# How do you graph 4x^2 + 4y^2 +32x +16y +71 = 0?

##### 1 Answer
May 24, 2016

The graphic is the circumference ${\left(x + 4\right)}^{2} + {\left(y + 2\right)}^{2} = {\left(\frac{3}{2}\right)}^{2}$

#### Explanation:

The circumference equation is
${\left(x - {x}_{0}\right)}^{2} + {\left(y - {y}_{0}\right)}^{2} = {r}^{2}$ so calculating
$4 {x}^{2} + 4 {y}^{2} + 32 x + 16 y + 71 - 4 \left({\left(x - {x}_{0}\right)}^{2} + {\left(y - {y}_{0}\right)}^{2} - {r}^{2}\right) = 0$ we get
$\left(32 + 8 {x}_{0}\right) x + \left(16 + 8 {y}_{0}\right) y + 71 + 4 {r}^{2} - 4 {x}_{0}^{2} - 4 {y}_{0}^{2} = 0$
solving the equations for ${x}_{0} , {y}_{0} , r$
$\left(\begin{matrix}32 + 8 {x}_{0} = 0 \\ 16 + 8 {y}_{0} = 0 \\ 71 + 4 {r}^{2} - 4 {x}_{0}^{2} - 4 {y}_{0}^{2} = 0\end{matrix}\right)$
we get
${x}_{0} = - 4 , {y}_{0} = - 2 , r = \frac{3}{2}$