# How do you find the equation of a circle center of (5,-2) and a radius of 2?

Apr 3, 2018

${\left(x - 5\right)}^{2} + {\left(y + 2\right)}^{2} = 4$

#### Explanation:

I'm sure there a better answer and explanation but from my little experience with geometry I know the following:
1. ${x}^{2} + {y}^{2} = {r}^{2}$ is the equation of a circle center of (0,0) and a radius of r.
2. to move the center you just need to move the points of x and y.

so a movement of 5 points in the positive direction of the x axis is translated to
${\left(x - 5\right)}^{2}$ instead of $\left({x}^{2}\right)$
a movement of 2 in the negative direction of the y axis is translated to
${\left(y + 2\right)}^{2}$ instead of $\left({y}^{2}\right)$

which then lead to the equation
${\left(x - 5\right)}^{2} + {\left(y + 2\right)}^{2} = {2}^{2}$

Apr 3, 2018

${\left(x - 5\right)}^{2} + {\left(y + 2\right)}^{2} = 4$

#### Explanation:

The generic equation of a circle with center $\left(h , k\right)$ and radius $r$ is ${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$.

With a center of $\left(5 , - 2\right)$ and a radius of $2$, you have
${\left(x - 5\right)}^{2} + \left(y - {\left(- 2\right)}^{2}\right) = {2}^{2}$
or
${\left(x - 5\right)}^{2} + {\left(y + 2\right)}^{2} = 4$.