Suppose we are given four circles $A,B,C,D$ in the Euclidean plane having radii $r_A,r_B,r_C,r_D$ such that $r_A=r_C,r_B=r_D$ and circles $A,C$ are tangent to each other and to $B,D$ but $B,D$ are only tangent to $A,C$. Suppose further that a given bounding square $E$ in the plane is tangent to each circle exactly twice. What is the value of the ratio
$$r_A\over r_B$$
Using alternate diagonals, it is possible to identify the following:
The center of $A(C)$ is $\sqrt{2r_A^2}$ distant from the closest corner of $E$, similarly for $B(D)$, and the distance from the center of $A$ to $C$ is simply $r_A+r_C=2r_A$, so we have diagonal length $d=2\sqrt2r_A+2r_A$. It is possible to arrive at a similar expression for the other diagonal, except that we must get a good value for the distance between the centers of $B$ and $D$. Since the four circle centers form a rhombus, the alternate centers cross perpendicularly, so we have $\sqrt{(r_A+r_B)^2-r_A^2}=\sqrt{r_B^2+2r_Ar_B}$ as the distance between centers of $B$ and $D$. Then we have alternate diagonal length
$$d=2\sqrt2r_B+2\sqrt{r_B^2+2r_Ar_B}=2\sqrt2r_A+2r_A$$
Rearranging, squaring and dividing by $r_A^2$ yields
$${r_B^2\over r_A^2}-2(\sqrt2+3){r_B\over r_A}+(1+\sqrt 2)^2=0$$
This quadratic yields
$${r_B\over r_A}=\sqrt2+3\pm\sqrt{4\sqrt2+8}\approx 0.718695432327948\dots$$
It is certainly the case that the negative branch is appropriate since $r_B\le r_A$ under the conditions and picture shown. Then the original requested ratio has value
$${r_A\over r_B}=\frac 1{\sqrt2+3-2\sqrt{\sqrt2+2}}$$
which does not appear to look any better upon conjugate multiplication and simplification.