Equation of Sphere/Rectangular Coordinates

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Theorem

The equation of a sphere with radius $R$ and center $\tuple {a, b, c}$ expressed in Cartesian coordinates is:

$\paren {x - a}^2 + \paren {y - b}^2 + \paren {x - c}^2 = R^2$


Corollary

The equation of a sphere with radius $R$ whose center is at the origin expressed in Cartesian coordinates is:

$x^2 + y^2 + z^2 = R^2$


Proof

Let the point $\tuple {x, y, z}$ satisfy the equation:

$(1): \quad \paren {x - a}^2 + \paren {y - b}^2 + \paren {x - c}^2 = R^2$

By the Distance Formula in 3 Dimensions, the distance between this $\tuple {x, y, z}$ and $\tuple {a, b, c}$ is:

$\sqrt {\paren {x - a}^2 + \paren {y - b}^2 + \paren {z - c}^2}$

But from equation $(1)$, this quantity equals $R$.

Therefore the distance between points satisfying the equation and the center is constant and equal to the radius.

Thus $\tuple {x, y, z}$ lies on the surface of a sphere with radius $R$ and center $\tuple {a, b, c}$.


Now suppose that $\tuple {x, y, z}$ does not satisfy the equation:

$\paren {x - a}^2 + \paren {y - b}^2 + \paren {z - c}^2 = R^2$

Then by the same reasoning as above, the distance between $\tuple {x, y, z}$ and $\tuple {a, b, c}$ does not equal $R$.

Therefore $\tuple {x, y, z}$ does not lie on the surface of a sphere with radius $R$ and center $\tuple {a, b, c}$.


Hence it follows that the points satisfying $(1)$ are exactly those points which are the sphere in question.

$\blacksquare$


Sources