question_answer

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

A)  \[\sqrt{3}\]                                       

B)                   \[\sqrt{5}\]                       

C)                   \[\sqrt{7}\]                                       

D)                   \[\sqrt{11}\]

Correct Answer: C

Solution :

The centre of the sphere \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-2y-4x-11=0\]is \[(0,1,2)\]and radius                    \[r=\sqrt{{{0}^{2}}+{{1}^{2}}+{{2}^{2}}+11}\] \[\Rightarrow \]               \[r=OP=4\] Given equation of plane is \[x+2y+2z-15=0\] \[\therefore \]Distance from centre \[(0,1,2)\]to the plane, \[ON=\frac{|0+2+4-15|}{\sqrt{{{1}^{2}}+4+4}}=\frac{9}{3}=3\]                 In \[\Delta ONP,\] \[N{{P}^{2}}=O{{P}^{2}}-O{{N}^{2}}\] \[={{4}^{2}}-{{3}^{2}}\] \[=7\]                 \[\Rightarrow \]               \[NP=\sqrt{7}\]                 \[\therefore \]Radius of the circle is \[\sqrt{7}.\]