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

Nov 24, 2016

${x}^{2} + 2 x + {y}^{2} + 6 y + 6 = 0$

Gather up the terms in $x$, and the terms in $y$ and complete the square by combing half the $x$ and $y$ coefficient for both as follows;

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

A circle of centre $\left(a , b\right)$ and radius $r$ has equation

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

Comparing with the above equation we can see that

${\left(x + 1\right)}^{2} + {\left(y + 3\right)}^{2} = {2}^{2}$

represents a circle of centre (-1,-3) and radius 2

graph{(x+1)^2 +(y+3)^2 =2^2 [-11.17, 8.83, -7.3, 2.7]}