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 concentric with the circle $x^2+y^2-6 x+12 y+15=0$ and of double its area is
$x^2+y^2-6 x+12 y-15=0$
$x^2+y^2-6 x+12 y-30=0$
$x^2+y^2-6 x+12 y-25=0$
$x^2+y^2-6 x+12 y-20=0$
Solution
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 $x^2+y^2+2(-3) x+2(6) y+(-15)=0$
$x^2+y^2-6 x+12 y-15=0$