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\[=\sqrt{7}\]You need to login to perform this action.
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