Three circles in a pentagon

Let the centers of center and left circles be $O$ and $P$ respectively, and $OR$ and $PS$ be perpendicular to $AB$ and $EA$ respectively. Let $\angle FAB = 2\theta$ . Then we note that $\tan \theta = \dfrac {OR}{AR} = \dfrac r4$ and:

$\begin{aligned} ES + SA & = EA \\ PS \cot \angle PES + PS \cot \angle PAS & = EA \\ r \cot 54^\circ + r \cot (54^\circ - \theta) & = 8 \\ \tan 36^\circ + \tan (36^\circ + \theta) & = \frac 8r = \frac 2{\tan \theta} \\ \implies \tan \theta & \approx 0.624775638 \\ r & = 4\tan \theta \approx 2.49910255 \\ \implies \lfloor 10^4r \rfloor & = \boxed{24991} \end{aligned}$