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

Jan 28, 2016

Sort the $x$ terms and $y$ terms so that they are near one another.

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

Now, complete the square for each part by adding constants. Remember that a constant added on the left side should also be added on the right to keep the equation balanced.

$\left({x}^{2} + 4 x + \textcolor{red}{4}\right) + \left({y}^{2} + 6 y + \textcolor{b l u e}{9}\right) = - 4 + \textcolor{red}{4} + \textcolor{b l u e}{9}$

This simplifies to be

${\left(x + 2\right)}^{2} + {\left(y + 3\right)}^{2} = 9$

Since the standard equation of a circle with a vertex $\left(h , k\right)$ and radius $r$ is

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

This tells us the circle has a vertex at $\left(- 2 , - 3\right)$ and a radius of $3$.

Graph the point $\left(- 2 , - 3\right)$ and travel $3$ in every direction. You should have the points:

graph{((x-1)^2+(y+3)^2-0.03)((x+5)^2+(y+3)^2-0.03)((x+2)^2+(y+3)^2-0.03)((x+2)^2+(y)^2-0.03)((x+2)^2+(y+6)^2-0.03)=0 [-10.665, 9.335, -7.76, 2.24]}

Now, draw the circle surrounding the center and passing through each exterior point:

graph{(x+2)^2+(y+3)^2=9 [-10.665, 9.335, -7.2, 2.8]}