# How do you find the center and radius for x^2+y^2 +2x-4y=4?

May 30, 2016

Put the equation into standard form to find it to be a circle with radius $3$ and center $\left(- 1 , 2\right)$

#### Explanation:

In order to put the equation of a conic section into standard form (at which point we can see most of its properties), we use a process called completing the square on both variables.

${x}^{2} + {y}^{2} + 2 x - 4 y = 4$

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

$\implies {\left(x + 1\right)}^{2} + {\left(y - 2\right)}^{2} = 4 + 4 + 1 = 9$

The standard form of the equation of a circle with radius $r$ centered at $\left({x}_{0} , {y}_{0}\right)$ is

${\left(x - {x}_{0}\right)}^{2} + {\left(y - {y}_{0}\right)}^{2} = {r}^{2}$

Thus, as our equation may be written as

${\left(x - \left(- 1\right)\right)}^{2} + {\left(y - 2\right)}^{2} = {3}^{2}$

it represents a circle of radius $3$, centered at $\left(- 1 , 2\right)$