# How do you find the equation of the circle with centre (2,5) which touches the x axis?

Jan 14, 2016

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

#### Explanation:

The general form for the equation of a circle is:
$\textcolor{w h i t e}{\text{XXX}} {\left(x - {x}_{c}\right)}^{2} + {\left(y - {y}_{c}\right)}^{2} = {r}^{2}$
where $\left({x}_{c} , {y}_{c}\right)$ is the center of the circle
and $r$ is the circle's radius.

We are told that $\left({x}_{c} , {y}_{c}\right) = \left(2 , 5\right)$
and that the circle touches the x-axis.
$\rightarrow$ the distance from the center of the circle to the x-axis is ${y}_{c}$
$\rightarrow$ the radius $r = {y}_{c} = 5$

Substituting
$\textcolor{w h i t e}{\text{XXX}} 2 \rightarrow {x}_{c}$,
$\textcolor{w h i t e}{\text{XXX}} 5 \rightarrow {y}_{c}$, and
$\textcolor{w h i t e}{\text{XXX}} 5 \rightarrow r$
in the general equation:
$\textcolor{w h i t e}{\text{XXX}} {\left(x - 2\right)}^{2} + {\left(y - 5\right)}^{2} = {5}^{2}$