Question

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?

Answer 1

It is a circle.

You graph it by:

1. set your compass to a radius of `sqrt365/2`
2. put the center at the point `(-13/2, -3)`
3. draw the circle.

Explanation:

I suspect that the intended equation is:

`x^2+y^2+6y+13x=40" [1]"`

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

The reference [Conic Section](https://en.wikipedia.org/wiki/Conic_section`General_Cartesian_form) tells us that, because the coefficient of the`x^2`term is equal to the coefficient of the`y^2`term, we know that the equation describes a`color(red)("circle")#. Here is a graph to prove it:

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 `(h,k)` is the center and r is the radius.

By completing the squares:

`x^2+13x+h^2+y^2+6y+k^2=40+h^2+k^2" [1.1]"`

We know, from their respective binomial expansions, that:

`-2hx = 13x` and `-2ky = 6y`

Solve for h and k:

`h=-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+(-13/2)^2+ (-3)^2`

`r^2 = 365/4`

`r = sqrt365/2`

Substituting these values into equation [2]:

`(x - (-13/2))^2 + (y-(-3))^2 = (sqrt365/2)^2" [2]"`

This is the standard Cartesian form with its center at `(-13/2,-3)` and a radius of `sqrt365/2`