What is the smallest circle such that an arbitrary set of circles can be placed on the circumference without overlapping? I have a set of circles of arbitrary radii: $r_1, r_2, r_3, ... r_n$. 
I wish to arrange them around an inner circle so that they are all touching the perimeter of the inner circle, and do not overlap each other.
What I don't know how to do is figure out the inner radius $r_{inner}$. 
I can figure out the angle each circle will use given an inner radius: $\theta_i  = 2\sin^{-1} \frac{r_i}{r_i+r_{inner}}$, so I can test whether an inner radius is correct.
My first guess was to solve $2\pi = \sum_i (2\sin^{-1} \frac{r_i}{r_i+r_{inner}})$ for $r_{inner}$, but that's beyond my skills. 
I also considered whether the sum of the diameters would equal the circumference of the circle, but that's a set of line segments rather than a smooth arc, and correcting that is also beyond me.
How do I figure out $r_{inner}$?
Numbers of circles around a circle is related, but assumes the circles are identical, which mine are not.
 A: Using the cosine rule, the angle at the centre of the inner circle between lines through the centres of two adjacent touching circles is given by
$$
\varphi(i,{i+1}) \;=\;
\cos^{-1}\left(\frac{R^2+R r_i+R r_{i+1}-r_i r_{i+1}}{(R+r_i) (R+r_{i+1})}\right)
$$
where I've used $R$ instead of $r_{inner}$ for simplicity.
So, we need 
$$
\qquad\qquad\qquad
\qquad\qquad\qquad
\sum\limits_{i=1}^n\varphi(i,{i+1})\;=\;2\pi
\qquad\qquad\qquad
\qquad\qquad\qquad(1)
$$
where we consider $r_{n+1}=r_1$.
Using trigonometric identities, for a given $n$, it is possible to convert this into a very complex expression full of square roots, but there seems to be no easy way of solving the result in general (although for $n=3$ and $r_i=1,2,3$, we somewhat surprisingly get $R=\frac{6}{23}$).
Numerical solution is probably the best approach. (The diagram below was generated using Mathematica to solve numerically.)
Moreover, although this is a necessary condition, it is not sufficient in general because a sequence of small circles between two large circles may not fit tightly, causing the solution above to give an incorrect result (with the large circles overlapping):
$\qquad\qquad\qquad\qquad$ 
So for all $i\neq j$, for the above solution to be correct we also need 
$$
\varphi(i,j) \;\geqslant\; \sum_{k=i}^{j-1}\varphi(k,{k+1})
$$
with appropriate wrapping around of indexes if $i>j$. If this does not hold for some $i$ and $j$, then the correct solution for $R$ is obtained from $(1)$ by discarding circles $i+1,\ldots,j-1$. Due to the possibility of nested overlapping, the testing and discarding needs to be done iteratively, considering increasing values of $|i-j|$ in order.
A: In the following diagram, $R$ is the radius of the center circle and $r_1$ and $r_2$ are two of the surrounding circles.
$\hspace{3cm}$
The Law of Cosines says
$$
(r_1+r_2)^2=(R+r_1)^2+(R+r_2)^2-2(R+r_1)(R+r_2)\cos(\varphi_{12})\tag{1}
$$
which implies that
$$
\begin{align}
\cos(\varphi_{12})
&=\frac{(R+r_1)^2+(R+r_2)^2-(r_1+r_2)^2}{2(R+r_1)(R+r_2)}\\[6pt]
&=\frac{(R+r_1)(R+r_2)-2r_1r_2}{(R+r_1)(R+r_2)}\\[6pt]
&=1-2\frac{r_1}{R+r_1}\frac{r_2}{R+r_2}\tag{2}
\end{align}
$$
We need to find $R$ so that
$$
\pi=\sum_{i=1}^n\cos^{-1}\left(1-2\frac{r_i}{R+r_i}\frac{r_{i+1}}{R+r_{i+1}}\right)\tag{3}
$$
where $r_{n+1}=r_1$. To this end, define
$$
f(x)=\pi-\sum_{i=1}^n\cos^{-1}\left(1-2\frac{r_i}{x+r_i}\frac{r_{i+1}}{x+r_{i+1}}\right)\tag{4}
$$
and then
$$
f'(x)=\sum_{i=1}^n\frac{\frac1{x+r_i}+\frac1{x+r_{i+1}}}
{\sqrt{\frac{x+r_i}{r_i}\frac{x+r_{i+1}}{r_{i+1}}-1}}\tag{5}
$$
So we can use Newton's method to get the recursion
$$
x_{k+1}=x_k-
\frac{\displaystyle\pi-\sum_{i=1}^n\cos^{-1}\left(1-2\frac{r_i}{x_k+r_i}\frac{r_{i+1}}{x_k+r_{i+1}}\right)}
{\displaystyle\sum_{i=1}^n\frac{\frac1{x_k+r_i}+\frac1{x_k+r_{i+1}}}
{\sqrt{\frac{x_k+r_i}{r_i}\frac{x_k+r_{i+1}}{r_{i+1}}-1}}}
$$
