How do you graph x^2+y^2+x-6y+9=0x2+y2+x−6y+9=0?

1 Answer
Steve M
Nov 24, 2016

Circle of radius 1/212 and centre (-1/2,3)(−12,3)

Explanation:

Complete the square for both xx and yy

x^2 + y^2 + x - 6y + 9 = 0 x2+y2+x−6y+9=0
x^2 + x + y^2 - 6y = -9 x2+x+y2−6y=−9
(x + 1/2)^2 - (1/2)^2 + (y - 3)^2 - (-3)^2 = -9 (x+12)2−(12)2+(y−3)2−(−3)2=−9
(x + 1/2)^2 - 1/4 + (y - 3)^2 - 9 = -9 (x+12)2−14+(y−3)2−9=−9
(x + 1/2)^2 + (y - 3)^2 = 1/4 (x+12)2+(y−3)2=14
(x + 1/2)^2 + (y - 3)^2 = (1/2)^2 (x+12)2+(y−3)2=(12)2

So we see that it is a circle of radius 1/212 and centre (-1/2,3)(−12,3)

graph{x^2 + y^2 + x - 6y + 9 = 0 [-6, 6, -1, 5]}

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