Question

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

Answer 1

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

Explanation:

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]}