The radius of nine point circle of the triangle whose vertices are $(4, 6), (0, 4), (6, 2)$ is

$$\begin{gathered} \mathrm{Given} \mathrm{vertices} \mathrm{of} \mathrm{triangle} \mathrm{are} \\ \mathrm{let} \mathrm{A}=(4,6), \mathrm{B}=(0,4), \mathrm{C}=(6,2) \\ \mathrm{Now} \mathrm{AB}=\sqrt{16+4}=\sqrt{20}\Rightarrow {\mathrm{AB}}^2=20 \\ \mathrm{BC}=\sqrt{36+4}=\sqrt{40}\Rightarrow {\mathrm{BC}}^2=40 \\ \mathrm{CA}=\sqrt{4+16}=\sqrt{20}\Rightarrow {\mathrm{CA}}^2=20 \\ \therefore \mathrm{Triangle} \mathrm{is} \mathrm{right} \mathrm{angled} \mathrm{isosceles} \\ \therefore \mathrm{Circum} \mathrm{radius} \left(\mathrm{R}\right)=\frac{\mathrm{hypotenuse}}{2}=\frac{2\sqrt{10}}{2}=\sqrt{10} \\ \mathrm{Now} \mathrm{radius} \mathrm{of} \mathrm{nine} \mathrm{point} \mathrm{circle} = \frac{\mathrm{R}}{2} \\ =\sqrt{\frac{5}{2}} \end{gathered}$$