How do you graph # x² + y² + 4x - 4y - 17 = 0#?

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Answer 1 · 2016-04-24T21:19:42

This is a circle with centre #(-2, 2)# and radius #5#

#0 = x^2+y^2+4x-4y-17#

#=x^2+4x+4+y^2-4y+4-25#

#=(x+2)^2+(y-2)^2-5^2#

Add #5^2# to both ends and transpose to get:

#(x+2)^2+(y-2)^2=5^2#

This is (almost) in the form:

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

which is the standard form of the equation of a circle centre #(h,k)# and radius #r#.

Hence we find #(h,k) = (-2,2)# and #r=5#.

This is a circle with centre #(-2, 2)# and radius #5#.

graph{((x+2)^2+(y-2)^2-5^2)((x+2)^2+(y-2)^2-0.03) = 0 [-11, 11, -3.7, 7.2]}