Equation of Sphere/Rectangular Coordinates

Theorem
The equation of a sphere with radius $R$ and center $\left({a, b, c}\right)$ expressed in Cartesian coordinates is:
 * $\left({x - a}\right)^2 + \left({y - b}\right)^2 + \left({x - c}\right)^2 = R^2$

Proof
Let the point $\left({x, y, z}\right)$ satisfy the equation:
 * $(1): \quad \left({x - a}\right)^2 + \left({y - b}\right)^2 + \left({x - c}\right)^2 = R^2$

By the Distance Formula in 3 Dimensions, the distance between this $\left({x, y, z}\right)$ and $\left({a, b, c}\right)$ is:
 * $\sqrt {\left({x - a}\right)^2 + \left({y - b}\right)^2 + \left({z - c}\right)^2}$

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 $\left({x, y, z}\right)$ lies on the surface of a sphere with radius $R$ and center $\left({a, b, c}\right)$.

Now suppose that $\left({x, y, z}\right)$ does not satisfy the equation:
 * $\left({x - a}\right)^2 + \left({y - b}\right)^2 + \left({z - c}\right)^2 = R^2$

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

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

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