A sphere is a 3dimensional circle, which retains many of the properties and characteristics of a 2dimensional 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 3variable 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 3dimensional 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 yexpression can remain asis, since there is only one yvariable 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 (z2)^2. The equation now reads
(x+2)^2 + y^2 + (z2)^2 = 8. 
Find the square root for the nonvariable 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 (z2)^2=0, the equation becomes z2=0 and z=2. The center of the sphere is those 3 coordinates and is written (2,0,2).
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