A sphere is a 3-dimensional circle, which retains many of the properties and characteristics of a 2-dimensional circle. One shared property is that the radius and center of the sphere are interrelated. You can find the sphere's radius and center through a standard 3-variable form equation. Learning to correctly and efficiently find the sphere's center and radius can help you to better understand the sphere's properties and the general properties of 3-dimensional geometry.
Rearrange the order of the terms, so terms with the same variable are together. For example, if the equation is x^2 + y^2 + z^2 + 4x - 4z = 0, then rearranging the terms would result in x^2 + 4x + y^2 + z^2 - 4z = 0.
Add parenthesis around the terms with the same variables to make them separate. For the example, change x^2 + 4x + y^2 + z^2 - 4z = 0 to (x^2 + 4x) + y^2 +( z^2 - 4z) = 0.
The y-expression can remain as-is, since there is only one y-variable term.
Complete the squares of the parenthesized terms. Completing the square means adding numbers to both sides of the equation so that the term can be factored as a binomial, or a polynomial to the power of 2. For the example, (x^2 + 4x) + y^2 +( z^2 - 4z) = 0 becomes (x^2 + 4x + 4) + y^2 +(z^2 - 4z + 4) = 0 + 4 + 4.
Factor the parenthesized expressions. For the example, the expression x^2 + 4x + 4 can be factored into (x+2)^2 and the expression z^2 - 4z + 4 can be factored into (z-2)^2. The equation now reads (x+2)^2 + y^2 + (z-2)^2 = 8.
Find the square root for the non-variable side of the equation. For the example, the square root of 8 is 2√2. This is the radius of the sphere.
Set each variable term equal to zero and solve. For (x+2)^2=0, the equation becomes x+2=0 and x=-2. For y^2=0, y=0. For (z-2)^2=0, the equation becomes z-2=0 and z=2. The center of the sphere is those 3 coordinates and is written (-2,0,2).
- Yagi Studio/Photodisc/Getty Images