The 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?

Gwendolynkristine
May 5, 2017

#1**+4 **

x^2 + y^2 + 2x - 6y + 1 = 0

Rearrange and subtract 1 from both sides.

x^2 + 2x + y^2 - 6y = -1

Add (2/2)^2 = 1 and (6/2)^2 = 9 to both sides.

x^2 + 2x + 1 + y^2 - 6y + 9 = -1 + 1 + 9

Factor.

(x + 1)^2 + (y - 3)^2 = 9

Now that the equation of the circle is in this form...we can see that the center is located at (-1, 3)

hectictar
May 5, 2017

#1**+4 **

Best Answer

x^2 + y^2 + 2x - 6y + 1 = 0

Rearrange and subtract 1 from both sides.

x^2 + 2x + y^2 - 6y = -1

Add (2/2)^2 = 1 and (6/2)^2 = 9 to both sides.

x^2 + 2x + 1 + y^2 - 6y + 9 = -1 + 1 + 9

Factor.

(x + 1)^2 + (y - 3)^2 = 9

Now that the equation of the circle is in this form...we can see that the center is located at (-1, 3)

hectictar
May 5, 2017