question_answer

Find the radius of the circle in which the sphere \[{{x}^{2}}+\text{ }{{y}^{2}}+\text{ }{{z}^{2}}+2x-2y-4z=19\]is cut by the plane\[x+2y+2z+7=0\].

A)  \[2\]               

B)  \[3\]

C)  \[1\]               

D)  \[4\]

Correct Answer: B

Solution :

Given equation of sphere is \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2x-2y-4z-19=0\] ?.(i) Its centre \[=(-1,1,2),\] radius \[=\sqrt{1+1+4+19}\] \[=\sqrt{25}=5\] and equation f plane is \[x+2y+2z+7=0\] ?.(ii) Length of the perpendicular from point \[(-1,\,1,\,2)\] to the plane (ii) is \[\left| \frac{-1+2+4+7}{\sqrt{1+4+4}} \right|=\frac{12}{3}=4\] \[\therefore \]Radius of circle                        \[=\sqrt{{{(5)}^{2}}-{{(4)}^{2}}}=\sqrt{25-16}=3\]