Quadrants touching a circle in a square

Nice, fun problem!

Let $r$ be the radius of the red circle with center $C$ . Consider the right triangle with its vertices at $C$ , the left lower corner, and the midpoint of the lower edge. Pythagoras gives us $(32+r)^2=16^2+(32-r)^2$ , which solves to $r=\boxed{2}$ .

Let s be the side of the square and r be the radius of the red circle,

Then, since the red circle and the orange quadrant are circles which share the same tangent,

(s/2)^2 = (s + r)^2 - (s-r)^2. = 4s*r

Solving, we get

r = s/16,

In our case, since s = 32, r = 2

same analysis

(32+r)^2 = (32-r)^2 + 16^2

1024 + 64r + r^2 = 1024 - 64r + r^2 + 256

128r = 256

r=2