question_answer

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

Correct Answer: B

Solution :

[b] Center of the sphere is \[\left( -1,\text{ }1,\text{ }2 \right)\] and its radius\[=\sqrt{1+1+4+19}=5\] CL, perpendicular distance of C from plane, is \[\left| \frac{-1+2+4+7}{\sqrt{1+4+4}} \right|=4\] Now \[A{{L}^{2}}=C{{A}^{2}}-C{{L}^{2}}=25-16=9\] Hence, radius of the circle\[=\sqrt{9}=3\]