Question 932792
{{{x^2 - 12x + y^2 - 16y + 84=0}}}

{{{(x^2 - 12x+__) + (y^2 - 16y +__)+ 84=0}}}

since we need {{{b^2}}} to have {{{a^2+2ab+b^2}}}  and we are given {{{(x^2 - 12x+_b^2_)}}} , where {{{a=1}}}, {{{2ab=12}}},then =>{{{2*1*b=12}}}=>{{{b=6}}} and {{{b^2=36}}}

and  {{{(y^2 - 16y +_b^2_)}}}, {{{a=1}}}, {{{2ab=-16}}},then =>{{{2*1*b=-16}}}=>{{{b=-8}}} and {{{b^2=64}}}

so, to complete the square, first write {{{84}}} as {{{36+64-16}}}}, add {{{36}}} to {{{(x^2 - 12x+_b^2_) -b^2}}} and subtract it so the equation doesn't change

{{{(x^2 - 12x+36) -36+ (y^2 - 16y +64)-64+ 84=0}}}

{{{(x^2 - 6)^2 + (y^2 -8)^2-36-64+ 84=0}}}

{{{(x^2 - 6)^2 + (y^2 -8)^2-100+ 84=0}}}

{{{(x^2-6)^2+(y^2-8)^2-16 = 0}}}

{{{(x^2-6)^2+(y^2-8)^2=16 }}}

so, you have a circle with center at ({{{6}}},{{{-8}}}) and radius {{{r=4}}}


{{{drawing( 600, 600, -5, 10, -5, 15,locate(6,8,C(6,8)),circle(6,8,.12), circle(6,8,4), graph( 600, 600, -5, 10, -5, 15,0)) }}}