More Circles and Lines

Let the center of the bottom circle be $O$ ; $OP$ , $OQ$ , and $JR$ be perpendicular $DA$ , $AB$ , and $LB$ respectively; and $\angle LBA = \theta$ . Then we have:

$\begin{aligned} AQ + QB & = AB \\ PO + OQ \cdot \cot \frac \theta 2 & = s & \small \blue{\text{Let }t = \tan \frac \theta 2} \\ 1 + \frac 1t & = s \end{aligned}$

Note that $JR = 2$ and

$\begin{aligned} JB \cdot \cos \theta & = JR \\ (BC - JC) \cos \theta & = 2 \\ (s - s \tan \theta) \cos \theta & = 2 \\ s(\cos \theta - \sin \theta) & = 2 \\ \left(1+\frac 1t\right)\left(\frac {1-t^2 - 2t}{1+t^2}\right) & = 2 \\ -t^3 - 3t^2 - t + 1 & = 2t + 2t^3 \\ 3t^3 + 3t^2 + 3t - 1 & = 0 \\ \implies t & = \sqrt[3]{\frac 8{27}+\frac {\sqrt 8}9} + \sqrt[3]{\frac 8{27}-\frac {\sqrt 8}9} - \frac 13 \\ & \approx 0.253076587 \\ s & = 1 + \frac 1t \approx 4.951373036 \\ \implies \lfloor 10^4s \rfloor & = \boxed{49513} \end{aligned}$

Reference: Cardano's method

$2a=\angle ABL$

$2b=\angle FBL$

Wolframalpha solution

Let s be the side of the square. {2 s - x - √((s - x)^2 + s^2) = 2, (s - x)/s =√(x^2 - 4)/2, A = (10^4)(s)