Squares inside circle

Label the diagram as follows, and let the side of the square on the left be $a$ , let the side of the square on the right be $b$ , and let $DH = x$ .

Then by the Pythagorean Theorem from $\triangle DCG$ we have $a^2 + b^2 = 25$ , from $\triangle ABH$ we have $a^2 + (a + x)^2 = R^2$ , and from $\triangle GHF$ we have $b^2 + (b - x)^2 = R^2$ .

Then $R^2 = a^2 + (a + x)^2 = b^2 + (b - x)^2$ , which solves to $x = b - a$ (for positive values of $a$ and $b$ ).

Substituting $x = b - a$ into $a^2 + (a + x)^2 = R^2$ gives $a^2 + b^2 = R^2$ , and since $a^2 + b^2 = 25$ we have $a^2 + b^2 = R^2 = 25$ , which solves to $R = \boxed{5}$ .