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.