + 6 = 0$

$ \Rightarrow x - 4x - 2 + 6 = 0$

$ \Rightarrow - 3x + 4 = 0$

$ \Rightarrow x = {4 \over 3}$

$\therefore$ $y = 2 \times {4 \over 3} + 1 = {{11} \over 3}$

$\therefore$ Point $N = \left( {{4 \over 3},{{11} \over 3}} \right)$

Now area of the triangle ANB

$ = {1 \over 2}\left| {\matrix{ {{x_1}} & {{y_1}} & 1 \cr {{x_2}} & {{y_2}} & 1 \cr {{x_3}} & {{y_3}} & 1 \cr } } \right|$

$ = {1 \over 2}\left| {\matrix{ 1 & 3 & 1 \cr 2 & 4 & 1 \cr {{4 \over 3}} & {{{11} \over 3}} & 1 \cr } } \right|$

$ = {1 \over 2}\left[ {1\left( {4 - {{11} \over 3}} \right) - 3\left( {2 - {4 \over 3}} \right) + 1\left( {{{22} \over 3} - {{16} \over 3}} \right)} \right]$

$ = {1 \over 2}\left[ {{1 \over 3} - {6 \over 3} + {6 \over 3}} \right]$

$ = {1 \over 6}$

$\tan 2\theta = 2 \Rightarrow {{2\tan \theta } \over {1 - {{\tan }^2}\theta }} = 2$

$\tan \theta = {{\sqrt 5 - 1} \over 2}$ (as $\theta$ is acute)

Area $ = {1 \over 2}{L^2}\sin 2\theta = {1 \over 2}.\,{5 \over {{{\tan }^2}\theta }}.\,2\sin \theta \cos \theta $

$ = {{5\sin \theta \cos \theta } \over {{{\sin }^2}\theta }}.\,{\cos ^2}\theta $

$ = 5\cot \theta .\,{\cos ^2}\theta $

$ = 5.\,{2 \over {\sqrt 5 - 1}}.\,{1 \over {1 + {{\left( {{{\sqrt 5 - 1} \over 2}} \right)}^2}}}$

$ = {{10} \over {\sqrt 5 - 1}}.\,{4 \over {4 + 6 - 2\sqrt 5 }}$

$ = {{40} \over {2\sqrt 5 {{(\sqrt 5 - 1)}^2}}} = {{4\sqrt 5 } \over {6 - 2\sqrt 5 }}$

$ = {{4\sqrt 5 (6 + 2\sqrt 5 )} \over {16}}$

$ = {{\sqrt 5 (3 + \sqrt 5 )} \over 2}$