Question

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

Answer 1

This is a circle with centre `(-2, 2)` and radius `5`

Explanation:

`0 = x^2+y^2+4x-4y-17`

`=x^2+4x+4+y^2-4y+4-25`

`=(x+2)^2+(y-2)^2-5^2`

Add `5^2` to both ends and transpose to get:

`(x+2)^2+(y-2)^2=5^2`

This is (almost) in the form:

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

which is the standard form of the equation of a circle centre `(h,k)` and radius `r`.

Hence we find `(h,k) = (-2,2)` and `r=5`.

This is a circle with centre `(-2, 2)` 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]}