Overlapping area between a circle and a square

Question

I have a circle and a square. They are aligned to their center. The radius of the given circle is less then half the value of diameter of the square.

How to find the overlapped area?

Answer 1

Let $r$ be the radius of the circle and $a$ be half the length of a side of the square.

If $r \leq a$ then the overlapping area is the area of the whole circle $\pi r^2$.

If $r \geq a \sqrt{2}$ then the overlapping area is the area of the whole square $a^2$.

Now to the more interesting case where $a < r < a\sqrt{2}$.

Exploiting the symmetry of the problem we divide the plane to 8 parts as shown in the picture, so that the overlapping area is 8 times the colored area.

Let $\theta = \arccos(a/r)$ the angle shown in the picture. The blue area is a circular sector of angle $\frac{\pi}{4}-\theta$ so its area equals $(\frac{\pi}{4}-\theta)\frac{r^2}{2}$. The red triangle has area $\frac{1}{2}ar\sin\theta$.

Putting all these together, the total overlapping area is $(\pi-4\theta)r^2+4ar\sin\theta$.