Question

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

• A. $2$
• B. $3$
• C. $4$
• D. $1$

Answer 1

The correct option is B $3$

Centre and radius of the given sphere are $(-1,1,2)$ and $\sqrt{1^2+1^2+2^2+19}=5$ respectively
Now the perpendicular distance between centre of sphere to the given plane is given by,
$d=\left|\frac{-1+2\left(1\right)+2\left(2\right)+7}{\sqrt{1^2+2^2+2^2}}\right|=4$
Therefore radius of the circle in which the given sphere and plane cut is
$=\sqrt{5^2-4^2}=3$