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View attachment 4533
Let 1/2b radius of the big circle, let r radius of the smaller circle
Let $c=1/2a=1/2b−r,
b=2c+2r,
a=2c.$
So $\frac{a}{b}=\frac{2c}{2c+2r}=\frac{c}{c+r}$
Small circles respectively tangential to the large circles so
$z=c+2r,t=a−r=2c−r$
Use Pythagoras in the triangles $CXY$, $DXY$ (where $Y$ is the centre of one of the footballs) to find two expressions for $XY^2$ in terms of $a$, $b$ and $r$. Putting those expressions equal to each other will give you an equation connecting $a$, $b$ and $r$.
You already know that $r = \frac12(b-a)$ (from your equation $c = \frac12a = \frac12b-r$). Substitute that value of $r$ into your equation, and it will give you the connection between $a$ and $b$.