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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 \(\frac{2r\sqrt{2}-2r}{2}=\sqrt{2}r-r=r(\sqrt{2}-1)\).

Let the inner circle's radius be x. We are solving for x in terms of r.

So connect the centers of two adjacent circles with radii r and the center of the small circle with radius x.

This is a isoceles right triangle with sides r+x, r+x, and 2r. 

We know that through the Pythagorean Theorem, the ratio of a side to the hypotenuse is \(1:\sqrt 2\).

So \(\sqrt 2(r+x)=2r\). Divide both sides by \(\sqrt 2\) and rationalize to get \(r+x=\sqrt 2 r\). Minus both sides by r and factor out r to get our answer: \(x = (\sqrt 2-1)r\)