The circles are sticky

The three circles can be pictured as all sitting on a flat surface with the radius $4$ circle "sandwiched" between the two larger radius $r$ circles, which are touching each other at point $B.$ Then with the center of the leftmost larger circle being $A$ and the center of the radius $4$ circle being $C,$ we see that, by symmetry, $\Delta ABC$ is right-angled at $B$ with

$|AB| = r, |AC| = r + 4$ and $|BC| = r - 4.$

Then by Pythagoras we see that

$|AC|^{2} = |AB|^{2} + |BC|^{2} \Longrightarrow (r + 4)^{2} = r^{2} + (r - 4)^{2}$

$\Longrightarrow r^{2} + 8r + 16 = r^{2} + r^{2} - 8r + 16 \Longrightarrow r^{2} = 16r \Longrightarrow r(r - 16) = 0.$

Now as we want $r \gt 0$ we can conclude that the radius of the two congruent circles is $\boxed{16}$ cm.