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

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Answer 1 · 2017-10-01T15:25:59

We have:
#x^2 + y^2 + 4x - 4y - 1 = 0#

We can gather terms in #x# and terms in #y#

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

Now we can complete the square fro the #'x# and #y# terms independently:

# (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 #3# centred at #(-2,2)#

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