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.

Solution 1:

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.