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

Jan 19, 2016

Rearrange into the standard form of the equation of a circle with centre $\left(2 , - 3\right)$ and radius $5$.

Explanation:

$0 = {x}^{2} + {y}^{2} - 4 x + 6 y - 12$

$= \left({x}^{2} - 4 x + 4\right) + \left({y}^{2} + 6 y + 9\right) - 25$

$= {\left(x - 2\right)}^{2} + {\left(y + 3\right)}^{2} - {5}^{2}$

Add ${5}^{2}$ to both ends and transpose to get:

${\left(x - 2\right)}^{2} + {\left(y - \left(- 3\right)\right)}^{2} = {5}^{2}$

This is in the form:

${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$

the standard form of the equation of a circle with centre $\left(h , k\right) = \left(2 , - 3\right)$ and radius $r = 5$

graph{(x^2+y^2-4x+6y-12)((x-2)^2+(y+3)^2-0.02)=0 [-15.12, 16.92, -11.33, 4.69]}