Question 661589
standard form of equation for a circle: 
(x-h)^2+(y-k)^2=r^2, (h,k)=(x,y) coordinates of center, r=radius
..
2x^2+2y^2+16x+8y−32=0
complete the square:
2x^2+16x+2y^2+8y−32=0
2(x^2+8x+16)+2(y^2+4y+4)=32+32+8
2(x+4)^2+2(y+2)^2=72
divide by 2
(x+4)^2+(y+2)^2=36 (equation of given circle)
h=-4
k=-2
r=6