Question

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

Answer 1

`(x - 3/2)^2 + (y - 2)^2 = 9/4`
graph{(x-3/2)^2 + (y-2)^2 - 9/4 = 0 [-2.205, 5.55, -0.087, 3.954]}

Explanation:

`x^2 + y^2 + 3x + 4y + 4 = 0`

Step 1. Group x and y terms together.
`(x^2 + 3x + alpha) + (y^2 + 4y + beta) = -4 + alpha + beta`
Now the goal to build a perfect by setting
`alpha = (3/2)^2 = 9/4; beta = (4/2)^2 = 4`

Now the circle equation in standard form is:
`(x-h)^2 + (y-k)^2 = r^2`
`h = sqrt(alpha) = 3/2; k = sqrt(beta) = 2`

And the circle equation in standard form is
`(x - 3/2)^2 + (y - 2)^2 = 9/4=(3/2)^2`