question_answer

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

A)  \[\sqrt{3}\]                                       

B) \[\sqrt{5}\]                        

C)  \[\sqrt{7}\]                       

D)        \[3\]

E)  \[\sqrt{2}\]

Correct Answer: C

Solution :

Since,     the     centre     of     sphere \[{{x}^{2}}+{{y}^{2}}+\text{ }{{z}^{2}}-2y-4z-11=0\]is (0, 1, 2) and radius is 4. \[\therefore \]Distance of a plane\[x+2y+2z-15=0\] from (0, 1, 2) \[=\frac{|0+2+4-15|}{\sqrt{1+4+4}}\]                 \[=\frac{9}{3}=3\] Now, \[NP=\sqrt{O{{P}^{2}}-O{{N}^{2}}}\]                 \[=\sqrt{{{4}^{2}}-{{3}^{2}}}=\sqrt{16-9}=\sqrt{7}\] \[\therefore \]Radius of circle\[=\sqrt{7}\]