The equation of the circle, concentric with the circle \(2 x^2+2 y^2-6 x+8 y+1=0\) and double of its area is

\(\begin{aligned}
& 2 x^2+2 y^2-6 x+8 y+1=0 \\
& \Rightarrow x^2+y^2-3 x+4 y+\frac{1}{2}=0
\end{aligned}\)
Equation of circle concentric to given circle is
\(x^2+y^2-3 x+4 y+k=0\)
Since area of required circle
\(\begin{aligned}
& =2(\text { area of given circle }) \\
& \Rightarrow \pi r_2^2=2\left(\pi r_1^2\right) \\
& \Rightarrow r_2=\sqrt{2} r_1 \\
& \Rightarrow \sqrt{\left(-\frac{3}{2}\right)^2+2^2-\mathrm{k}}=\sqrt{2} \sqrt{\left(\frac{-3}{2}\right)^2+2^2-\frac{1}{2}} \\
& \Rightarrow \sqrt{\frac{25}{4}-\mathrm{k}}=\sqrt{2} \sqrt{\frac{23}{4}}
\end{aligned}\)
\(\begin{aligned} & \Rightarrow \frac{25}{4}-\mathrm{k}=\frac{23}{2} \\ & \Rightarrow \mathrm{k}=\frac{-21}{4}\end{aligned}\)
Hence, the required equation is
\(\begin{aligned}
& x^2+y^2-3 x+4 y-\frac{21}{4}=0 \\
& \Rightarrow 4 x^2+4 y^2-12 x+16 y-21=0
\end{aligned}\)