Question

How do you write the standard form of the following equation then graph x² + y² - 4x + 14y - 47 = 0 ?

Answer 1

We do a double completion of square--one with the `x` terms and one with the y terms.

`1(x^2 - 4x + 4- 4) + 1(y^2 + 14y + 49 - 49) - 47 = 0`

`1(x^2 - 4x + 4) - 4 + 1(y^2 + 14y + 49) - 49 - 47 = 0`

`(x - 2)^2 - 4 + (y + 7)^2 - 49 - 47 = 0`

`(x -2)^2 + (y + 7)^2 = 100`

Now recall that when a circle is in the form `(x -a)^2 + (y - b)^2 = r^2`, the centre is given by `(a, b)` and the radius `r`.

Therefore this is a circle with centre `(2, -7)` and radius `10`. It gives the following graph:

Hopefully this helps!