$10$ 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:


Let the centres of the green circles be $O_1$ (lower) and $O_2$ (upper).
Let $P$ be the midpoint of $O_1O_2$ which is also the centre of the square.
Let the uppermost two orange circles above the line $O_1O_2$ have centres $C_1$ (lower) and $C_2$ (upper).
Let $Q$ be the point where the circle centre $C_2$ touches the line $O_1O_2$.
Now if we write $O_1P=d$, then $O_2P=d$, and since $\triangle O_1C_1P\equiv\triangle O_2C_2Q$, then $O_2Q=d\implies O_1Q=3d$.
Applying Pythagoras to $\triangle O_1C_1P\implies (R-r)^2=d^2+r^2\implies d^2=R^2-2Rr$
Applying Pythagoras to $\triangle O_1C_2Q\implies (R+r)^2=9d^2+r^2\implies 9d^2=R^2+2Rr$
Solving these gives $$8R^2=20Rr\implies R=\frac{5r}{2}$$
From this we can get $$d=\frac15R^2\implies d=\frac{R}{\sqrt{5}}\implies d=\frac{r\sqrt{5}}{2}$$
Now let $B$ be the centre of the lower blue circle of radius $t$, and let $BP=y$.
Applying Pythagoras in $\triangle PC_1B\implies (R+t)^2=d^2+y^2$
$$\implies(R+t)^2-\frac{R^2}{5}=y^2$$
$$\implies \frac45R^2+2Rt+t^2=y^2$$
Hence, $$y^2=5r^2+5rt+t^2$$
Now consider the lengths of the diagonals of the square: we have $$2(y+t\sqrt{2})=2d+2R\sqrt{2}$$
$$\implies y+t\sqrt{2}=d+\frac52r\sqrt{2}=r\left(\frac{\sqrt{5}+5\sqrt{2}}{2}\right)$$
Substituting for $y$ we now have
$$5r^2+5rt+t^2=\left[r\left(\frac{\sqrt{5}+5\sqrt{2}}{2}\right)-t\sqrt{2}\right]^2$$
So for a given value of $t$ you can obtain $r$ by solving a quadratic equation:
$$r^2\left(\frac{35+10\sqrt{10}}{4}\right)-rt(15+\sqrt{10})+t^2=0$$
