A geometry problem by Nahom Assefa

Note that the center of the quadrant (corner of the square), center of the circle and their point they are tangent to each other is a straight line. The center of the circle is at the midpoint of the square base. Let the radius of the circle be $r$ . By Pythagorean theorem , we have:

$\begin{aligned} r^2 + \left(\frac 12\right)^2 & = (1-r)^2 \\ r^2 + \frac 14 & = 1 - 2r + r^2 \\ 2r & = 1 - \frac 14 \\ \implies r & = \boxed{\frac 38} \end{aligned}$

Let the radius of the smallest circle be $r$ . Then $(1-r)^2=r^2+\left (\dfrac{1}{2}\right) ^2\implies 2r=1-\dfrac{1}{4}=\dfrac{3}{4}\implies r=\boxed {\dfrac {3}{8}}$