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Q.
The radius of the circle in which the sphere $x^{2}+y^{2}+z^{2}+2x-2y-4z-19=0$ is cut by the plane $ x + 2y + 2z + 7 = 0$ is
Three Dimensional Geometry
Solution:
Centre of sphere $= (-1,1,2)$
Radius of sphere $=\sqrt{1+1+4+19}=5$
Perpendicular distance from centre to the plane
$OC=d=\left|\frac{-1+2+4+7}{\sqrt{1+4+4}}\right|=\frac{12}{3}=4$
$AC^{2}=AO^{2}-OC^{2}=5^{2}-4^{2}=9$
$\Rightarrow AC=3$
