# What is the standard form of x^2 - 2x + y^2 + 4y = 11?

Dec 5, 2015

${\left(x - 1\right)}^{2} + {\left(y - \left(- 2\right)\right)}^{2} = {\left(\sqrt{6}\right)}^{2}$

#### Explanation:

Recognizing the given equation ${x}^{2} - 2 x + {y}^{2} + 4 y = 11$
as the equation of a circle
and
knowing that the standard form of a circle with center $\left(a , b\right)$ and radius $r$ is
$\textcolor{w h i t e}{\text{XXX}} {\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}$

We need to complete the squares for each of $x$ and $y$ in the given form:
$\textcolor{w h i t e}{\text{XXX}} \left({x}^{2} - 2 x + 1\right) - 1 + \left({y}^{2} + 4 y + 4\right) - 4 = 11$

$\textcolor{w h i t e}{\text{XXX}} {\left(x - 1\right)}^{2} + {\left(y + 2\right)}^{2} - 5 = 11$

$\textcolor{w h i t e}{\text{XXX}} {\left(x - 1\right)}^{2} + {\left(y + 2\right)}^{2} = 6$

Normally I would consider the above "close enough" to standard form;
however to be completely accurate:
$\textcolor{w h i t e}{\text{XXX}} {\left(x - \left(1\right)\right)}^{2} + {\left(y - \left(- 2\right)\right)}^{2} = {\left(\sqrt{6}\right)}^{2}$
with center at $\left(1 , - 2\right)$ and radius $\sqrt{6}$