Source: rec.puzzles, citing Ingenious Mathematical Problems and Methods, L.A. Graham, Dover, 1959.
A/P = (1-(4-pi)r^2) / ( 4-(8-2pi)r )
After differentiating and setting equal to zero, we find:
r = [A/P](r) = 1/[2+sqrt(pi)]
Both the radius and the ratio have the same value.
This scales with the size of the square. So if the side length is s,
r = [A/P](r) = s/[2+sqrt(pi)]