If a circular equation is written in the form:

#color(white)("XXX")(x-color(red)(a))^2+(y-color(blue)(b))^2=color(green)(r)^2#

then it has a center at #(color(red)(a),color(blue)(b))# and a radius of #color(green)(r)#

We will want to manipulate the given: #x^2+y^2+8x+4y+16=0#

into this form.

First separating the #x# terms, the #y# terms and the constant as

#color(white)("XXX")(color(red)(x^2+8x))+(color(blue)(y^2+4y))=color(green)(-16)#

Completing the square for each of the #x# and #y# sub-expressions:

#color(white)("XXX")(color(red)(x^2+8x+16))+(color(blue)(y^2+4y+4))=color(green)(-16)color(red)(+16)color(blue)(+4)#

#color(white)("XXX")(color(red)(x+4))^2+(color(blue)(y+2))^2= color(green)(4#

#color(white)("XXX")(x-(color(red)(-4)))^2+(y-(color(blue)(-2)))^2=color(green)(2)^2#

with center at #(color(red)(-4),color(blue)(-2))# and radius #color(green)(2)#

A graph of the original equation helps verify this result:

graph{x^2+y^2+8x+4y+16=0 [-9.184, 1.916, -4.89, 0.66]}