# How do you write the equation of the circle in standard form x^2 + y^2 + 2x + 4y - 4091 = 0?

Mar 13, 2018

Completing the square in $x$ and $y$. See below
${x}^{2} + {y}^{2} + 2 x + 4 y - 4091 = {\left(x + 1\right)}^{2} + {\left(y + 2\right)}^{2} - 1 - 4 - 4091 = {\left(x + 1\right)}^{2} + {\left(y + 2\right)}^{2} - 4096 = {\left(x + 1\right)}^{2} + {\left(y + 2\right)}^{2} - {64}^{2}$
${\left(x + 1\right)}^{2} + {\left(y + 2\right)}^{2} = {64}^{2}$ this is the equation of a circle with center in $\left(- 1 , - 2\right)$ and radius $r = 64$