# How do you graph  x² + y² + 4x - 4y - 17 = 0?

Apr 24, 2016

This is a circle with centre $\left(- 2 , 2\right)$ and radius $5$

#### Explanation:

$0 = {x}^{2} + {y}^{2} + 4 x - 4 y - 17$

$= {x}^{2} + 4 x + 4 + {y}^{2} - 4 y + 4 - 25$

$= {\left(x + 2\right)}^{2} + {\left(y - 2\right)}^{2} - {5}^{2}$

Add ${5}^{2}$ to both ends and transpose to get:

${\left(x + 2\right)}^{2} + {\left(y - 2\right)}^{2} = {5}^{2}$

This is (almost) in the form:

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

which is the standard form of the equation of a circle centre $\left(h , k\right)$ and radius $r$.

Hence we find $\left(h , k\right) = \left(- 2 , 2\right)$ and $r = 5$.

This is a circle with centre $\left(- 2 , 2\right)$ and radius $5$.

graph{((x+2)^2+(y-2)^2-5^2)((x+2)^2+(y-2)^2-0.03) = 0 [-11, 11, -3.7, 7.2]}