An equation of sphere with centre at origin and radius $r$ can be represented as

x^{2} + y^{2} + z^{2} = r

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x^{2} + y^{2} + z^{2} = r^{2}

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x^{2} + y^{2} + z^{2} = 2 r^{2}

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None of the above

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Solution

The correct option is B $x^{2} + y^{2} + z^{2} = r^{2}$
Sphere is locus of a point in 3D whose distance from a fixed point(center) is constant (radius)
$\Rightarrow \sqrt{(x - 0)^{2} + (y - 0)^{2} + (z - 0)^{2}} = | r |$
$\Rightarrow x^{2} + y^{2} + z^{2} = r^{2}$ , square both sides

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Similar questions

x^{2} + y^{2} + z^{2} = r^{2}

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x^{2} - y^{2} + z^{2} = r^{2}

z^{2} + y^{2} - x^{2} = r^{2}

If $x = r s i n α c o s β, y = r s i n α s i n β$ and $z = r c o s α$ , then

d

x^{2} + y^{2} + z^{2} + 2 u x + 2 v y + 2 w z + 4 = 0

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(1, - 2, 2)

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u = f (r)

r^{2} = x^{2} + y^{2} + z^{2}

\frac{\partial^{2} u}{\partial x^{2}} + \frac{\partial^{2} u}{\partial y^{2}} + \frac{\partial^{2} u}{\partial z^{2}} = f^{''} (r) + \frac{2}{r} f (r)

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