1. Given equations

The line is $y=x+1$ and the ellipse is $\frac{x^2}{4}+\frac{y^2}{2}=1.$

The line meets the ellipse at points $P$ and $Q$.

We need the radius $r$ of the circle having $PQ$ as diameter.

Since diameter $=PQ$, we have $r=\frac{PQ}{2}.$ So, $3r=\frac{3PQ}{2} \quad \Rightarrow \quad (3r)^2=\frac{9}{4}PQ^2.$

So our main task is to find $PQ^2$.

2. Find the intersection points

Substitute $y=x+1$ into the ellipse: $\frac{x^2}{4}+\frac{(x+1)^2}{2}=1.$

Multiply by $4$: $x^2+2(x+1)^2=4.$

Expand: $x^2+2(x^2+2x+1)=4$ $x^2+2x^2+4x+2=4$ $3x^2+4x-2=0.$

Let the roots be $x_1$ and $x_2$. These correspond to the $x$-coordinates of $P$ and $Q$.

For the quadratic $3x^2+4x-2=0,$ we have $x_1+x_2=-\frac{4}{3}, \qquad x_1x_2=-\frac{2}{3}.$

Now, $y_1=x_1+1, \qquad y_2=x_2+1.$

3. Find the distance $PQ$

Since both points lie on the line $y=x+1$, the difference in $y$-coordinates equals the difference in $x$-coordinates: $y_1-y_2=(x_1+1)-(x_2+1)=x_1-x_2.$

Hence, $PQ^2=(x_1-x_2)^2+(y_1-y_2)^2=(x_1-x_2)^2+(x_1-x_2)^2=2(x_1-x_2)^2.$

Now, $(x_1-x_2)^2=(x_1+x_2)^2-4x_1x_2.$

So,