How do you graph x^2+y^2-8x+6y+16=0?

##0 = x^2+y^2-8x+6y+16 = (x-4)^2 + (y+3)^2 – 3^2##
is a circle of radius ##3## with centre ##(4 -3)##
The equation of a circle of radius ##r## centred at ##(a b)## can be written:
##(x-a)^2+(y-b)^2 = r^2##
We are given:
##0 = x^2+y^2-8x+6y+16##
##=x^2-8x+16 + y^2+6y+9 – 9##
##=(x-4)^2 + (y+3)^2 – 3^2##
So:
##(x-4)^2+(y+3)^2 = 3^2##
which is in the form of the equation of a circle of radius ##3## centre ##(4 -3)##
graph{(x-4)^2+(y+3)^2 = 3^2 [-7.875 12.125 -7.8 2.2]}