[math-fun] http://en.wikipedia.org/wiki/Steiner_chain n=4

If the four curvatures are k1, k2, k3, k4, and the bounding circles are k0 and k5, then empirically, k1 + k3 = k2 + k4 = k0 + k5 for some assignment of signs to k0 and k5. --rwg