If the inscribed square has sides of 8in, the diameter of the circle is equal to the diagonal of the square.
d^2=x^2+x^2
d^2=2x^2
d=√(2x^2)
Since d=2r, r=d/2 so
r=(1/2)√(2x^2)
r=√((2x^2)/4)
r=√(x^2/2), since x=8
r=√(64/2)
r=√32
r=√(16*2)
r=4√2 in (exact)
r≈5.66 in (to nearest hundredth of an inch)