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

The given circle is $x^2-y^2-6x+12y+15=0$

Equation of any circle, concentric with the above circle is $x^2+y^2-6x+12y+k=0$

The area of the given circle is 

      $\begin{array}{rcl}A& =& \mathrm{\pi }\left({\mathrm{r}}^2\right)\\ & =& \mathrm{\pi }\left(9+36-15\right)\\ & =& 30\mathrm{\pi }\end{array}$

SInce, the area of the required circle is double of its area, hence the area of the required circle is $60\mathrm{\pi }$

Hence, 

     $\begin{array}{rcl}60\mathrm{\pi }& =& \mathrm{\pi }\left(9+36-\mathrm{k}\right)\\ 45-\mathrm{k}& =& 60\\ \mathrm{k}& =& -15\end{array}$

Therefore, the equation of  the required circle is $\boxed{{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{-}\mathbf{6}\mathbf{x}\mathbf{+}\mathbf{12}\mathbf{y}\mathbf{-}\mathbf{15}\mathbf{=}\mathbf{0}}$

(or)

$$\begin{gathered} {\mathrm{x}}^2+{\mathrm{y}}^{\mathrm{e}}-6\mathrm{x}+12\mathrm{y}+15=0 \\ \mathrm{c}=(3, -6) \\ \mathrm{r}=\sqrt{9+36-15} \\ \mathrm{r}=\sqrt{30} \\ \sqrt{9+36-\mathrm{c}} \\ {\mathrm{\pi R}}^2=2\mathrm{\pi }\left(30\right) \\ {\mathrm{R}}^2=60 \\ By verification circle is \\ {\mathrm{x}}^2+{\mathrm{y}}^2-6\mathrm{x}+12\mathrm{y}-15=0 \end{gathered}$$