Question 727644


{{{
16x^2+16y^2-32x+8y=0
}}}


{{{
16(x^2-2x)+16(y^2+(1/2)y)=0
}}}


{{{
16(x^2-2x+1)+16(y^2+(1/2)y+1/16)=0+16*1+16(1/16)
}}}


{{{
(16(x^2-2x+1)+16(y^2+(1/2)y+1/16))/16=(0+16*1+16(1/16))/16
}}}



{{{(x-1)^2+(y+1/4)^2=17/16
}}}




:)