# How do you convert x^2+y^2+6x-2y-3=0 to standard form?

Dec 19, 2015

${\left(x - \left(- 3\right)\right)}^{2} + {\left(y - 1\right)}^{2} = {\left(\sqrt{13}\right)}^{2}$
is the standard form for a circle with center $\left(- 3 , 1\right)$ and radius $\sqrt{13}$

#### Explanation:

Given ${x}^{2} + {y}^{2} + 6 x - 2 y - 3 = 0$

Re-group the terms as
${x}^{2} + 6 x \textcolor{w h i t e}{\text{XX")+color(white)("XX")y^2-2ycolor(white)("XX")-3 color(white)("XXX}} = 0$

Move the constant to the right side and complete the squares
${x}^{2} + 6 x + 9 \textcolor{w h i t e}{\text{XX")+color(white)("XX")y^2-2y+1color(white)("XX}} = 3 + 9 + 1$

Rewrite as squared terms
${\left(x + 3\right)}^{2} + {\left(y - 1\right)}^{2} = 13$

graph{x^2+y^2+6x-2y-3=0 [-8.684, 7.12, -2.95, 4.95]}