Question

The equation of a circle passing through the points of intersection of the circles $x^2 + y^2 - 4x - 6y - 12 = 0 , x^2 + y^2 + 6x + 4y - 12 = 0$ and having radius $\sqrt{13}$ is

• A. $x^2 + y^2 - 2x - 12 = 0$
• B. $x^2 + y^2 + 2y - 12 = 0$
• C. $x^2 + y^2 - 2y - 13 = 0$
• D. $x^2 + y^2 + 2x - 12 = 0$

Answer 1

Answer: (D)

The required circle is
$\left(x^{2}+y^{2}-4 x-6 y-12\right)+\lambda\left(x^{2}+y^{2}+6 x\right.$
$+4 y-12=0$
$\left[\right.$ using $\left.S_{1}+\lambda S_{2}=0\right]$
$\Rightarrow x^{2}(1+\lambda)+y^{2}(1+\lambda)+x(6 \lambda-4)+y(4 \lambda-6)$
$-12 \lambda-12=0$
$\Rightarrow x^{2}+y^{2}+\frac{x(6 \lambda-4)}{1+\lambda}+\frac{y(4 \lambda-6)}{1+\lambda}-\frac{12(\lambda+1)}{(\lambda+1)}=0$
$\Rightarrow x^{2}+y^{2}+\frac{x(6 \lambda-4)}{(1+\lambda)}+\frac{y(4 \lambda-6)}{(1+\lambda)}-12=0$
Given that, $r=\sqrt{13}$
Here, $g=\frac{3 \lambda-2}{\lambda+1}, f=\frac{2 \lambda-3}{\lambda+1}$ and $c=-12$
Therefore, $13=g^{2}+f^{2}-c$
$\Rightarrow 13=\frac{(3 \lambda-2)^{2}}{(\lambda+1)^{2}}+\frac{(2 \lambda-3)^{2}}{(\lambda+1)^{2}}+12$
$\Rightarrow l 3(\lambda+ l )^{2}=(3 \lambda-2)^{2}+(2 \lambda-3)^{2}+ l 2(\lambda+ l )^{2}$
$\Rightarrow 13\left(\lambda^{2}+2 \lambda+1\right) $
$= 9 \lambda^{2}+4-12 \lambda+4 \lambda^{2}+9-12 \lambda+12 \lambda^{2}+12+24 \lambda$
$ \Rightarrow 13 \lambda^{2}+13+26 \lambda=25 \lambda^{2}+25 $
$ \Rightarrow 12 \lambda^{2}+12-26 \lambda=0 $
$ \Rightarrow 12 \lambda^{2}-26 \lambda+12=0 $
$ \Rightarrow 6 \lambda^{2}-13 \lambda+6=0 $
$ \Rightarrow 6 \lambda^{2}-9 \lambda-4 \lambda+6=0 $
$ \Rightarrow 3 \lambda(2 \lambda-3)-2(2 \lambda-3)=0$
$ \Rightarrow (2 \lambda-3)(3 \lambda-2)=0$
$ \Rightarrow \lambda=\frac{3}{2}, \frac{2}{3}$
Hence, required equation, when $\lambda=\frac{3}{2}$
$\Rightarrow x^{2}\left(1+\frac{3}{2}\right)+y^{2}\left(1+\frac{3}{2}\right)+x\left(6 \times \frac{3}{2}-4\right) $
$+y\left(4 \times \frac{3}{2}-6\right)-12 \times \frac{3}{2}-12=0$
$ \Rightarrow \frac{5}{2} x^{2}+\frac{5}{2} y^{2}+x(5)+y(0)-30=0 $
$ \Rightarrow 5 x^{2}+5 y^{2}+10 x-60=0 $
$ \Rightarrow x^{2}+y^{2}+2 x-12=0$
and required equation when $\lambda=\frac{2}{3}$
$\Rightarrow x^{2}\left(1+\frac{2}{3}\right)+y^{2}\left(1+\frac{2}{3}\right)+x\left(6 \times \frac{2}{3}-4\right)$
$+y\left(4 \times \frac{2}{3}-6\right)-12 \times \frac{2}{3}-12=0$
$\Rightarrow x^{2}\left(\frac{5}{3}\right)+y^{2}\left(\frac{5}{3}\right)+x(0)+y\left(\frac{8}{3}-6\right)-20=0$
$\rightarrow \frac{5 x^{2}}{3}+\frac{5 y^{2}}{3}-\frac{10 y}{3}-20=0$
$\Rightarrow 5 x^{2}+5 y^{2}-10 y-60=0$
$\Rightarrow x^{2}+y^{2}-2 y-12=0$