# How do you write the equation x^2+x+y^2-6/5y=1 in standard form and find the center and radius?

Apr 8, 2017

${\left(x + \frac{1}{2}\right)}^{2} + {\left(y - \frac{3}{5}\right)}^{2} = \frac{161}{100}$

#### Explanation:

The standard form of the $\textcolor{b l u e}{\text{equation of a circle}}$ is.

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{{\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
where (a ,b) are the coordinates of the centre and r, the radius.

$\text{Rearrange the given equation into this form}$

$\text{Using the method of "color(blue)"completing the square}$

$\left({x}^{2} + x \textcolor{red}{+ \frac{1}{4}}\right) + \left({y}^{2} - \frac{6}{5} y \textcolor{m a \ge n t a}{+ \frac{9}{25}}\right) = 1 \textcolor{red}{+ \frac{1}{4}} \textcolor{m a \ge n t a}{+ \frac{9}{25}}$

$\Rightarrow {\left(x + \frac{1}{2}\right)}^{2} + {\left(y - \frac{3}{5}\right)}^{2} = \frac{161}{100}$

$\text{here " a=-1/2,b=3/5" and } {r}^{2} = \frac{161}{100}$

$\Rightarrow \text{centre } = \left(- \frac{1}{2} , \frac{3}{5}\right)$

$\text{and " r=sqrt(161/100)=sqrt161/10~~1.27" to 2 dec. places}$
graph{(x+1/2)^2+(y-3/5)^2-161/100=0 [-10, 10, -5, 5]}