What is the diameter of the sphere $x^{2} + y^{2} + z^{2} - 4 x + 6 y - 8 z - 7 = 0 ?$

4 units

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5 units

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6 units

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12 units

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Solution

The correct option is D 12 units
General equation of the circle is ${(x - a)}^{2} + {(y - b)}^{2} + {(z - c)}^{2} = r^{2}$ where $(a, b, c)$ is center of circle and $r$ is radius of circle.
So, $x^{2} + y^{2} + z^{2} - 4 x + 6 y - 8 z - 7 = 0$ can be written as
$(x^{2} - 2 \times 2 x + 2^{2}) + (y^{2} + 2 \times 3 y + 3^{2}) + (z^{2} - 2 \times 4 z + 4^{2}) - 2^{2} - 3^{2} - 4^{2} - 7 = 0$
$\Rightarrow {(x - 2)}^{2} + {(y + 3)}^{2} + {(z - 4)}^{2} = 4 + 9 + 16 + 7 = 36 \Rightarrow {(x - 2)}^{2} + {(y + 3)}^{2} + {(z - 4)}^{2} = 6^{2}$
Which is the equation of sphere with center $(2, - 3, 4)$ and radius $6$ unit.
Thus, the diameter is $12$ units.

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