# How do you graph (x-3)^2 + (y-2)^2 = 9?

Aug 16, 2015

It's a circle with centre $\left(3 , 2\right)$ and radius $3$.

#### Explanation:

The equation of a circle can be expressed as:
${\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {R}^{2}$
where $\left(a , b\right)$ is the co-ordinate of the centre of the circle and $R$ is the radius.

Here $a = 3$ and $b = 2$, so it's a circle with centre $\left(3 , 2\right)$ and radius $\sqrt{9} = 3$.

graph{(x-3)^2+(y-2)^2 = 9 [-6 10, -2, 6]}