Question 1008532
the formula of the circle is:

 {{{(x-h)^2+(y-k)^2=r^2}}}

    where {{{h}}} and {{{k}}} are the {{{x}}} and {{{y}}} coordinates of the center of the circle 


{{{2x^2+2y^2-2x+8y=1/2 }}}......both sides divide by {{{2}}}

{{{x^2+y^2-x+4y=1/4 }}}

{{{(x^2-x)+(y^2+4y)=1/4 }}}......complete square

{{{(x^2-x+b^2)-b^2+(y^2+4y+b^2)-b^2=1/4 }}}....recall that {{{(a+-b)^2}}}={{{a^2+2ab+b^2}}} or {{{a^2-2ab+b^2}}}

{{{(x^2-x+b^2)-b^2+(y^2+4y+b^2)-b^2=1/4 }}}...so, in your case {{{a=1}}}, and {{{2ab=1}}} in {{{x}}} part, and {{{a=1}}}, and {{{2ab=4}}} in {{{y}}} part

then,

{{{2ab=1}}}=>{{{2*1*b=1}}}{{{b=1/2}}} in {{{x}}} part

and {{{2ab=4}}}=>{{{2*1*b=4}}}=>{{{b=4/2}}}=> {{{b=2}}} in {{{y}}} part


so, we have


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


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


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


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

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

{{{(x-1/2)^2+(y+2)^2=1/2+8/2 }}}

{{{(x-1/2)^2+(y+2)^2=9/2 }}}

so, it is a circle and 

{{{h=1/2}}}
{{{k=-2}}}
{{{r=sqrt(9/2)=sqrt(9)/sqrt(2)=3/sqrt(2)}}}

the center is at:({{{1/2}}},{{{-2}}}) and radius is {{{r=3/sqrt(2)}}}

{{{drawing( 600, 600, -5, 5, -5, 5,
circle(1/2,-2,.03),locate(1/2,-2,C(1/2,-2)),
 graph( 600, 600, -5, 5, -5, 5,-sqrt(-(x-1/2)^2+9/2)-2 ,sqrt(-(x-1/2)^2+9/2)-2)) }}}