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

##### 1 Answer
Apr 1, 2016

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

#### Explanation:

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

${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$

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

${\left(x - 2\right)}^{2} + {\left(y - \left(- 3\right)\right)}^{2} = {2}^{2}$

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]}