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

Feb 18, 2015

Hello,

Use canonical form :

${x}^{2} + 8 x = {\left(x + 4\right)}^{2} - 16$.

Indeed, you have ${\left(x + 4\right)}^{4} = {x}^{2} + 8 x + 16$.

Same thing with $y$ :

${y}^{2} - 6 y = {\left(y - 3\right)}^{2} - 9$.

So, your equation has a good looking form :

${\left(x + 4\right)}^{2} - 16 + {\left(y - 3\right)}^{2} - 9 + 16 = 0$

or better : ${\left(x + 4\right)}^{2} + {\left(y - 3\right)}^{2} = {3}^{2}$

You recognize the circle of center $\setminus \Omega \left(- 4 , 3\right)$, of radius $r = 3$.

graph{(x+4)^2+(y-3)^2=9 [-13.87, 6.13, -2.22, 7.78]}