# How do you graph 2x^2+8x+2y^2=0?

Feb 13, 2016

See the accompanying graph
graph{2x^2+8x+2y^2=0 [-5, 5, -2.5, 2.5]}

#### Explanation:

Equation of circle in Standard form is
${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$
Here $\left(h , k\right)$ are the coordinates of the centre of the circle and $r$ is its radius.

Given equation is
$2 {x}^{2} + 8 x + 2 {y}^{2} = 0$

We see that 2 can be factored out both sides of the equation. Thus we obtain

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

This can be written in the standard form by adding 4 to the both sides and by using the identity ${\left(a + b\right)}^{2} = {a}^{2} + 2 a b + {b}^{2}$.

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

${\left(x + 2\right)}^{2} + {\left(y - 0\right)}^{2} = {2}^{2}$
or ${\left(x - \left[- 2\right]\right)}^{2} + {\left(y - 0\right)}^{2} = {2}^{2}$
This gives us $\left(- 2 , 0\right)$ coordinates of the centre. Also $r = 2 \text{ units}$
We also observe that origin $\left(0 , 0\right)$ lies on the circle.