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

Jan 25, 2016

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

#### Explanation:

${x}^{2} + {y}^{2} + 3 x + 4 y + 4 = 0$

Step 1. Group x and y terms together.
$\left({x}^{2} + 3 x + \alpha\right) + \left({y}^{2} + 4 y + \beta\right) = - 4 + \alpha + \beta$
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:
${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$
h = sqrt(alpha) = 3/2; k = sqrt(beta) = 2

And the circle equation in standard form is
${\left(x - \frac{3}{2}\right)}^{2} + {\left(y - 2\right)}^{2} = \frac{9}{4} = {\left(\frac{3}{2}\right)}^{2}$