question_answer

The radius of the circle\[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-2y-4z-11=0\],\[x+2y+2z-15=0\]is:

A) \[\sqrt{3}\]                                        

B) \[\sqrt{5}\]

C) \[\sqrt{7}\]                                        

D) \[3\]

Correct Answer: C

Solution :

Key Idea: Radius of a circle in a sphere \[=\sqrt{\begin{align}   & {{(Radius\,\,of\,\,sphere)}^{2}}- \\  & {{(\bot distance\,\,from\,\,origin\,\,to\,\,sphere)}^{2}} \\ \end{align}}\] Given equation of sphere is                 \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-2y-4z-11=0\] Whose centre is \[O(0,\,\,1,\,\,2)\] and radius \[OP\] is\[4\]. \[\therefore \]Distance of a plane \[x+2y+2z-15=0\]from\[(0,\,\,1,\,\,2)\]is                 \[ON=\frac{|0+2+4-15|}{\sqrt{1+4+4}}\]                        \[=\frac{9}{3}=3\] In\[\Delta ONP\],                 \[O{{P}^{2}}=O{{N}^{2}}+N{{P}^{2}}\] \[\Rightarrow \]               \[N{{P}^{2}}={{(4)}^{2}}-{{(3)}^{2}}=7\] \[\Rightarrow \]               \[NP=\sqrt{7}\]