# How do you graph the circle #x^2 + y^2 + 4x - 4y - 1 = 0#?

##### 1 Answer

Oct 1, 2017

We have:

We can gather terms in

# (x^2 + 4x) + (y^2 - 4y) = 1 #

Now we can complete the square fro the

# (x+2)^2-2^2 + (y - 2)^2-2^2 = 1 #

# :. (x+2)^2-4 + (y - 2)^2-4 = 1 #

Leading to:

# (x+2)^2-4 + (y - 2)^2 = 9 #

# :. (x+2)^2-4 + (y - 2)^2 = 3^2 #

Which if we compare to the standard equations of the conics represents circle of radius

graph{x^2 + y^2 + 4x - 4y - 1 = 0 [-9.26, 4.784, -1.27, 5.75]}