Question 463746
Let s = the length of the edge of the square, and let r = radius of the smaller circle.  ==> the radius of the circle inscribed in the square is s/2.

Then {{{(s/2)^2 + (s/2)^2 = (s/2 + r + sqrt(2)*r)^2 = (s/2 + (1+sqrt(2))r)^2}}}
by the Pythagorean theorem.

<==> {{{(1+sqrt(2))^2r^2  + s(1+sqrt(2))r - s^2/4 = 0}}}, after simplification.

Applying the quadratic formula, we get

{{{r = (-s(1+sqrt(2)) +- sqrt( 2s^2(1+sqrt(2))^2 ))/(2*(1+sqrt(2) )^2) }}}.

r = {{{(s/2)((sqrt(2) - 1)/(sqrt(2) + 1))}}}, or {{{-s/2}}}.

Eliminate the second value for r.  (Why?)

From here just apply the formulas {{{A = pi*r^2}}} and {{{C = 2*pi*r}}} to get the area and circumference of the small circle, respectively.