Find the radius and center of a circle with the equation `x^2 + y^2 - 8x + 2y + 8 = 0` ?

Given
`color(white)("XXX")x^2+y^2-8x+2y+8=0`

Noting that if we can convert this into the standard form for a circle:
`color(white)("XXX")(x-a)^+(y-b)^2=c^2`
we will have a circle with radius `c` and center `(a,b)`

`x^2+y^2-8x+2y+8=0`

`rarr (color(blue)(x^2-8x))+(color(red)(y^2+2y))=-8`

`rarr (color(blue)(x^2-8x+4^2))+(color(red)(y^2+2y+1^2))=-8color(blue)(+4^2)color(red)(+1^2)`

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

`rarr (x-4)^2+(y-(-1))^2=3^2`
a circle with center `(4,-1)` and radius `3`

~~~~~~~~~~ Process of "Completing the Squares" ~~~~~~~~~~~~~
If we want to convert an expression in the form:
`color(white)("XXX")x^2+ax`
into a squared binomial with the form:
`color(white)("XXX")(x+b)^2`
Since `(x+b)^2=x^2+2bx+b^2`
we will need to add some amount `b^2` to the initial expression `x^2+ax`
where `2bx=ax`
(this amount will, of course, need to also be added to the other side of the equation).

For the given equation of this problem, let's simplify by dropping the `color(red)y` term and consider only what we would have with
`color(white)("XXX")color(blue)(x^2-8x=-8)`
To convert `(x^2-8x)` into a squared binomial we will need to add `color(magenta)(((-8)/2)^2)` (to both sides) to get
`color(white)("XXX")color(blue)(x^2-8x)color(magenta)(+4^2)=color(blue)(-8)color(magenta)(+4^2)`

`color(white)("XXX")(x-4)^2= -8 +16`

Similarly to convert `color(red)(y^2+2y)` into a squared binomial, we will need to add `color(magenta)((color(red)(+2/)2)^2)` to both sides.