Question

How do you graph `x^2+y^2+3x+4y+4=0`?

Answer 1

The graph is the circumference

`(x-3/3)^2+(y-2)^2=(3/2)^2`

Explanation:

`x^2+y^2+3x+4y+4 = 0` can be represented in the form
`(x-x_c)^2+(y-y_c)^2-r^2=0` which is the equation of the circumference with center at `{x_c, y_c}` and radius `r`.
Equating coefficients we get

`x_c = 3/2,y_c=2,x_c^2+y_c^2-r^2=4->r=3/2`

The graph is the circumference

`(x-3/3)^2+(y-2)^2=(3/2)^2`