How do you graph #x^2 + y^2 – 4x + 22y + 61 = 0#?

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Answer 1 · 2016-03-11T00:00:55

This is a circle, centre #(2, -11)#, with radius #8#

#0 = x^2+y^2-4x+22y+61#

#=x^2-4x+4+y^2+22y+121-64#

#=(x-2)^2+(y+11)^2-8^2#

Add #8^2# to both ends and transpose to get:

#(x-2)^2+(y+11)^2 = 8^2#

which is (almost) in the standard form:

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

the equation of a circle with centre #(h, k) = (2, -11)# of radius #r=8#.

graph{(x^2+y^2-4x+22y+61)((x-2)^2+(y+11)^2-0.06) = 0 [-17.8, 22.2, -19.96, 0.04]}