What is the equation to the sphere whose centre is at $(- 2, 3, 4)$ and radius is $6$ units?

x^{2} + y^{2} + z^{2} + 4 x - 6 y - 8 z = 7

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x^{2} + y^{2} + z^{2} + 4 x - 6 y - 8 z = 7

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x^{2} + y^{2} + z^{2} + 4 x - 6 y - 8 z = 4

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x^{2} + y^{2} + z^{2} + 4 x + 6 y + 8 z = 4

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Solution

The correct option is A $x^{2} + y^{2} + z^{2} + 4 x - 6 y - 8 z = 7$
Equation of Sphere is $(x - (- 2))^{2} + (y - 3)^{2} + (z - 4)^{2} = 6^{2} \Rightarrow x^{2} + y^{2} + z^{2} + 4 x - 6 y - 8 z + 4 + 9 + 16 = 36 \Rightarrow x^{2} + y^{2} + z^{2} + 4 x - 6 y - 8 z = 36 - 29 \Rightarrow x^{2} + y^{2} + z^{2} + 4 x - 6 y - 8 z = 7$

Therefore, Answer is $A$

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x^{2} + y^{2} + z^{2} - 4 x + 6 y - 8 z - 7 = 0 ?

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