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

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Answer 1 · 2017-10-01T15:25:59

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

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

$(x^2 + 4x) + (y^2 - 4y) = 1$

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

$(x+2)^2-2^2 + (y - 2)^2-2^2 = 1$

$:. (x+2)^2-4 + (y - 2)^2-4 = 1$

Leading to:

$(x+2)^2-4 + (y - 2)^2 = 9$

$:. (x+2)^2-4 + (y - 2)^2 = 3^2$

Which if we compare to the standard equations of the conics represents circle of radius $3$ centred at $(-2,2)$

graph{x^2 + y^2 + 4x - 4y - 1 = 0 [-9.26, 4.784, -1.27, 5.75]}

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