Question 730930
{{{2x^2+2y-12x+4y-15=0}}}......here you missing {{{y^2}}} to have a circle

I guess {{{2y}}} should be {{{2y^2}}}, but it also might be {{{4y^2}}}

let's do it with {{{2y^2}}} 

{{{2x^2+2y^2-12x+4y-15=0}}}....write it in the standard form

The standard equation of a circle with center C({{{h}}},{{{k}}}) and radius {{{r}}} is as follows:

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

{{{2x^2+2y^2-12x+4y-15=0}}}..put all terms with {{{x}}} and {{{x^2}}} together and all terms with {{{y}}} and {{{y^2 }}} together using brackets

{{{(2x^2-12x)+(2y^2+4y)-15=0}}}......complete the square within each bracket

{{{(2x^2-12x+18)-18+(2y^2+4y+2)-2-15=0}}}

{{{2(x^2-6x+9)+2(y+2x+1)-35=0}}}

{{{2(x-3)^2+2(y+1)^2-35=0}}}

{{{2(x-3)^2+2(y+1)^2=35}}} ....both sides divide by {{{2}}}


{{{2(x-3)^2/2+2(y+1)^2/2=35/2}}} 

{{{(x-3)^2+(y+1)^2=17.5}}} 


so,the center and radius of the circle are:

C({{{3}}},{{{-1}}})

{{{r=4.18}}}

{{{ drawing(600, 600, -6, 10, -10, 10,circle( 3, -1, 4.18 ),circle(3,-1,0.1 ),graph( 600, 600, -6, 10, -10, 10, 0)) }}}