The equation of a circle concentric with the circle $x^2+y^2-6x+12y+15=0$ and having area that is twice the area of the given circle is.

For the circle $x^2+y^2-6x+125+15=0$ $\text{ Centre }C=(3,-6),\text{ radius }r=\sqrt{9+36-15}=\sqrt{30}$

Area of this circle is $\pi r^2=\pi \left(30\right)=30\pi$

Let s be the radius of required circle. Then $\text{ area }=\pi s^2\therefore \pi s^2=2\left(30\pi \right)\Rightarrow s^2=60\Rightarrow s=\sqrt{60}\text{. }$

Equation of the required circle is $(x-3{)}^2+(y+6{)}^2=(\sqrt{60}{)}^2\Rightarrow x^2+y^2-6x+12y-15=0$