The radius of the circle in which the sphere $x^2+y^2+z^2+2 x-2 y-4 z-19=0$ is cut by the plane $x+2 y+2 z+7=0$ is

• A

  4
• B

  1
• C

  2
• Correct answer
  D

  3

Solution

centre of sphere $=(-1,1,2)$ Radius of sphere $\sqrt{1+1+4+19}=5$ Perpendicular distance from centre to the plane $\mathrm{OC}=\mathrm{d}=\left|\frac{-1+2+4+7}{\sqrt{1+4+4}}\right|=\frac{12}{3}=4$ $\mathrm{AC}^2=\mathrm{AO}^2-\mathrm{OC}^2=5^2-4^2=9 \Rightarrow \mathrm{AC}=3$

Hence, the correct answer is option: 'D'.