+164 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?
+9479 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)
+9479 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)
