# How do you graph x^2 + y^2 + 8x - 6y + 16 = 0?

Apr 11, 2016

${x}^{2} + {y}^{2} + 8 x - 6 y + 16 = 0$
is a circle with radius $3$ and center $\left(- 4 , 3\right)$

#### Explanation:

${x}^{2} + {y}^{2} + 8 x - 6 y + 16 = 0$

rArr (x^2+8x+16)+(y^x-6ycolor(red)(+9)=0color(red)(+9)

$\Rightarrow {\left(x + 4\right)}^{2} + {\left(y - 3\right)}^{2} = 3 \wedge 2$
$\textcolor{w h i t e}{\text{XXX}}$which is the general form for a circle with center $\left(- 4 , 3\right)$ and radius $3$
graph{x^2+y^2+8x-6y+16=0 [-9.636, 4.41, -0.826, 6.196]}