Question

The equation of the sphere concentric with the sphere $ 2{{x}^{2}}+2{{y}^{2}}+2{{z}^{2}}-6x+2y-4z=1 $ and double its radius is:

• A. $ {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-x+y-z=1 $
• B. $ {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-6x+2y-4z=1 $
• C. $ 2{{x}^{2}}+2{{y}^{2}}+2{{z}^{2}}-6x+2y-4z-15=0 $
• D. $ {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-3x+y-2z=1 $
• E. $ 2{{x}^{2}}+2{{y}^{2}}+2{{z}^{2}}-6x+2y-4z-25=0 $

Answer 1

Answer: (E)

Equation of sphere is $ 2{{x}^{2}}+2{{y}^{2}}+2{{z}^{2}}-6x-2y-4z-1=0 $ Radius of sphere is
$ \sqrt{\frac{9}{4}+\frac{1}{4}+\frac{4}{4}+\frac{1}{2}}=2 $
Equation of family of concentric sphere is $ {{x}^{2}}+{{y}^{2}}+\text{ }{{z}^{2}}-3x+\text{ }y-2z+\lambda =0 $ ...(i)
$ \therefore $ According to question, $ \sqrt{\frac{9}{4}+\frac{1}{4}+1-\lambda }=4 $
$ \Rightarrow $ $ \frac{14}{4}-\lambda =16 $
$ \Rightarrow $ $ \lambda =\frac{14}{4}-16=-\frac{25}{2} $
$ \therefore $ From Eq. (i)
$ {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-3x+y-2z-\frac{25}{2}=0 $
$ \Rightarrow $ $ 2{{x}^{2}}+2{{y}^{2}}+2{{z}^{2}}-6x+2y-4z-25=0 $