Two Little Red Bombs

To simplify our calculation, we might as why assume that the radius of circle is $1$ unit and multiply the final answer by scale factor of $(\sqrt{2}-1)^2$ .

By $\textbf{Pythagorean Theorem}$ ,

$(a+b)^2=(1+a)^2+(1+b)^2$

$\Rightarrow a^2+2ab+b^2=1+2a+a^2+1+2b+b^2$

$\Rightarrow 2ab=1+2a+1+2b$

$\Rightarrow ab=1+a+b$

Let the area of rectangle $PQRS$ be $A$

$A=2(1+a)\cdot2(1+b)$

$A=4(1+a)(1+b)$

$A=4(1+a+b+ab)$

$\Rightarrow A=4(ab+ab)$

$A=8ab$

By $\textbf{AM-GM Inequality}$ ,

$\cfrac{a+b}{2} \geq \sqrt{ab}$

$a+b \geq 2\sqrt{ab}$

$a+b+1 \geq 2\sqrt{ab}+1$

$ab \geq 2\sqrt{ab}+1$

$(\sqrt{ab})^2-2\sqrt{ab}-1 \geq 0$

When $(\sqrt{ab})^2-2\sqrt{ab}-1= 0$ ,

$\sqrt{ab}=\cfrac{2 \pm \sqrt{4+4}}{2}=\cfrac{2 \pm \sqrt{8}}{2}=\cfrac{2 \pm 2\sqrt{2}}{2}=1\pm\sqrt{2}$

$\therefore (\sqrt{ab}-(1+\sqrt{2}))(\sqrt{ab}-(1-\sqrt{2}))\geq 0$

$\Rightarrow \sqrt{ab}\leq1-\sqrt{2}$ (impossible)

or

$\sqrt{ab}\geq 1+\sqrt{2}$

$\Rightarrow ab\geq (1+\sqrt{2})^2$

$\therefore A=8ab \geq 8(\sqrt{2}+1)^2$

Returning to our original question, the minimum area of rectangle $PQRS$ (reinstated) $=8(\sqrt{2}-1)^2(\sqrt{2}+1)^2=8(2-1)^2=8$

$\textbf{Clarification:}$ The equality holds if and only if $a=b$ , which asserts the fact that rectangle $PQRS$ is a square if and only if its area is minimum.