The equation of the circle concentric with the circle $x^2+y^2-6 x+12 y+15=0$ and of double its area is

The equation of the circle is $S \equiv x^2+y^2-6 x+12 y+15=0$ Let the equation of concentric circle of given circles, is $S_2=x^2+y^2-6 x+12 y+15=0$ On comparing the circle $S_1$ with, $x^2+y^2+2 g x+2 f y+c=0$ $\Rightarrow \quad g=-3, f=6, c=15$ Then, radius of circle is $=\sqrt{g^2+f^2-c}$ $=\sqrt{9+36-15}$ $=\sqrt{45-15}=\sqrt{30}$ units and centre is $(-g,-f)=(3,-6)$ Now, the area of the circle $S$ is $=\pi$ (radius) ${ }^2$ $=\pi(\sqrt{30})^2$ $=30 \pi$ Let the radius of the concentric circle is $r_2$. $r_2=\sqrt{g^2+f^2-c}$ $=\sqrt{9+36-c}$ $=\sqrt{45-c}$ Then, according to question, the area of concentric circle $=2 \times$ area of $S$ $=2 \times 30 \pi=60 \pi$ $\Rightarrow \quad \pi r_2^2=60 \pi$ $\Rightarrow \quad(\sqrt{45-c})^2=60$ $\Rightarrow \quad 45-c=60$ $\Rightarrow \quad c=-15$ Hence, the equation of concentric circle is $x^2+y^2+2(-3) x+2(6) y+(-15)=0$ $x^2+y^2-6 x+12 y-15=0$