Question

How do you find the center and radius for `x^2+y^2 +2x-4y=4`?

Answer 1

Put the equation into standard form to find it to be a circle with radius `3` and center `(-1, 2)`

Explanation:

In order to put the equation of a conic section into standard form (at which point we can see most of its properties), we use a process called completing the square on both variables.

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

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

`=> (x+1)^2 + (y-2)^2 = 4+4+1 = 9`

The standard form of the equation of a circle with radius `r` centered at `(x_0, y_0)` is

`(x-x_0)^2 + (y-y_0)^2 = r^2`

Thus, as our equation may be written as

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

it represents a circle of radius `3`, centered at `(-1,2)`