Notes
two circles inside a square solution

Two Circles Inside a Square

Solution by Lengths in a Square

With the points labelled as above, let $r$ be the radius of the smaller circle, $R$ of the larger, and let $x$ be the side length of the square. The unknown rectangle has sides $x - 2 R$ and $x - 2 r$. The radii satisfy $\pi r^2 = 2 \pi$ and $\pi R^2 = 8 \pi$ so $r^2 = 2$ and $R^2 = 8$.

Triangle $B C D$ is an isosceles right-angled triangle since the centres of the circles lie on the diagonal of the square. The hypotenuse has length $r + R$, so the length of $B C$ is $\frac{r + R}{\sqrt{2}}$ using the relationships between lengths in a square. The side length of the square is then $r + \frac{r + R}{\sqrt{2}} + R = \frac{1 + \sqrt{2}}{\sqrt{2}}(r + R)$.

The sides of the red rectangle are therefore:

The area of this rectangle is therefore:

Since $r^2 = 2$ and $R^2 = 8$, $(r R)^2 = 16$ and so $r R = 4$. This then simplifies to $7$.