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
$\left(x^2+y^2-4x-6y-12\right)+\lambda (x^2+y^2+6x$
$+4y-12=0$
$[$ using $S_1+\lambda S_2=0]$
$\Rightarrow x^2(1+\lambda )+y^2(1+\lambda )+x(6\lambda -4)+y(4\lambda -6)$
$-12\lambda -12=0$
$\Rightarrow x^2+y^2+\frac{x(6\lambda -4)}{1+\lambda }+\frac{y(4\lambda -6)}{1+\lambda }-\frac{12(\lambda +1)}{(\lambda +1)}=0$
$\Rightarrow x^2+y^2+\frac{x(6\lambda -4)}{(1+\lambda )}+\frac{y(4\lambda -6)}{(1+\lambda )}-12=0$
Given that, $r=\sqrt{13}$
Here, $g=\frac{3\lambda -2}{\lambda +1},f=\frac{2\lambda -3}{\lambda +1}$ and $c=-12$
Therefore, $13=g^2+f^2-c$
$\Rightarrow 13=\frac{(3\lambda -2{)}^2}{(\lambda +1{)}^2}+\frac{(2\lambda -3{)}^2}{(\lambda +1{)}^2}+12$
$\Rightarrow l3(\lambda +l{)}^2=(3\lambda -2{)}^2+(2\lambda -3{)}^2+l2(\lambda +l{)}^2$
$\Rightarrow 13\left({\lambda }^2+2\lambda +1\right)$
$=9{\lambda }^2+4-12\lambda +4{\lambda }^2+9-12\lambda +12{\lambda }^2+12+24\lambda$
$\Rightarrow 13{\lambda }^2+13+26\lambda =25{\lambda }^2+25$
$\Rightarrow 12{\lambda }^2+12-26\lambda =0$
$\Rightarrow 12{\lambda }^2-26\lambda +12=0$
$\Rightarrow 6{\lambda }^2-13\lambda +6=0$
$\Rightarrow 6{\lambda }^2-9\lambda -4\lambda +6=0$
$\Rightarrow 3\lambda (2\lambda -3)-2(2\lambda -3)=0$
$\Rightarrow (2\lambda -3)(3\lambda -2)=0$
$\Rightarrow \lambda =\frac{3}{2},\frac{2}{3}$
Hence, required equation, when $\lambda =\frac{3}{2}$
$\Rightarrow x^2\left(1+\frac{3}{2}\right)+y^2\left(1+\frac{3}{2}\right)+x\left(6\times \frac{3}{2}-4\right)$
$+y\left(4\times \frac{3}{2}-6\right)-12\times \frac{3}{2}-12=0$
$\Rightarrow \frac{5}{2}{x}^2+\frac{5}{2}{y}^2+x(5)+y(0)-30=0$
$\Rightarrow 5x^2+5y^2+10x-60=0$
$\Rightarrow x^2+y^2+2x-12=0$
and required equation when $\lambda =\frac{2}{3}$
$\Rightarrow x^2\left(1+\frac{2}{3}\right)+y^2\left(1+\frac{2}{3}\right)+x\left(6\times \frac{2}{3}-4\right)$
$+y\left(4\times \frac{2}{3}-6\right)-12\times \frac{2}{3}-12=0$
$\Rightarrow x^2\left(\frac{5}{3}\right)+y^2\left(\frac{5}{3}\right)+x(0)+y\left(\frac{8}{3}-6\right)-20=0$
$\to \frac{5x^2}{3}+\frac{5y^2}{3}-\frac{10y}{3}-20=0$
$\Rightarrow 5x^2+5y^2-10y-60=0$
$\Rightarrow x^2+y^2-2y-12=0$