Three Dimensional Geometry Question 42
Question: The equation of the sphere concentric with the sphere $ 2x^{2}+2y^{2}+2z^{2}-6x+2y-4z=1 $ and double its radius is
[Kerala (Engg.) 2005]
Options:
A) $ x^{2}+y^{2}+z^{2}-x+y-z=1 $
B) $ x^{2}+y^{2}+z^{2}-6x+2y-4z=1 $
C) $ 2x^{2}+2y^{2}+2z^{2}-6x+2y-4z-15=0 $
D) $ x^{2}+y^{2}+z^{2}-3x+y-2z=1 $
E) $ 2x^{2}+2y^{2}+2z^{2}-6x+2y-4z-25=0 $
Show Answer
Answer:
Correct Answer: E
Solution:
Equation of sphere is, $ x^{2}+y^{2}+z^{2}-3x+y-2z-\frac{1}{2}=0 $ \ Centre of sphere $ =( \frac{3}{2},\frac{-1}{2},1 ) $ Radius $ =\sqrt{( \frac{9}{4} )+( \frac{1}{4} )+1+\frac{1}{2}}=2 $ Now, radius of required sphere = 4, which is concentric with the given sphere. Hence, equation of required sphere is, $ {{( x-\frac{3}{2} )}^{2}}+{{( y+\frac{1}{2} )}^{2}}+{{(z-1)}^{2}}=16 $ i.e., $ 2x^{2}+2y^{2}+2z^{2}-6x+2y-4z-25=0 $ .
BETA
