If you connect the radii of each circle to its tangent points with the other circles, you have a square with side lenth 2r.
The diagonal of this square is \(2r\sqrt{2}\) by the pythagorean theorem.
The diagonal is r+small circle diameter+r = \(2r\sqrt{2}\)
Subtracting both sides by 2r, the diameter of the samll circle should be \(2r\sqrt{2}-2r\) which means the radius should be \(\frac{2r\sqrt{2}-2r}{2}=\sqrt{2}r-r=r(\sqrt{2}-1)\).
