Each outer circle is tangent to the central circle, so its center is at distance $1+r$ from the center $O$. The eight centers form a regular octagon centered at $O$ (central angles $45^\circ$).
Neighboring outer circles are tangent, so adjacent centers are $2r$ apart. A chord subtending a central angle $45^\circ$ in a circle of radius $1+r$ has length $2(1+r)\sin22.5^\circ$, so $2r=2(1+r)\sin22.5^\circ$.
Solving, $r=\dfrac{\sin22.5^\circ}{1-\sin22.5^\circ}=\dfrac1{\csc22.5^\circ-1}$. Since $\sin22.5^\circ=\dfrac{\sqrt{2-\sqrt2}}2$, we have $\csc22.5^\circ=\sqrt{4+2\sqrt2}$, so $$r=\boxed{\dfrac1{\sqrt{4+2\sqrt2}-1}}\approx0.62.$$
