Interesting geometry problem (square and two circles)

Let the square be of side $2a$. Clearly $a > 6$. Then with the origin at the centre of the square, the inside circle can be expressed as $(x-b)^2+y^2=(a-b)^2$

So we know that the intercept of this circle with the positive $Y$ axis must be $a-4$ and the intercept with the $-X$ axis is $6-a$. Hence $$b^2+(a-4)^2=(a-b)^2, \qquad (6-a-b)^2=(a-b)^2$$

$$\implies a \in \{3, 8\} \implies a = 8$$

So the area required is $256$.

Let $A$ denote the leftmost point (shown) on the small circle and $C$ the rightmost point shown. Let $B$ denote the "top" point shown on the small circle. Let $O$ be the center of the square.

Now let $x$ be the length of $AO$. Then the length of $BO$ is $2+x$ and the length of $OC$ is $6+x$.

An elementary theorem from geometry tells us $\angle ABC$ is $90^\circ$. It follows that triangles $\triangle ABC$ and $\triangle AOB$ are similar. By similar triangles we have $$ {6+2x\over \sqrt {x^2+(2+x)^2}}= {{\sqrt {x^2+(2+x)^2}}\over x} $$ whence $x=2$.

The side length of the square is thus $16$.

Inside the small circle are two intersecting chords. Call the lengths of their segments $h$, $h$, $\ell$ and $L$. I hope it's obvious the two $h$'s correspond to the two pieces of the vertical chord. Clearly $L=h+4=6+\ell$, since all three are the radius of the large circle. By the Intersecting Chords Theorem,

$$ h^2=\ell L=(h-2)(h+4)=h^2+2h-8$$

from which $h=4$ follows easily. This implies the large circle has radius $8$, which means the square has sides of length $16$, hence area $256$.