$A = 2 \pi r^2 - r^2 (2 \arccos(d/2r) - \sin(2 \arccos(d/2r)))$
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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$\begingroup$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
$$\sin2x=2\sin x\cos x$$
and
$$\sin x=\sqrt{1-\cos^2 x}$$
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.
$\endgroup$ 1 $\begingroup$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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