Question

How do you graph `x^2 + y^2 - 4x + 6y - 12 = 0`?

Answer 1

Rearrange into the standard form of the equation of a circle with centre `(2, -3)` and radius `5`.

Explanation:

`0 = x^2+y^2-4x+6y-12`

`=(x^2-4x+4)+(y^2+6y+9)-25`

`=(x-2)^2+(y+3)^2-5^2`

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

`(x-2)^2+(y-(-3))^2 = 5^2`

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, -3)` and radius `r=5`

graph{(x^2+y^2-4x+6y-12)((x-2)^2+(y+3)^2-0.02)=0 [-15.12, 16.92, -11.33, 4.69]}