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\[\Delta ONP\], \[O{{P}^{2}}=O{{N}^{2}}+N{{P}^{2}}\] \[\Rightarrow \] \[N{{P}^{2}}={{(4)}^{2}}-{{(3)}^{2}}=7\] \[\Rightarrow \] \[NP=\sqrt{7}\]You need to login to perform this action.
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