# How do you write the equation x^2 + y^2 – x + 2y + 1 = 0 into standard form?

Dec 10, 2015

${\left(x - \frac{1}{2}\right)}^{2} + {\left(y + 1\right)}^{2} = \frac{1}{4}$

#### Explanation:

Sort the $x$-terms and $y$-terms.

$\textcolor{b l u e}{{x}^{2} - x} + \textcolor{red}{{y}^{2} + 2 y} = - 1$

Now, complete the square for each variable.

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

Don't forget to balance both sides of the equation. (If you add something to one side, add it to the other side as well.)

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

This is in the standard form of a circle

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

Where:
$\left(h , k\right)$ is the center and $r$ is the length of the radius.

Thus, this is a circle with a center at $\left(\frac{1}{2} , - 1\right)$ and a radius of $\frac{1}{2}$. Here's a graph. (Notice the center at $\left(0.5 , - 1\right)$ and a diameter of $1$.)