a way to find the overlap of 2 circles(not same radius) with $x$ distance apart

Question

There are 2 circles, with radii $R_1$ and $R_2$ and they are $x$ distance apart. I want to find the area of the overlap without using al-Kashi law of cosines.

Answer 1

I put together a small diagram in Paint to illustrate the method.

Note that the area between the circles is the sum of the area of the two sectors, minus the area of the two triangles.

So, the triangle which consists of the two radii, and the line segment between the centers (not drawn) has side lengths $R_1, R_2, x$.

So, call the angle opposite $R_1$ $\frac12\theta_2$, and the angle opposite $R_2$ $\frac12\theta_1$.

Also, call $$\psi_1=\frac{R_1^2+x^2-R_2^2}{2R_1x},\;\psi_2=\frac{R_2^2+x^2-R_1^2}{2R_2x}$$

So, we have $\theta_1=2\cos^{-1}\left(\frac{R_1^2+x^2-R_2^2}{2R_1x}\right)=2\cos^{-1}(\psi_1)$, and likewise, $\theta_2=2\cos^{-1}\left(\frac{R_2^2+x^2-R_1^2}{2R_2x}\right)=2\cos^{-1}(\psi_2)$

Note that the area of a sector of a circle is $\frac\theta2r^2$.

So, the area of both sectors is $\frac{\theta_1}2R_1^2+\frac{\theta_2}2R_2^2$, which is $$R_1^2\cos^{-1}(\psi_1)+R_2^2\cos^{-1}(\psi_2)$$

Now, let's compute $\sin\theta_1$ and $\sin\theta_2$.

By the relation $\sin^2\theta+\cos^2\theta=1$, we have $\sin\frac{\theta_1}2=\sqrt{1-\psi_1^2}$, $\sin\frac{\theta_2}2=\sqrt{1-\psi_2^2}$.

So, $\sin\theta_1=2\psi_1\sqrt{1-\psi_1^2}$, $\sin\theta_2=2\psi_2\sqrt{1-\psi_2^2}$.

So, the area of the two triangles is $\frac12R_1^2\sin\theta_1+\frac12R_2^2\sin\theta_2$, which is equal to $$R_1^2\psi_1\sqrt{1-\psi_1^2}+R_2^2\psi_2\sqrt{1-\psi_2^2}$$

So, the answer is $$R_1^2\cdot\left(\cos^{-1}(\psi_1)-\psi_1\sqrt{1-\psi_1^2}\right)+R_2^2\cdot\left(\cos^{-1}(\psi_2)-\psi_2\sqrt{1-\psi_2^2}\right)$$