Question

The equation of a circle is `x²+y²+2x-4y-8=0`, what are the co-ordinates of the center and the length of the radius of the circle?

Answer 1

Let's first convert to standard form by doing a double completion of square. This form is much easier to work with.

Explanation:

`x^2 + 2x + y^2 - 4y = 8`

`1(x^2 + 2x + 1 - 1) + (y^2 - 4y + 4 - 4) = 8`

`(x + 1)^2 - 1 + (y - 2)^2 - 4 = 8`

`(x + 1)^2 + (y - 2)^2 = 13`

In standard form `a = (x - p)^2 + (y - q)^2`, the center is at `(p, q)` and the radius is at `a`.

Therefore, the center is at (-1, 2) and the radius measures 13 units.

Hopefully this helps!