The equation of the concentric circle, with the circle $C_1$ having equation $x^2+y^2-6 x-4 y-12=0$ and…
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
$x^2+y^2-6 x-4 y=27$
$x^2+y^2-6 x-4 y=13$
$x^2+y^2-6 x-4 y=50$
$x^2+y^2-6 x-4 y=37$
Solution
$\begin{array}{ll}
& x^2+y^2-6 x-4 y-12=0 \\
\therefore \quad & \left(x^2-6 x+9-9\right)+\left(y^2-4 y+4-4\right)-12=0 \\
\therefore \quad & (x-3)^2+(y-2)^2=25
\end{array}$
$\therefore \quad$ 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. $x^2+y^2-6 x-4 y=37$