Question

The equation of the sphere whose centre is $ (6,-1,\text{ }2) $ and which touches the plane $ 2x-y+2z-2=0, $ is:

• 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 $

Answer 1

Answer: (C)

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 $