smc

Draw lines from points $A$ and $G$ to the center of the circle $K$. As shown from the diagram, $r$ is the radius of the circumscribed circle. Set $\overline{EK}$ to $x$, so as $\overline{EF}$ is $1+1=2$, $\overline{FK}$ is $2-x$. Then, $△FGK$ and $△AEK$ are both right, so an equation can be formed with the Pythagorean Theorem. The radii (hypotenuses) are equal, so the following equation can be made and solved: $$\begin{array}{rl}{\frac{1}{2}}^2+(2-x{)}^2& =1^2+x^2\\ \frac{1}{4}+(-x+2{)}^2& =1+x^2\\ \frac{1}{4}+x^2-4x+4& =1+x^2\\ \frac{1}{4}-4x+4& =1\\ -4x& =-3-\frac{1}{4}\\ 4x& =\frac{13}{4}\\ x& =\frac{13}{16}\end{array}$$ $x$ is solved, so plugging it in to the original formula yields: $$\begin{array}{r}\sqrt{1^2+{\frac{13}{16}}^2}\\ =\sqrt{1+\frac{169}{256}}\\ =\sqrt{\frac{425}{256}}\\ =\frac{\sqrt{425}}{\sqrt{256}}\\ =\frac{5\sqrt{17}}{16}\\ =\boxed{\frac{5\sqrt{17}}{16}}.\end{array}$$