Question

How do you graph `x^2 + y^2 + 4x + 6y + 4 =0`?

Answer 1

Sort the `x` terms and `y` terms so that they are near one another.

`x^2+4x+y^2+6y=-4`

Now, complete the square for each part by adding constants. Remember that a constant added on the left side should also be added on the right to keep the equation balanced.

`(x^2+4x+color(red)4)+(y^2+6y+color(blue)9)=-4+color(red)4+color(blue)9`

This simplifies to be

`(x+2)^2+(y+3)^2=9`

Since the standard equation of a circle with a vertex `(h,k)` and radius `r` is

`(x-h)^2+(y-k)^2=r^2`

This tells us the circle has a vertex at `(-2,-3)` and a radius of `3`.

Graph the point `(-2,-3)` and travel `3` in every direction. You should have the points:

graph{((x-1)^2+(y+3)^2-0.03)((x+5)^2+(y+3)^2-0.03)((x+2)^2+(y+3)^2-0.03)((x+2)^2+(y)^2-0.03)((x+2)^2+(y+6)^2-0.03)=0 [-10.665, 9.335, -7.76, 2.24]}

Now, draw the circle surrounding the center and passing through each exterior point:

graph{(x+2)^2+(y+3)^2=9 [-10.665, 9.335, -7.2, 2.8]}