# How do you write the equation of the circle in standard form, identify the center and radius of x^2+y^2-2x+6y+9=0?

Mar 16, 2018

${\left(x - 1\right)}^{2} + {\left(y + 3\right)}^{2} = 1$

#### Explanation:

$\text{the equation of a circle in standard form is }$

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{{\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

$\text{where "(a,b)" are the coordinates of the centre and r}$
$\text{is the radius}$

$\text{to obtain this form use "color(blue)"completing the square}$
$\text{on both x and y terms}$

${x}^{2} - 2 x + {y}^{2} + 6 y + 9 = 0$

$\Rightarrow \left({x}^{2} + 2 \left(- 1\right) x \textcolor{red}{+ 1} \textcolor{red}{- 1}\right) + \left({y}^{2} + 2 \left(3\right) y \textcolor{b l u e}{+ 9} \textcolor{b l u e}{- 9}\right) + 9 = 0$

$\Rightarrow {\left(x - 1\right)}^{2} + {\left(y + 3\right)}^{2} = 0 - 9 \textcolor{b l u e}{+ 9} \textcolor{red}{+ 1}$

$\Rightarrow {\left(x - 1\right)}^{2} + {\left(y + 3\right)}^{2} = 1 \leftarrow \textcolor{red}{\text{in standard form}}$

$\Rightarrow \text{centre "=(1,-3)" and radius } = 1$