<s>[INST] $10$ circles ($2$ large of radius $R$, $6$ small of radius $r$ and 2 small of radius $t$) are enclosed in a square. How we find $r$ in terms of $t$? Let us embed $2$ large intersecting circles of radius $R$ into a square as depicted by the figure below. These two circles are highlighted green. Into these $2$ circles we embedd $6$ smaller ones of equal radius $r$ (highlighted orange).
Finally we have $2$ circles of the same radius $t$, which both touch the large (green) circles and the square.
How we can find a formula that calculates the radius $r$ of the $6$ small circles when inputting the radius $t$?

One idea to proceed might be to define a distance $d$ from the center of one of the large (green) circles to the point at which the large radius $R$ is touching one of the small (orange) circles. Then at least we would get the Pythagorean triangle equation $r^2+d^2=(R − r)^2$ as a possibly useful starting point:

 [/INST] 
The blue triangle is easy and has already been illustrated by the OP.  Without loss of generality, suppose the square is the unit square in the coordinate plane.  Then the common point of tangency of the two small circles is $(1/2,1/2)$, and the distance between this point and a center of the larger circle is simply $d_1 = \sqrt{2}(1/2 - R)^2$.  Therefore, the blue triangle corresponds to the equation $$2 (1/2 - R)^2 + r^2 = (R - r)^2.$$
Now let $d_2$ be the distance from $(1/2,1/2)$ to the point of tangency of the small circle that is outside one of the larger circles, with the diagonal line.  Then we must have the simultaneous conditions $$(d_2 + d_1)^2 + r^2 = (R + r)^2, \\ (d_2 - d_1)^2 + r^2 = (R - r)^2.$$  The solution to this system is left as an exercise for the reader; we have
$$\begin{align}
r &= \frac{10 - \sqrt{10}}{45}, \\
R &= \frac{10 - \sqrt{10}}{18}, \\
d_1 &= \frac{\sqrt{10} - 1}{9\sqrt{2}}, \\
d_2 &= \frac{2 \sqrt{5} - \sqrt{2}}{9}.
\end{align}$$
Note this uniquely fixes all of the circles in the square.  This means the two externally tangent circles with radius $t$ are also uniquely determined, and their common radius is easily shown to obey the relationship $$(1-R-t)^2 + (R-t)^2 = (R+t)^2,$$ or $$t = 1 - 2 \sqrt{R} + R = \frac{28 - \sqrt{10} - 6\sqrt{20 - 2 \sqrt{10}}}{18}.$$
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