# How do you find the equation of a circle with center at the point (-8,5) and tangent to the x-axis?

Jan 16, 2016

${\left(x + 8\right)}^{2} + {\left(y - 5\right)}^{2} = {5}^{2}$

#### Explanation:

A circle with center $\left({x}_{c} , {y}_{c}\right)$ tangent to the x-axis has a radius, $r$, equal to the distance from the center to the x-axis, namely ${y}_{c}$.

The general formula for a circle with center $\left({x}_{c} , {y}_{c}\right)$ and radius $r$ is
$\textcolor{w h i t e}{\text{XXX}} {\left(x - {x}_{c}\right)}^{2} + {\left(y - {y}_{c}\right)}^{2} = {r}^{2}$

For the given values $\left({x}_{c} , {y}_{c}\right) = \left(- 8 , 5\right)$
and derived value $r = 5$
this gives us:
$\textcolor{w h i t e}{\text{XXX}} {\left(x + 8\right)}^{2} + {\left(y - 5\right)}^{2} = {5}^{2}$
graph{(x+8)^2+(y-5)^2=25 [-16.86, 5.64, -1.03, 10.22]}