Sending the point where all three outer circles meet to infinity you get two parallel lines (they only touch at infinity) and one orthogonal line since inversion preserves angles and it's clear that the two semicircles are orthogonal. Then you are left with two circles that meet all three lines and it becomes easy to figure out their exact position and radius.

Sadly that is where my knowledge of inversion stops, but if I had to I suppose I could reconstruct the equations from first principles.

If you know that the lines must go through the small circle's center, it becomes a fairly simple geometrical problem with three Pythagorean equations and three variables (x, y, r).

https://projecteuler.net/problem=143

"Anyway, I’m not going to let him Liebniz my Newton here."

I like the idea of a whole series of Sangaku puzzles done in stained glass (but, yeah, these mathematicians need to refine their stained glass skills — I suggest trying to make the next piece 2 × s). I propose that the element you are trying to find an answer for (here, the small circle here whose radius is a mystery) be the only element in color in order to draw attention to the unknown. (I can already see a small red circle floating in a sea of white shards … not unlike the Japanese flag I suppose if the flag were square.)

Let the square be a unit square with vertices at (0, 0), (1, 0), (1, 1), (0, 1), and let the center of the small circle be (x, y) and the radius be r. Each of the 3 circular arcs can be used to form an equation relating to the distance from the centers of the circular arcs to the center of the small circle:

(0.5 - x)^2 + y^2 = (0.5 + r)^2

x^2 + (y - 0.5)^2 = (0.5 - r)^2

(1 - x)^2 + y^2 = (1 - r)^2

These equations can be rewritten as:

x^2 + y^2 - r^2 = r + x

x^2 + y^2 - r^2 = y - r

x^2 + y^2 - r^2 = 2x - 2r

Equating the linear expressions on the right sides of those equations gives x = 3r and y = 5r, and substituting this into the quadratic expression gives 33r^2 = 4r.