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

Oct 1, 2017

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

We can gather terms in $x$ and terms in $y$

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

Now we can complete the square fro the $' x$ and $y$ terms independently:

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

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

${\left(x + 2\right)}^{2} - 4 + {\left(y - 2\right)}^{2} = 9$
$\therefore {\left(x + 2\right)}^{2} - 4 + {\left(y - 2\right)}^{2} = {3}^{2}$
Which if we compare to the standard equations of the conics represents circle of radius $3$ centred at $\left(- 2 , 2\right)$