# How do you graph 8x ^ { 2} + 72x + 8y ^ { 2} + 16y - 342= 0?

Mar 9, 2018

It is equation of a circle with center $\left(- \frac{9}{2} , - 1\right)$ and radius $8$.

#### Explanation:

Observe that

1. $8 {x}^{2} + 72 x + 8 {y}^{2} + 16 y - 342 = 0$ is a quadratic equation,
2. in which coefficients of ${x}^{2}$ and ${y}^{2}$ are equal and
3. their is no term containing $x y$.

Hence, it represents te equation of a circle. Let us convert it to appropriate form. Diving each term by $8$, we get

${x}^{2} + 9 x + {y}^{2} + 2 y - \frac{342}{8} = 0$ and grouping it makes it

$\left({x}^{2} + 9 x + {\left(\frac{9}{2}\right)}^{2}\right) + \left({y}^{2} + 2 y + 1\right) = {\left(\frac{9}{2}\right)}^{2} + 1 + \frac{171}{4}$

or ${\left(x + \frac{9}{2}\right)}^{2} + {\left(y + 1\right)}^{2} = \frac{256}{4} = 64 = {8}^{2}$

i.e. ${\left(x + \frac{9}{2}\right)}^{2} + {\left(y + 1\right)}^{2} = {8}^{2}$

Hence, this is equation of a point whose distance from $\left(- \frac{9}{2} , - 1\right)$ is always $8$ i.e.

It is equation of a circle with center $\left(- \frac{9}{2} , - 1\right)$ and radius $8$.

graph{(8x^2+72x+8y^2+16y-342)((x+9/2)^2+(y+1)^2-0.08)=0 [-20, 20, -10, 10]}