Closed form solution for a simple-looking Euclidean geometry problem? 
The diagram above is distilled from an engineering problem I'm working on. The goal is to compute $R$ in terms of $r$ and $\theta$, which are known.
The best I have been able to get so far is an equation that combines all $3$ variables, and is therefore numerically solvable for $R$. However it does not seem that it can be rearranged in closed form to isolate $R$. When I posed it to Mathematica, it returned a hideous monstrosity.
I am wondering if there is a more elegant geometrical approach that I'm missing to isolate $R$ in terms of $r$ and $\theta$.

Here is my work so far:
If we knew $\alpha$, the problem would be trivial. Drop the red perpendicular and observe that:
$$
\cos\alpha=\frac{R-r}{R},
$$
which can be rearranged to give:
$$
R = \frac{r}{1-\cos\alpha}.
$$
We do not know $\alpha$ directly, but we know that $\alpha = \theta-\beta$, and we can compute $\beta$ in terms of $R$ and $r$ from the isosceles triangle:
$$
\beta = 2\arcsin\left(\frac{r}{2R}\right)
$$
Now we plug our new expression for $\alpha$ into the above expression for $R$ and we get a nasty-looking equation with $R$ on both sides:
$$
R = \frac{r}{1-\cos\left(\theta - 2\arcsin\left(\frac{r}{2R}\right) \right)}.
$$

Edit: Numerical Solutions
Here is some data from numerical solutions. I have plotted $R$ vs $r$ with $\theta=\pi/3$ and $R$ vs $\theta$ with $r=1$.
It seems pretty clear that:


*

*$R$ and $r$ have a directly proportional relationship for fixed $\theta$

*$R$ is inversely proportional to $\theta$ for fixed $R$ (a log-log plot looks straight)




 A: I got that,
$0=R^2\sin(\frac{\theta}{2})^2-R(\frac{\sqrt{2r}}{2})^2+2\sqrt{R}\cos(\frac{\theta}{2})(\frac{\sqrt{2r}}{2})(\frac{r}{2})-(\frac{r}{2})^2$
This is a quartic polynmial from which one could in theory get the solution in a closed form by looking at the roots, I hope you could see if you agree with it.
First you need to notice that $\sin(\frac{\alpha}{2})=\frac{\sqrt{2Rr}}{2R},\ \sin(\frac{\beta}{2})=\frac{r}{2R}$, I add a new image to see if it helps visualize it, from there the rest is just calculations.

$\cos(\frac{\theta}{2})=\cos(\frac{\alpha+\beta}{2})$
$\cos(\frac{\theta}{2})=-\sin(\frac{\alpha}{2})\sin(\frac{\beta}{2})+\cos(\frac{\alpha}{2})\cos(\frac{\beta}{2})$
$\cos(\frac{\theta}{2})=-\sin(\frac{\alpha}{2})\sin(\frac{\beta}{2})+\sqrt{\cos(\frac{\alpha}{2})^2\cos(\frac{\beta}{2})^2}$
$\cos(\frac{\theta}{2})=-\sin(\frac{\alpha}{2})\sin(\frac{\beta}{2})+\sqrt{(1-\sin(\frac{\alpha}{2})^2)(1-\sin(\frac{\beta}{2})^2)}$
$\cos(\frac{\theta}{2})=(\frac{\sqrt{2Rr}}{2R})(\frac{r}{2R})+\sqrt{(1-(\frac{\sqrt{2Rr}}{2R})^2)(1-(\frac{r}{2R})^2)}$
$(\cos(\frac{\theta}{2})-(\frac{\sqrt{2Rr}}{2R})(\frac{r}{2R}))^2=(1-(\frac{\sqrt{2Rr}}{2R})^2)(1-(\frac{r}{2R})^2)$
$\cos(\frac{\theta}{2})^2-2\cos(\frac{\theta}{2})(\frac{\sqrt{2Rr}}{2R})(\frac{r}{2R})+(\frac{\sqrt{2Rr}}{2R})^2(\frac{r}{2R})^2=1+(\frac{\sqrt{2Rr}}{2R})^2(\frac{r}{2R})^2-(\frac{\sqrt{2Rr}}{2R})^2-(\frac{r}{2R})^2$
$\cos(\frac{\theta}{2})^2-2\cos(\frac{\theta}{2})(\frac{\sqrt{2Rr}}{2R})(\frac{r}{2R})=1-(\frac{\sqrt{2Rr}}{2R})^2-(\frac{r}{2R})^2$
$0=1-\cos(\frac{\theta}{2})^2-(\frac{\sqrt{2Rr}}{2R})^2-(\frac{r}{2R})^2+2\cos(\frac{\theta}{2})(\frac{\sqrt{2Rr}}{2R})(\frac{r}{2R})$
$0=\sin(\frac{\theta}{2})^2-(\frac{\sqrt{2Rr}}{2R})^2-(\frac{r}{2R})^2+2\cos(\frac{\theta}{2})(\frac{\sqrt{2Rr}}{2R})(\frac{r}{2R})$
$0=R^2\sin(\frac{\theta}{2})^2-(\frac{\sqrt{2Rr}}{2})^2-(\frac{r}{2})^2+2\cos(\frac{\theta}{2})(\frac{\sqrt{2Rr}}{2})(\frac{r}{2})$
$0=R^2\sin(\frac{\theta}{2})^2-R(\frac{\sqrt{2r}}{2})^2+2\sqrt{R}\cos(\frac{\theta}{2})(\frac{\sqrt{2r}}{2})(\frac{r}{2})-(\frac{r}{2})^2$
A: Probably there isn't a nice closed form.
On substituting $z = \frac{r}{2R}$, we get the final expression:
$ \cos (\theta - 2\arcsin(z)) = 1 - 2z $
A query on WolframAlpha, solve cos(x - 2arcsin(z)) = 1 - 2z for z, generates an ugly closed form for $z$ .
