smc

Let $\alpha$ be the angle that subtends the arc $AB$. By the law of cosines, $6^2=5^2+5^2-2\cdot 5\cdot 5\cos (\alpha )$ implies $\cos (\alpha )=7/25$. The half-angle formula says that $\cos (\alpha /2)=\frac{\sqrt{1+\cos (\alpha )}}{2}=\sqrt{\frac{32/25}{2}}=\sqrt{\frac{16}{25}}=\frac{4}{5}$. The law of cosines tells us $AC=\sqrt{5^2+5^2-2\cdot 5\cdot 5\cdot \frac{4}{5}}=\sqrt{50-50\frac{4}{5}}=\sqrt{10}$, which is answer choice $\boxed{\sqrt{10}}$.