Question

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

Answer 1

Hello,

Use canonical form :

`x^2+8x = (x+4)^2 - 16`.

Indeed, you have `(x+4)^4 = x^2 + 8x + 16`.

Same thing with `y` :

`y^2-6y = (y-3)^2 - 9`.

So, your equation has a good looking form :

`(x+4)^2 -16 + (y-3)^2 - 9 + 16 = 0`

or better : `(x+4)^2 + (y-3)^2 = 3^2`

You recognize the circle of center `\Omega(-4,3)`, of radius `r=3`.

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