To find the exact starting curvatures, let the outer enclosing circle have a radius of \(1\), giving it a curvature of \(k_0=-1\). The three inner circles are congruent, so let their curvatures be \(k_1=k_2=k_3=k\). Plugging this into Descartes’ Theorem, we get: \[(-1+3k)^2=2(1+3k^2) \Longleftrightarrow 3k^2-6k-1=0.\] Solve for \(k\) to get our beginning curvatures. Next, Descartes’ Theorem also shows how to find the curvature of a fourth circle if we already know the curvatures of three mutually tangent circles: \[k_4=k_1+k_2+k_3\pm2\sqrt{k_1k_2+k_2k_3+k_3k_1}.\] Finally, we just need to implement the recursion to fill in the gaps.