jmc

The triangle is isosceles, so the perpendicular bisector of its base is also an axis of symmetry, which thus passes through the center of the circle in which the triangle is inscribed: By the Pythagorean theorem, the altitude shown is $\sqrt{5^2-3^2}=4$.

Now we can draw and label a radius of the circle: The triangle shown in bold is a right triangle, so we apply the Pythagorean theorem to get the equation $$3^2+(4-r{)}^2=r^2.$$Expanding gives $$25-8r+r^2=r^2$$and thus $$25-8r=0;$$the solution is $r=\frac{25}{8}=\boxed{3\frac{1}{8}}$.