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?

Question

Sadie Alexander · Accepted Answer

It is a circle.

You graph it by:

set your compass to a radius of #sqrt365/2#
put the center at the point #(-13/2, -3)#
draw the circle. 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://tutor.hix.ai) 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#

Sadie Alexander · Answer

undefined To identify the type of conic section represented by the equation $x^2 + y^2 + 6y + 13 = 40$, we need to rearrange it into a standard form. Completing the square for the $y$-terms yields:

$$x^2 + y^2 + 6y + 13 = 40 $$
$$x^2 + (y^2 + 6y + 9) + 13 - 9 = 40 $$
$$x^2 + (y + 3)^2 = 36 $$

Now, the equation is in the form $x^2 + (y - k)^2 = r^2$, which represents a circle with center $(0, -3)$ and radius $6$. Therefore, the given equation represents a circle.

To graph it:
1. Plot the center of the circle, which is at $(0, -3)$.
2. Use the radius of $6$ to plot points on the circle. You can plot points $(0, 3)$, $(6, 0)$, $(0, -9)$, and $(-6, 0)$ to help draw the circle.
3. Sketch the circle through these points, ensuring it is symmetrical around the center $(0, -3)$.