# Vector calc center and radius of circle problem

• Feb 4th 2009, 06:38 PM
Frostking
Vector calc center and radius of circle problem
I am studying for a test and one of the review questions is stumping me.

I am suppose to find the center and radius of a circle with the following equation:

x^2 + 4x + y^2 - 6y + z^2 + 12z = 0

I know that the equation for a sphere is x^2 + y^2 + z^2 = radius ^2

Also I think that you would take (x - x at center)^2 + (y- y at center)^2 + (z -- z at center)^2 and this is equal to the radius squared. But I do not know how to deal with 4x, -6y and 12z.
Thanks to all you math enlightened folks for all the help that I have received for the past year!
• Feb 4th 2009, 06:41 PM
skeeter
complete the square for x , y, and z ...

$x^2 + 4x + y^2 - 6y + z^2 + 12z = 0
$

$(x^2 + 4x + 4) + (y^2 - 6y + 9) + (z^2 + 12z + 36) = 4 + 9 + 36$

can you finish?
• Feb 4th 2009, 06:54 PM
Frostking
Vector calc center and radius answer response
Thanks so much for your quick assistance.

I see that by completing the squares we would have the square of the radius as 49 and therefore the radius is 7. I also see that the center coordinate would involve 2,3,6 but I think it would be (2, -3, 6) and the answer says it should be (-2,3, -6)????
• Feb 4th 2009, 07:05 PM
skeeter
the book answer is correct

$(x - a)^2 + (y - b)^2 + (z - c)^2 = r^2
$

center at $(a, b, c)$