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]}

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