SOLUTION: Find the center and radius of a sphere, x^2+y^2+z^2-4x+2y-6z+10=0

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Question 619259: Find the center and radius of a sphere, x^2+y^2+z^2-4x+2y-6z+10=0
Found 3 solutions by Alan3354, Theo, ikleyn:
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website!
Find the center and radius of a sphere, x^2+y^2+z^2-4x+2y-6z+10=0
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Complete the squares, same as in 2-space.

Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website!
x^2+y^2+z^2-4x+2y-6z+10=0 becomes:
(x^2 - 4x) + (y^2 + 2y) + (z^2 - 6z) = 0 which becomes:
(x-2)^2 + (y+1)^2 + (z-3)^2 = 4 + 1 + 9 which becomes:
(x-2)^2 + (y+1)^2 + (z-3)^2 = 14
center is (2,-1,3)
radius = sqrt(14)
here's a reference:
https://pantherfile.uwm.edu/ericskey/www/TANOTES/Ageometry/node11.html
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once you group into x's and y's and z's, then you solve each by completing the squares.
here's a reference on how to do that:
http://www.algebrahelp.com/lessons/equations/completingthesquare/

Answer by ikleyn(53299)   (Show Source):
You can put this solution on YOUR website!
.
Find the center and radius of a sphere, x^2+y^2+z^2-4x+2y-6z+10=0
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        In the post by @Theo, the center is defined correctly, while the radius is incorrect.
        I came to fix it.

x^2+y^2+z^2-4x+2y-6z+10 = 0 becomes:
(x^2 - 4x) + (y^2 + 2y) + (z^2 - 6z) = -10 which becomes:
(x-2)^2 + (y+1)^2 + (z-3)^2 = -10 + 4 + 1 + 9 which becomes:
(x-2)^2 + (y+1)^2 + (z-3)^2 = 4.
center is (2,-1,3).
radius = sqrt(4) = 2.

Solved.