2008 AMC 12B Problem 9

Let $O$ be the center and $D$ the point where $\overline{OC}$ meets $\overline{AB}.$ Since $C$ is the midpoint of arc $AB,$ $\overline{OC}$ is the perpendicular bisector of the chord, so $AD = 3.$ In right triangle $ADO,$ $OD = \sqrt{5^2 - 3^2} = 4,$ so $DC = OC - OD = 5 - 4 = 1.$ Then in right triangle $ADC,$ $AC = \sqrt{AD^2 + DC^2}$ $= \sqrt{3^2 + 1^2}$ $= \sqrt{10}.$ Thus, the correct answer is A.