What is the center and radius of `x^2+y^2=4y-11-8x`?

Notice that although this is a circle, it isn't in the familiar form that we know:

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

To get it into this form, we will complete the square for both `x` and `y`.

`x^2 + y^2 = 4y - 11 - 8x`

`x^2 + 8x + y^2 - 4y = - 11`

`x^2 + 8x + color(red)16 + y^2 - 4y + color(blue)4 = - 11 + color(red)16 + color(blue)4`

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

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

`therefore` the centre of the circle is `(-4, 2)` and the radius is `3`.

In order to not make this problem painful and difficult for ourselves, we should get it into the standard form for a circle, which is

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

How do we do this, you ask? Let's first get everything on one side EXCEPT the number:

`x^2+y^2+8x-4y=-11`

Some rearranging...

`x^2+8x+y^2-4y=-11`

....and we are ready to complete the square! If you need a refresher, we're going to create perfect square binomials. The two in our formula above are perfect square binomials. We will create them by dividing the "middle" coefficient for each variable by 2 and then squaring the result:

`x^2+8x+?`
`8/2=4`
`4^2=16`
`x^2+8x+16=(x+4)^2`

`y^2-4y+?`
`-4/2=-2`
`(-2)^2=4`
`y^2-4y+4=(y-2)^2`

We're not done, though. Completing the square requires that `16` and `4` be added to `-11` as well:

`x^2+8x+16+y^2-4y+4=-11+16+4`
`(x+4)^2+(y-2)^2=9`

Hooray! We're in standard form. Let's find our answers. The center will be `(-4,2)` since the equation includes the opposites of the center's coordinates. The radius is the square root of `9`, which is `3`.

Hope this helped you!