A) \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-12x+2y-4z-16=0\]
B) \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-12x+2y-4z=0\]
C) \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-12x+2y-4z+16=0\]
D) \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-12x+2y-4z+6=0\]
E) \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-12x+2y-4z-5=0\]
Correct Answer: C
Solution :
Radius of sphere is J- distance of\[(6,-1,2)\] from \[2x-y+2z-2=0\] i.e., \[\left| \frac{12+1+4-2}{\sqrt{4+1+4}} \right|=\left| \frac{15}{3} \right|=5\] \[\therefore \]Equation of sphere is \[{{(x-6)}^{2}}+{{(y+1)}^{2}}+{{(z-2)}^{2}}=25\] \[\Rightarrow \] \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-12x+2y-4z\] \[+36+1+4-25=0\] \[\Rightarrow \] \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-12x+2y-4z+16=0\]You need to login to perform this action.
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