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$A = 2 \pi r^2 - r^2 (2 \arccos(d/2r) - \sin(2 \arccos(d/2r)))$

Given $A$ and $r$ I would like to solve for $d$. However, I get stuck breaking the $d/2r$ out the trig functions.

For context this is the area of two overlapping circles minus the overlapping region. Given a radius and desired area I'd like to be able to determine how far apart they should be. I know $A$ should be bounded below by $0 (d = 0)$ and above by $2 \pi r^2 (d \le 2r)$.

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This is a transcendental equation for $d$ which can't be solved for $d$ in closed form. You can get rid of the trigonometric functions in the last term using

$$\sin2x=2\sin x\cos x$$


$$\sin x=\sqrt{1-\cos^2 x}$$

and thus

$$\sin\left(2\arccos\frac d{2r}\right)=\frac dr\sqrt{1-\left(\frac d{2r}\right)^2}\;,$$

but that still leaves $d$ both in the argument of the arccosine and outside.

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Thank you for the fast and thorough response, as well as the formatting help. If anyone else has thoughts on how to determine the distance between the centers of the circles given a desired radius and area I'm open to other angles of attack. Otherwise, I'll look into an algorithmic method for finding the solution. – Jeff H Nov 3 '11 at 23:37

You can simplify your algebra a little bit by working in terms of angles. If $\theta$ is the half of the central angle of the arc overlapping with the other circle, then $d=2r\cos\theta$ and your equation is further simplified into $$ A=2\pi r^2+r^2(\sin(2\theta)-2\theta). $$ Of course, this is still implicit; one cannot really solve this equation for $\theta$ without a numerical method.

In addition, you may be interested to know that there are various "tricks", ways to write an explicit solutions to transcendental equation like yours in terms of custom special functions, basically, integrals that would still need to be computed numerically. An early paper in this area by E.E. Burniston & C.E. Siewert is called "Exact analytical solution of the transcendental equation $\alpha\sin\zeta=\zeta$" and was published in SIAM J. Appl. Math., Vol.24, No.4 (1973). It can be downloaded from C.E. Siewert's web page.

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Interesting paper, thank you! The equation does look a bit like the well-studied Kepler equation... – J. M. Nov 4 '11 at 18:48
Oh, great, thank you, I did not realise this equation has a name. In my line of work equations like $x\pm\sin x=0$ occur much more often. But now that we know that this equation has a name, I am pretty sure it is worth doing a search through citations to Burniston & Siewert with a keyword "Kepler". – Aleksey Pichugin Nov 4 '11 at 19:01
Heh, you'd be surprised at the sheer amount of papers in the astronomy literature devoted to numerical/symbolic methods for solving Kepler. :) There's even a book! – J. M. Nov 4 '11 at 19:05
Sounds like a cool book to read (esp. check out p.109). I just read what Wikipedia has to say about Kepler's equation. Essentially, if we introduce $M=2\pi-A/r^2$ and $x=(6M)^{1/3}$, then solution to the equation above is given by the infinite series $$2\theta=x+\frac{x^3}{60}+\frac{x^5}{1400}+\frac{x^7}{25200}+\dots$$ (check out the Wikipedia for the general expression for this series). It does not converge very well. – Aleksey Pichugin Nov 4 '11 at 19:22
Thank you both for finding this connection to what seems to be a well-studied problem. You've rekindled my hope of finding an elegant solution! – Jeff H Nov 5 '11 at 16:32

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