Ineed finding the center and the radius

So you solve for the standard form of an circle which is ((a^2+b^2)=radius^2 or ((a-c)^2)+((b+-d)^2)
x^2+10x+y^2+8y+32=0
so complete the square
x^2+10+- z can be written in (a-c)^2 or (a+c)^2 form so
we notice that 5+5=10 and 5*5 =25 so we put it in
x^2+10+25=(x+5)(x+5) or (x+5)^2
then we subtract 25 from the side that we are on
(x+5)^2+y^2+8y+7=0
factor the y^2+8y+t=0
two numbers that add to 8 are 4 and 4 so t must be 16
y^2+8t+16=(y+4)^2
so we subtract 16 from the equation
(x+5)^2+(y+4)^2-9=0
add 9 to both sides
(x+5)^2+(y+4)^2=9
so in standard form
(x+5)^2+(y+4)^2=3^2
the radius is 3
we plot the points
by trial and error
(x,y)
(-2,-4)
(-8,-4)
(-5,-1)
(-5,-7)
so the total height is 6 units or the radius is 3 and the width is 6 so the center is
-8+3=-5 -7+3=-4
center is (-5,-3)
radius is 3