# How would you find the center and radius of x^2 + y^2 - 6x=0?

Jun 26, 2018

Center: $\left(3 , 0\right)$
Radius: $3$

#### Explanation:

The standard form for the equation of a circle with center $\left(\textcolor{red}{a} , \textcolor{b l u e}{b}\right)$ and radius $\textcolor{g r e e n}{r}$ is
$\textcolor{w h i t e}{\text{XXX}} {\left(x - \textcolor{red}{a}\right)}^{2} + {\left(y - \textcolor{b l u e}{b}\right)}^{2} = {\textcolor{g r e e n}{r}}^{2}$

We need to convert the given equation ${x}^{2} + {y}^{2} - 6 x = 0$
into this standard form.

Grouping the terms containing $x$
${x}^{2} - 6 x \textcolor{w h i t e}{\text{xxx")+color(white)("xxx")y^2color(white)("xxx}} = 0$

The ${\left(y - \textcolor{b l u e}{b}\right)}^{2}$ term in the standard form can be written directly as ${\left(y - \textcolor{b l u e}{0}\right)}^{2}$
but the writing the ${\left(x - \textcolor{red}{a}\right)}^{2}$ will require some manipulation of ${x}^{2} - 6 x$; to do this we will "complete the square:
${x}^{2} - 6 \textcolor{m a \ge n t a}{+ {3}^{2}} \textcolor{w h i t e}{\text{x")+color(white)("x")(y-color(blue)0)^2color(white)("xx}} = 0 \textcolor{m a \ge n t a}{+ {3}^{2}}$

${\left(x - \textcolor{red}{3}\right)}^{2} \textcolor{w h i t e}{\text{xxx")+color(white)("x")(y-color(blue)0)^2color(white)("x}} = {\textcolor{g r e e n}{3}}^{2}$
$\textcolor{w h i t e}{\text{XXX}}$which is the required form for a circle with
$\textcolor{w h i t e}{\text{XXX}}$ center at $\left(\textcolor{red}{3} , \textcolor{b l u e}{0}\right)$ and
$\textcolor{w h i t e}{\text{XXX}}$ radius $\textcolor{g r e e n}{3}$