Generalisation : n circles with radius r an an n+1th with radius R.
Of course n≥3 ! And n = 5 gives obviously r = R (the classical 6 equal circles around a 7th).
Let O' the center of the other circle with radius R.
Let A1, A2... An the centers of the n circles with radius r.
We get at once OAi = R+r, OO' = 2R and AiAi+1 = 2r
A1 and An are on the common tangent to the two circles of radius R, by symmetry (OA1 = O'A1).
The full 360░ angle is then made of copies of triangle OA1U2, with angle θ = arcsin(r/(R+r)), and of two copies of triangle OTA1, with angle φ = arccos(R/(R+r)).
We deduce then :
(n-1)arcsin(r/(R+r)) + arccos(R/(R+r)) = π
That is, with u = R/(R+r) = 1/(1+r/R), or inversely r/R = (1-u)/u :
(n-1)arcsin(1-u) + arccos(u) = π
Which can be solved numerically, using Newton method for instance,
the derivative of arccos(x) being -1/√1 - x² and that of arcsin(x) being 1/√1 - x²
A script solving for any n: