How do you graph #x^2 + y^2 + 4x - 4y - 1 = 0#?
1 Answer
Feb 15, 2016
circle: centre =(-2 ,2) , r = 3
Explanation:
The general equation of a circle is
#x^2 + y^2 + 2gx + 2fy + c = 0# the equation here :
#x^2 + y^2 + 4x - 4y - 1 = "0 is in this form"# by comparison: 2g = 4 → g = 2 , 2f = -4 → f=-2 and c = -1
centre = (-g , -f) = (-2 , 2)
and r
# = sqrt(g^2 + f^2 - c) =sqrt(2^2+(-2)^2+1 )=sqrt9 = 3#
