# How do you graph x^2 + y^2 + 4x - 4y - 1 = 0?

Jan 22, 2016

This is the equation of a circle centre $\left(- 2 , 2\right)$ and radius $3$

#### Explanation:

$0 = {x}^{2} + {y}^{2} + 4 x - 4 y - 1$

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

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

Add ${3}^{2}$ to both ends and transpose to get:

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

This is in the form:

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

the standard form of the equation of a circle with centre $\left(h , k\right) = \left(- 2 , 2\right)$ and radius $r = 3$

graph{(x^2+y^2+4x-4y-1)((x+2)^2+(y-2)^2-0.01)=0 [-12.33, 7.67, -3.08, 6.92]}