Two Circles and a Square: Find the area of the square

Let the side length of the square be $2a$ . Consider the two intersecting chords in the smaller circle. By intersecting chords theorem ,

$\begin{aligned} (a-12)a & = (a-8)^2 \\ a^2 - 12 a & = a^2 - 16a + 64 \\ 4a & = 64 \\ a & = 16 \end{aligned}$

Therefore, the area of the square is $(2a)^2 = \boxed{1024}$ .

Coordinate geometry seems the easiest way to go: put an origin at the centre of the diagram, and let the large circle have radius $R$ and the small one radius $r$ . From the diagram, the centre of the smaller circle is at $(R-r,0)$ so its equation is $(x-R+r)^2+y^2=r^2$ .

This circle intersects the negative $x$ -axis at $(R-2r,0)$ , so that $R-2r=-R+12$

At the intersection with the positive $y$ -axis we have $(R-r)^2+(R-8)^2=r^2$

From the first equation, we get $R-r=6$ . The second equation then becomes $6^2+(R-8)^2=r^2$

Rearranging, we have $R^2-r^2=16R-100$

Dividing by $R-r=6$ gives $R+r=\frac16 (16R-100)$

Adding $R-r=6$ again we get $2R=6+\frac16 (16R-100)$

which is easily solved to find $R=16$ , so the side of the square is $32$ and its area is $\boxed{1024}$ .

Let radius of larger circle $= R$ , radius of smaller circle $= r$

$\begin{aligned} 2R - 2r &= 12 \\ \implies R &= r+6 \\ \\ (R-r)^2 + (R-8)^2 &= r^2 \\6^2 + (r-2)^2 &= r^2 \\ \\ r&= 10 \\ \implies R &= 16 \end{aligned}$

Side of the square $= 2R = 32$ . Area of square $= 32^2 = \boxed{1024}$

Let the quarter's side length be x+12, then the right triangle inside the smaller circle will have a height of x+4 (that's x+12-8) with a left base x and a right base x+12. Using similar right triangle,
x/(x+4) = (x+4)/(x+12).
x(x+12) = (x+4)².
x = 4.
Answer.
= [2(x+12)]².
= 1024