SOLUTION: what is the center of a circle with the equation {{{x^2+y^2-4x+6y+1=0}}}

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Question 8724: what is the center of a circle with the equation x%5E2%2By%5E2-4x%2B6y%2B1=0
Answer by FSAN85(6) About Me  (Show Source):
You can put this solution on YOUR website!
OKay, this is what to do
first rearrange the equation:
x^2 - 4x + y^2 + 6y +1 = 0
take the 1 to ther side
x^2 - 4x + y^2 + 6y = -1
now expand ur equation
(x^2 - 4x +4) + (y^2 + 6y +9) = -1 + 4 + 9
break down
(x - 2)^2 + (y + 3)^2 = 12
Take the radical form of all the parts
(x-2) + (y+3) = (square root)12
set the x ans ys to 0, that will give u ur center which turn out to be:
A circle with a center at ( 2 , -3 ) with a radius 3.46 (squareroot of 12)