Area Where Two Circles Overlap

Finds the area shared by two circles. The overlap is two circular segments joined along a line, so each sector has its triangle subtracted before the two are added. The length of the common chord is also shown.

Two circles that partly overlap produce a lens-shaped region. Its area is two circular segments, one cut from each circle, joined along their common chord.

S = r_1^2 \theta_1 + r_2^2 \theta_2 - \dfrac{1}{2}\sqrt{(-d + r_1 + r_2)(d + r_1 - r_2)(d - r_1 + r_2)(d + r_1 + r_2)}

\theta_1 = \arccos\dfrac{d^2 + r_1^2 - r_2^2}{2 d r_1} \qquad \theta_2 = \arccos\dfrac{d^2 + r_2^2 - r_1^2}{2 d r_2}

$r_1$ and $r_2$ are the radii
$d$ is the distance between the centres
$\theta_1$ and $\theta_2$ are the half-angles each centre subtends to the common chord

The angles come from the law of cosines. Joining the two centres to one of the intersection points forms a triangle with sides $r_1$ , $r_2$ and $d$ , and its angles are exactly $\theta_1$ and $\theta_2$ .

The square root subtracted at the end is four times the area of that same triangle, written out from Heron's formula. Adding the two sectors counts the triangle twice, and this removes the duplicate.

Example

Take the defaults: circles of radius 5 and 4 whose centres are 6 apart.

$\theta_1$ is $\arccos(45 \div 60)$ , about 0.7227, and $\theta_2$ is $\arccos(27 \div 48)$ , about 0.9728, both in radians.

Then $25 \times 0.7227$ is about 18.0683, $16 \times 0.9728$ is about 15.5647, and half the square-root term is about 19.8431. Combining them gives an overlap of about 13.7995.

The common chord is about 6.6144 long. The smaller circle has an area of about 50.2655, so roughly 27.45% of it lies in the overlap.

Three cases

Centres too far apart mean no overlap at all. That is when $d$ is at least $r_1 + r_2$ , and the area is 0. At exactly that distance the circles touch externally, which still encloses nothing.

Centres too close mean the smaller circle sits entirely inside the larger one. That is when $d$ is at most the difference of the radii, and the overlap is simply the smaller circle. There are no intersection points, so the common chord is reported as 0.

Only between those two limits does the formula above apply. It is the calculation behind questions such as how much two transmitter or sensor coverage areas overlap.