Question

Okay so like for an equation like x^2+y^2+4y-12=0...how would i convert that to the standard form of a circle, (x-h)^2+(y-k)^2=r^2?

Answer 1

Please see below.

Explanation:

.

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

We need to do what is called "Completing the Square" as follows:

Since there is no `x` term, we leave `x^2` as it is. But we will complete the square for `y` because we have a `y` term:

`x^2+(y^2+4y+4)-4-12=0`

What we did was adding a `4` and subtracting a `4`. As such, the net change to the equation is `0`. Doing this allows us to complete the square for `y`:

`x^2+(y+2)^2-16=0`

Now, we can rewrite the `x^2` term as `(x-0)^2`

`(x-0)^2+(y+2)^2-16=0`

`(x-0)^2+(y+2)^2=16`

This equation is now in the form of:

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

If you compare the two equations you will find:

`h=0, k=-2, and r=4`

It is the equation of a circle whose center is at `(0,-2)` and its radius is `=4` as you can see in its graph below: