# How do you find an equation of the sphere with center (2, -6, 4) and radius 5?

Jun 16, 2016

The equation can be written in the form:

${\left(x - 2\right)}^{2} + {\left(y - \left(- 6\right)\right)}^{2} + {\left(z - 4\right)}^{2} = {5}^{2}$

#### Explanation:

Given any two points $\left({x}_{1} , {y}_{1} , {z}_{1}\right)$ and $\left({x}_{2} , {y}_{2} , {z}_{2}\right)$, the distance between them is given by the formula:

$d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2} + {\left({z}_{2} - {z}_{1}\right)}^{2}}$

So for a sphere with centre $\left(2 , - 6 , 4\right)$ and radius $5$, every point $\left(x , y , z\right)$ on the surface satisfies:

$5 = \sqrt{{\left(x - 2\right)}^{2} + {\left(y - \left(- 6\right)\right)}^{2} + {\left(z - 4\right)}^{2}}$

Squaring both sides and transposing this becomes:

${\left(x - 2\right)}^{2} + {\left(y - \left(- 6\right)\right)}^{2} + {\left(z - 4\right)}^{2} = {5}^{2}$

This is in the form:

${\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} + {\left(z - c\right)}^{2} = {r}^{2}$

with $\left(a , b , c\right) = \left(2 , - 6 , 4\right)$ and $r = 5$.

Note the similarity with the equation of a circle with centre $\left(h , k\right)$ and radius $r$, namely:

${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$