The centre and radius of the sphere given by $x^{2} + y^{2} + z^{2} - 6 x + + 8 y - 10 z + 1 = 0$ are:

(- 3, 4, - 5), 49

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(- 6, 8, - 10), 1

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(3, - 4, 5), 7

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(6, - 8, 10), 7

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Solution

The correct option is D $(3, - 4, 5), 7$
The sphere can be re-written as
$(x^{2} - 6 x) + (y^{2} + 8 y) + (z^{2} - 10 z) + 1 = 0$
$(x - 3)^{2} - 9 + (y + 4)^{2} - 16 + (z - 5)^{2} - 25 + 1 = 0$
$(x - 3)^{2} + (y + 4)^{2} + (z - 5)^{2} - 50 + 1 = 0$
$(x - 3)^{2} + (y + 4)^{2} + (z - 5)^{2} = 49$
$(x - 3)^{2} + (y + 4)^{2} + (z - 5)^{2} = 7^{2}$ .
Hence the centre of the sphere lies at $(3, - 4, 5)$ and the radius of the sphere is $7$ units.

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