# What is the equation of the circle with a center at (3 ,1 ) and a radius of 1 ?

Jan 20, 2016

${\left(x - 3\right)}^{2} + {\left(y - 1\right)}^{2} = 1$

#### Explanation:

The general form for the equation of a circle with a center at $\left(h , k\right)$ and radius $r$ is

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

We know that

$\left(h , k\right) \rightarrow \left(3 , 1\right) \implies h = 3 , k = 1$
$r = 1$

So the equation of the circle is

${\left(x - 3\right)}^{2} + {\left(y - 1\right)}^{2} = {1}^{2}$

or, slightly more simplified (squaring the $1$):

${\left(x - 3\right)}^{2} + {\left(y - 1\right)}^{2} = 1$

The circle graphed:

graph{((x-3)^2+(y-1)^2-1)((x-3)^2+(y-1)^2-.003)=0 [-2.007, 9.093, -1.096, 4.454]}