# How do you graph x^2 + y^2 + 6x - 4y - 12 = 0?

Mar 1, 2016

circle : centre (-3,2) , r = 5

#### Explanation:

The general form of the equation of a circle is :

${x}^{2} + {y}^{2} + 2 g x + 2 f y + c = 0$

centre = (-g,-f) and r$= \sqrt{{g}^{2} + {f}^{2} - c}$

the equation ${x}^{2} + {y}^{2} + 6 x - 4 y - 12 = 0 \text{ is in this form }$

and by comparison: 2g=6 →g=3, 2f=-4 →f=-2 and c = -12

centre = (-3,2) and r=$\sqrt{{3}^{2} + {\left(- 2\right)}^{2} + 12} = \sqrt{25} = 5$