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

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Answer 1 · 2016-01-22T22:11:39

This is the equation of a circle centre #(-2, 2)# and radius #3#

#0 = x^2+y^2+4x-4y-1#

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

#= (x+2)^2+(y-2)^2-3^2#

Add #3^2# to both ends and transpose to get:

#(x-(-2))^2+(y-2)^2 = 3^2#

This is in the form:

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

the standard form of the equation of a circle with centre #(h, k) = (-2, 2)# and radius #r=3#

graph{(x^2+y^2+4x-4y-1)((x+2)^2+(y-2)^2-0.01)=0 [-12.33, 7.67, -3.08, 6.92]}