If (2,3,5) is one end of a diameter of the sphere x2+y2+z2−6x−12y−2z+20=0 then co-ordinates of the other end of the diameter are
A
(4,3,5)
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B
(4,9,-3)
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C
(4,9,3)
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D
(4,3,-3)
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E
(4,9,5)
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Solution
The correct option is B (4,9,-3) Equation of sphere is, x2+y2+z2−6x−12y−2z+20=0
Centre of sphere = (3,6,1) and one end of circle = (2,3,5)
Let other end be (x,y,z) ∴2+x2=3,3+y2=6,5+z2=1 ⇒ x=6-2, y=12-3, z=2-5 ⇒ x=4, y=9, z=-3
Hence, other end is (4,9,-3)