What is the area of intersection of two unit circles whose centers are $d$ units apart ($1 \leq d \leq 2$)?
Similar cases are discussed at Expected area of the intersection of two circles Get the size of an area defined by 2 overlapping circles and https://mathworld.wolfram.com/Circle-CircleIntersection.html . I get a different formula (as below); I'm not sure if I've made an error, or if somehow my formula is the same, at least in the case I'm considering.
Solution: $$A = 4 \left [\frac \theta 2 – \sin 2 \theta \right ]$$ where $$\theta = \arccos {\frac d 2}.$$
Consider the diagram below:
The circles have centers $P,Q$ and points of intersection $X,X'$. The midpoint of $XX'$ (which is also the midpoint of $PQ$) is $M$. Line $PM$ intersects circle $Q$ at $S$. Let $\theta = \angle XPM = \arccos \frac d 2$.
By symmetry, the area of intersection is $4$ times the area bounded by $XMS$, and
$$\begin{align*}
A_{XMS} &= A_{\text{sector } XSP} – A_{\triangle XMP} \\
A_{\text{sector } XSP} &= \frac \theta 2 \\
A_{\triangle XMP} &= \frac 1 2 (PM) (MX) \\
&= \frac 1 2 \sin \theta \cos \theta \\
&= \sin 2 \theta.
\end{align*}$$
Therefore, the area of intersection $A = 4 \left [\frac \theta 2 – \sin 2 \theta \right ]$.
Is this correct? If not, where is the error? Will this formula be valid for $d < 1$?
Best Answer
The double angle formula for sine is:
$$\sin(2θ)=2\sin(θ)\cos(θ)$$ rather than $\frac12\sin(θ)\cos(θ)$ like you've written.
Hence in the end you have:
$$A=4\left(\frac{θ}{2}-\frac{\sin(2θ)}{4}\right)$$
So apart from this small mistake, the rest of your calculations are fine, and the final (corrected) formula agrees with the formula provided in the Wolfram link you posted, if you substitute $R=r=1$, and use the fact that $d=2\cos(θ)$.
And, as you pointed out, you can make the formula look a bit cleaner by setting $φ=2θ=2\operatorname{arccos\left(\frac{d}{2}\right)}$ to get:
$$A=φ-\sin(φ)$$