Overlapping Circles

In this example both circles have radius $r$, and each centre lies on the other circle. O and P are the centres; A and B are the intersection points.

Triangle OPA has three equal sides $r$, so it is equilateral and each angle is $\pi/3$. Triangle OPB is its reflection. The sector angle at each centre is therefore $2\pi/3$.

The overlap is the segment on one side of AB plus the matching segment on the other. Each uses the same radius and angle:

$A_{\text{overlap}}=2\left[\frac12r^2\left(\frac{2\pi}{3}-\sin\frac{2\pi}{3}\right)\right]$.

Since $\sin(2\pi/3)=\sin(\pi/3)=\sqrt3/2$, this becomes $r^2(2\pi/3-\sqrt3/2)$. These two angles add to 180° and have the same perpendicular height, hence the same sine.

Each segment needs its own radius and angle. For a crescent, the curved region inside one circle but outside the other, the two segments may lie on the same side of their common chord. Then subtract the smaller segment from the larger one.

For the overlap in the first diagram, add the segments on opposite sides of AB. For this crescent, subtract segments on the same side. Decide which region is required before calculating. If the circles do not meet, there is no common chord; if one is entirely inside the other, the overlap is the whole smaller circle.