How do you graph $(x-2)^2+(y+3)^2=4$ ?

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Answer 1 · 2016-04-01T20:55:22

This is a circle with centre $(2, -3)$ and radius $2$

The equation of a circle with centre $(h, k)$ and radius $r$ may be written:

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

Comparing this with the given equation, we find $(h, k) = (2, -3)$ and $r = 2$

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

graph{((x-2)^2+(y+3)^2-4)((x-2)^2+(y+3)^2-0.008) = 0 [-3.04, 6.96, -5.28, -0.28]}

How do you graph (x-2)^2+(y+3)^2=4(x-2)^2+(y+3)^2=4?