# How do you identity if the equation x^2+y^2+6y+13=40 is a parabola, circle, ellipse, or hyperbola and how do you graph it?

Jul 2, 2017

It is a circle.

You graph it by:

1. set your compass to a radius of $\frac{\sqrt{365}}{2}$
2. put the center at the point $\left(- \frac{13}{2} , - 3\right)$
3. draw the circle.

#### Explanation:

I suspect that the intended equation is:

${x}^{2} + {y}^{2} + 6 y + 13 x = 40 \text{ [1]}$

Otherwise, the 13 and 40 would have been combined into a single constant term.

graph{x^2+y^2+6y+13x=40 [-30, 30, -15, 15]}

We can make fit the general Cartesian form for the equation of a circle:

(x - h)^2 + (y-k)^2 = r^2" [2]

where $\left(h , k\right)$ is the center and r is the radius.

By completing the squares:

${x}^{2} + 13 x + {h}^{2} + {y}^{2} + 6 y + {k}^{2} = 40 + {h}^{2} + {k}^{2} \text{ [1.1]}$

We know, from their respective binomial expansions, that:

$- 2 h x = 13 x$ and $- 2 k y = 6 y$

Solve for h and k:

$h = - \frac{13}{2}$ and $k = - 3$

We can obtain a value for r by substituting these values to the right side of equation [1.1}:

${r}^{2} = 40 + {\left(- \frac{13}{2}\right)}^{2} + {\left(- 3\right)}^{2}$

${r}^{2} = \frac{365}{4}$

$r = \frac{\sqrt{365}}{2}$

Substituting these values into equation [2]:

${\left(x - \left(- \frac{13}{2}\right)\right)}^{2} + {\left(y - \left(- 3\right)\right)}^{2} = {\left(\frac{\sqrt{365}}{2}\right)}^{2} \text{ [2]}$

This is the standard Cartesian form with its center at $\left(- \frac{13}{2} , - 3\right)$ and a radius of $\frac{\sqrt{365}}{2}$