The line $y=x+1$ meet the ellipse $\frac{x^2}{4}+\frac{y^2}{2}=1$ at two points P and Q. If r is the radius of the circle with PQ as diameter then $(3r{)}^2$ is equal to

Ellipse $x^2+2y^2=4$

Line y=x+1

Point of intersection

$$\begin{gathered} x^2+2(x+1{)}^2=4 \\ 3x^2+4x-2=0 \\ \left|x_1-x_2\right|=\frac{\sqrt{∆}}{\left|a\right|}=\frac{\sqrt{40}}{3} \\ AB=2r=\left|x_1-x_2\right|\sqrt{1+m^2} \end{gathered}$$

m is slope of given line

$$\begin{gathered} AB=\frac{\sqrt{40}}{3}\sqrt{1+1} \\ 2r=\frac{\sqrt{80}}{3}\Rightarrow r=\frac{\sqrt{80}}{6} \\ (3r{)}^2={\left(3\times \frac{\sqrt{80}}{6}\right)}^2=\frac{80}{4}=20 \end{gathered}$$