Why these easy stuff?

$x^2 + y^2 = a^2$ is the equation of a circle centered at the origin with radius $a$ . WLOG let's say that the sides of the rectangle are parallel to the axes and that point $P$ in the first quadrant has coordinates $(x,y)$ , but since it's on the cicle, $y = \sqrt{a^2-x^2}$ , so $P = \left( x, \sqrt{a^2-x^2} \right)$

The axes split the rectangle into four congruent rectangles. To maximize the total area, we have to maximize the area of each of these rectangles, which is width times height, or $A = x \sqrt{a^2-x^2}$ . To do that, we set the derivative equal to $0$ and solve for $x$ .

$A^\prime = x \cdot \frac {-x}{\sqrt{a^2-x^2}} + \sqrt{a^2-x^2} \cdot 1 = \frac {-x^2+a^2-x^2}{\sqrt{a^2-x^2}} = \frac {a^2-2x^2}{\sqrt{a^2-x^2}} \stackrel{set}{=} 0$

$\frac {a^2-2x^2}{\sqrt{a^2-x^2}} = 0 \Leftrightarrow a^2-2x^2 = 0 \Leftrightarrow x^2 = \frac {a^2} 2 \Leftrightarrow x = \frac {\sqrt{2}a} 2$

Since $P$ lies on the cicle, we can now calculate its y-coordinate

$y = \sqrt{a^2-x^2} = \sqrt{a^2-\frac {a^2} 2} = \sqrt{\frac{a^2} 2} = \frac {\sqrt{2}a} 2 = x$

The area of the whole rectangle (that is actually a square) is $2x \cdot 2y$ .

$A = 2x \cdot 2y = 2 \frac {\sqrt{2}a} 2 \cdot 2 \frac {\sqrt{2}a} 2 = \boxed{2a^2}$