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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: $x = (\sqrt 2-1)r$