# How do you graph x^2+y^2+x-6y+9=0?

Nov 24, 2016

Circle of radius $\frac{1}{2}$ and centre $\left(- \frac{1}{2} , 3\right)$

#### Explanation:

Complete the square for both $x$ and $y$

${x}^{2} + {y}^{2} + x - 6 y + 9 = 0$
${x}^{2} + x + {y}^{2} - 6 y = - 9$
${\left(x + \frac{1}{2}\right)}^{2} - {\left(\frac{1}{2}\right)}^{2} + {\left(y - 3\right)}^{2} - {\left(- 3\right)}^{2} = - 9$
${\left(x + \frac{1}{2}\right)}^{2} - \frac{1}{4} + {\left(y - 3\right)}^{2} - 9 = - 9$
${\left(x + \frac{1}{2}\right)}^{2} + {\left(y - 3\right)}^{2} = \frac{1}{4}$
${\left(x + \frac{1}{2}\right)}^{2} + {\left(y - 3\right)}^{2} = {\left(\frac{1}{2}\right)}^{2}$

So we see that it is a circle of radius $\frac{1}{2}$ and centre $\left(- \frac{1}{2} , 3\right)$

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