lf $(2, 3, 5)$ is one end of a diameter of the sphere $x^{2} + y^{2} + z^{2} - 6 x - 12 y - 2 z + 20 = 0$ , then the coordinates of the other end of the diameter are

(4, 3, - 3)

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(4, 9, - 3)

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(4, - 3, 3)

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(4, 3, 5)

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Solution

The correct option is B $(4, 9, - 3)$
Given equation of sphere is $x^{2} + y^{2} + z^{2} - 6 x - 12 y - 2 z + 10 = 0$ .
Centre of the given sphere is $(3, 6, 1)$ .
Also coordinate of the one end of the diameter of the sphere is $(2, 3, 5)$ .
Mid point of the end point of the diameter is centre
$∴$ coordinate of the other end of the diameter is $(4, 9, - 3)$ .

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x^{2} + y^{2} + z^{2} - 6 x - 12 y - 2 z + 20 = 0

x^{2} + y^{2} + z^{2} - 6 x - 12 y - 2 z + 20 = 0

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