# How do you write  x^2 – x + y^2 + y = 0 in standard form and why type of conic is it?

Dec 3, 2015

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

#### Explanation:

Complete the square for both the $x$ and $y$ parts.

color(blue)((x^2-x+?))+color(red)((y^2+y+?))=0

$\textcolor{b l u e}{\left({x}^{2} - x + \frac{1}{4}\right)} + \textcolor{red}{\left({y}^{2} + y + \frac{1}{4}\right)} = \frac{1}{4} + \frac{1}{4}$

We add $\frac{1}{4}$ to both the $x$ and $y$ portions because they will factor into ${\left(x - \frac{1}{2}\right)}^{2}$ and ${\left(y + \frac{1}{2}\right)}^{2}$, respectively. Don't forget to add to the right side the exact same numbers you added to the left so the equation remains balanced.

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

This is a circle, since the standard form of a circle is:

${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$

There's a center at $\left(\frac{1}{2} , - \frac{1}{2}\right)$ and a radius of $\frac{\sqrt{2}}{2}$.