Question

How do you find the center and radius for `x^2 + y^2 - 8x - 2y - 8 = 0`?

Answer 1

The center is `(4, 1)` and the radius `3`.

Explanation:

`0 = x^2+y^2-8x-2y-8`

`= x^2-8x+16 + y^2-2y+1 - 9`

`= (x-4)^2+(y-1)^2-3^2`

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

`(x-4)^2+(y-1)^2 = 3^2`

This is in the form:

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

which is the general form of the equation of a circle with centre `(h, k) = (4, 1)` and radius `r=3`