[Math] distance between centres of two overlapping congruent circles

Question

If there are two overlapping congruent circles such that the area of intersection is 10% of the area of each circle, what is the distance between their centres in terms of the radius r cm?

Answer 1

Let $O$ and $O'$ be the centers and $P$ and $Q$ be the intersection points of both circles.

First, let's calculate the area of the circular segment $PQ$. We need the angle $\alpha=\angle POQ$.

Call $d=OO'/2$ and $r$ the radius of the circles. Then $$\alpha=2\arccos\frac dr$$

The area of the circular segment is obtained substracting the triangle $OPQ$ to the circular sector $OPQ$, and is $$\frac{\alpha r^2}{2}-d\sqrt{r^2-d^2}$$ being $\alpha$ expressed in radians (this makes the formulae clearer).

The area of intersection is twice the circular segment's. Therefore, you must solve this for $d$:

$$r^2\arccos\frac dr-d\sqrt{r^2-d^2}=0.05\pi r^2$$

I'd say that this is impossible to solve by algebraic methods, but any math program can do it with numerical methods, given the radius.

The equation becomes a bit cleaner if you call $k=d/r$ the ratio between $d$ and $r$. Dividing the equation by $r^2$ yields: $$\arccos k-k\sqrt{1-k^2}=0.05\pi$$ Sadly it still can't be solved by algebraic methods.

PS: Don't forget that the arccos must be in radians, and that $d$ is the distance between centers halved.