SOLUTION: he general form of the equation of a circle is x2+y2+2x−6y+1=0.
What are the coordinates of the center of the circle?
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-> SOLUTION: he general form of the equation of a circle is x2+y2+2x−6y+1=0.
What are the coordinates of the center of the circle?
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You can put this solution on YOUR website! rewrite in form of (x-h)^2+(y-k)^2=r^2
x^2+2x+y^2-6y=-1
complete the square of both
x^2+2x+1+y^2-6y+9=9, after adding 20 to both sides
(x+1)^2+(y-3)^2=3^2
The center is the additive inverse of both constants, so it is at (-1,3) and the radius is 3, the square root of 9.
The center is at (-1, 3)
