Circle inscribed between quarter circles - proving its center point

$\overline{AG}$ is perpendicular to $\bigcirc O$ (at $G$) and hence $\overline{EG}$ is one of its diameters. Likewise for $\overline{DF}$. Two distinct diameters always meet at the center. Q.E.D.

Note: FareedAbiFarraj's demonstration that $\overline{OF} = \overline{OG}$ does not prove that $O$ is the center:

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As mentioned in a comment, the centers of tangent circles lie on the perpendicular to the tangent line at the point of tangency. So, in OP's diagram, $O$, $A$, $F$ are collinear, as are $O$, $B$, $G$.

enter image description here

The length of the radius of $\bigcirc O$ follows from computing the power of point $A$ with respect to that circle in two ways:

$$|AM|^2 = |AE||AF| \quad\to\quad s^2 = 2(s-r)\cdot 2s \quad\to\quad 4r = 3s \quad\square$$

It's perhaps worth noting that $\triangle AOM$ is of the $3$-$4$-$5$ variety.

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Geometry

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