Do H là trung điểm MN \(\Rightarrow OH\perp MN\)
\(MH=\dfrac{1}{2}MN=5\left(cm\right)\Rightarrow OH=\sqrt{OM^2-MH^2}=\sqrt{R^2-MH^2}=\sqrt{39}\left(cm\right)\)
\(\Rightarrow CH=OC-OH=R-OH=8-\sqrt{39}\) (cm)
\(\Rightarrow MC=\sqrt{MH^2+CH^2}\approx5,3\left(cm\right)\)