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

Question

4372 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-22T22:11:39

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

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

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

$= (x+2)^2+(y-2)^2-3^2$

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

$(x-(-2))^2+(y-2)^2 = 3^2$

This is in the form:

$(x-h)^2 + (y-k)^2 = r^2$

the standard form of the equation of a circle with centre $(h, k) = (-2, 2)$ 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]}

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