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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

Best Answer 

 #1
avatar+7324 
+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
avatar+7324 
+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

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