The equation of the concentric circle, with the circle \(C_1\) having equation \(x^2+y^2-6 x-4 y-12=0\) and having double area compared to the area of \(\mathrm{C}_1\), is

\(x^2+y^2-6 x-4 y-12=0 \)
\( \therefore \quad \left(x^2-6 x+9-9\right)+\left(y^2-4 y+4-4\right)-\) \(12=0 \)
\( \therefore \quad (x-3)^2+(y-2)^2=25 \)
\(\therefore \quad\) for circle \(\mathrm{C}_1:\) Centre is \((3,2)\) and radius \(=5\)
\(\therefore \quad\) Area of \(\mathrm{C}_1=\pi \mathrm{r}^2=25 \pi\)
Let the radius of required circle be R .
Area of required circle \(=2\left(\right.\) Area of \(\left.\mathrm{C}_1\right)\)
\(\begin{array}{ll}
\therefore & \pi R^2=2(25 \pi) \\
\therefore & R^2=50 \\
\therefore & R=5 \sqrt{2} \text { units }
\end{array}\)
\(\therefore \quad\) Equation of the required circle is \((x-3)^2+(y-2)^2=50\) i.e. \(x^2+y^2-6 x-4 y=37\)