# How do you find the center and radius of (x-2)^2+(y+3)^2=4?

Jun 5, 2018

In order to find the center and radius, we must first understand how a standard circle equation is written.

#### Explanation:

The standard form for a circle is written as:

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

where $h$ is the x-value of the origin, $k$ is the y-value of the origin, and $r$ is the radius of the circle. Firstly, let's take out our origin from the equation. A way I like to imagine the origin is to switch the sign that $h$ and $k$ have. Therefore, our origin is $\left(2 , - 3\right)$. Let's plug in our origin back into the equation to make sure:

${\left(x - \left(2\right)\right)}^{2} + {\left(y - \left(- 3\right)\right)}^{2} = 4$
${\left(x - 2\right)}^{2} + {\left(y + 3\right)}^{2} = 4$

For our radius, we take the square root of ${r}^{2}$ value:

${r}^{2} = 4$
$r = 2$
Note that $- 2$ is not an answer because it is not possible to have a negative distance in a graph.