# How do you find center, radius, and intercepts of a circle 3x² + 3y² - 6x + 12y = 0?

Oct 23, 2016

Center: $\left(1 , - 2\right)$
Radius: $\sqrt{5}$
Intercepts: $\left(0 , 0\right) , \left(0 , - 4\right) , \left(2 , 0\right)$

#### Explanation:

Given
$\textcolor{w h i t e}{\text{XXX}} 3 {x}^{2} + 3 {y}^{2} - 6 x + 12 y = 0$

Let's start by simplifying by dividing everything by the common factor of $3$
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} + {y}^{2} - 2 x + 4 y = 0$

Our target will be to convert this into standard circle form:
$\textcolor{w h i t e}{\text{XXX}} {\left(x - \textcolor{red}{a}\right)}^{2} + {\left(y - \textcolor{b l u e}{b}\right)}^{2} = {\textcolor{g r e e n}{r}}^{2}$
for a circle with center $\left(\textcolor{red}{a} , \textcolor{b l u e}{b}\right)$ and radius $\textcolor{g r e e n}{r}$

Regrouping the $x$ and $y$ terms separately
color(white)("XXX")(x^2-2xcolor(white)("XX"))+(y^2+4ycolor(white)("XX"))=0

Completing the square for each
$\textcolor{w h i t e}{\text{XXX}} \left({x}^{2} - 2 x \textcolor{c y a n}{+ 1}\right) + \left({y}^{2} + 4 y \textcolor{p u r p \le}{+ 4}\right) = 0 \textcolor{c y a n}{+ 1} \textcolor{p u r p \le}{+ 4}$

Rewriting as squares in standard circle form
color(white)("XXX")(x-color(red)1)^2+(y-color(blue)(""(-2)))^2=color(green)(""(sqrt(5))^2

This gives us
$\textcolor{w h i t e}{\text{XXX}}$Center at $\left(\textcolor{red}{1} , \textcolor{b l u e}{- 2}\right)$
$\textcolor{w h i t e}{\text{XXX}}$Radius of $\textcolor{g r e e n}{\sqrt{5}}$

To get the intercepts, it is probably easier to work from the earlier equation:
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} + {y}^{2} - 2 x + 4 y = 0$

The Y-intercepts occur when $x = 0$
$\textcolor{w h i t e}{\text{XXX}} {y}^{2} + 4 y = 0$
color(white)("XXX")y(y+4)=0 color(white)("XXX")y=0 or y=--4#

The X-intercepts occur when $y = 0$
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} - 2 x = 0$
$\textcolor{w h i t e}{\text{XXX}} x \left(x - 2\right) = 0$
$\textcolor{w h i t e}{\text{XXX}} x = 0 \mathmr{and} x = 2$