SOLUTION: x^2+y^2+6x+4y+12=0

Question 575999: x^2+y^2+6x+4y+12=0
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
x^2+y^2+6x+4y+12=0
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Complete the square on the x-terms and on the y-terms.
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x^2+6x+9 + y^2+4y+4 + 12 -13 = 0
(x+3)^2 + (y+2)^2 = 1
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Circle with center at (-3,-2); radius = 1
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Cheers,
Stan H.
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