Question 810613
{{{x^2 + y^2 -12x-22y+ 142= 0}}}...first group like terms


{{{(x^2-12x)+ (y^2-22y)+ 142= 0}}}

{{{(x^2-12x+(_))-(_)+(y^2-22y+(_))-(_)+142= 0}}}....complete the square

{{{(x^2-12x+36)-36+ (y^2-22y+121)-121+ 142= 0}}}

{{{(x-6)^2+(y-11)^2-36-121+142 = 0}}}

{{{(x-6)^2+(y-11)^2-15 = 0}}}

{{{(x-6)^2+(y-11)^2=15}}}....=> so, this is a circle with center at ({{{6}}},{{{11}}}) and radius {{{sqrt(15)=3.9}}}


{{{ graph( 600, 600, -15, 15, -15, 15, sqrt(-(x-6)^2+15)+11,- sqrt(-(x-6)^2+15)+11) }}}