Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have two circles with radii $r$ and $R$. I need to know how far apart to draw them so that their intersection has area $A$. Basically, I'm solving for $d$ in the diagram seen in the MathWorld link.

It's not hard to solve for $A$ (see MathWorld),

$A = r^2\cos^{-1}\left(\frac{d^2+r^2-R^2}{2dr}\right) + R^2\cos^{-1}\left(\frac{d^2+R^2-r^2}{2dR}\right)-\frac{1}{2}\sqrt{(-d+r+R)(d+r-R)(d-r+R)(d+r+R)}$

but I've been struggling to find a solution for $d$.

share|cite|improve this question
You will not find a "formula" for $d$ in terms of $R$ and $r$; for sure I can't. – André Nicolas May 3 '12 at 15:56
I agree with André, there might not be an explicit formula. I tried calculating $A$ by some other method which gives a simpler result but at the price of two new intertwined parameters (two angles). Although different, this reminds me a bit of shapes called "lunules", which are like crescent moons. – Olivier Bégassat May 3 '12 at 17:00
Although there appears to be no analytical solution in terms of elementary functions, you can find code (in Python) for a numerical solution at this page – Tom Apr 23 at 8:44

Equations arising from the areas of circular segments are often not solvable with elementary functions, largely because the area of a circular segment with radius $r$ and central angle $\theta$ is $\frac{1}{2}r^2(\theta-\sin\theta)$, which involves both $\theta$ and $\sin\theta$ in a way that can't be pulled apart algebraically. It's quite likely, though, that given specific values of your parameters $r$, $R$, and $A$ it would be possible to find a numeric approximation for $d$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.