SOLUTION: Decide whether the graph of the equation is a line, a parabola, or a circle. x^2+4x+y^2-6y=3

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Decide whether the graph of the equation is a line, a parabola, or a circle. x^2+4x+y^2-6y=3      Log On


   

Question 134872: Decide whether the graph of the equation is a line, a parabola, or a circle.
x^2+4x+y^2-6y=3
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
x^2+4x+y^2-6y=3
Complete the square on the x-terms and the y-terms separately to get:
(x^2+4x+4) + (y^2-6y+9) = 3+4+9
(x-2)^2 + (y-3)^2 = 16
Circle with center at (2,3) and radius of 4
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Cheers,
Stan H.