Question 974306
3x2 + 3y2 - 21x + 6 y + 7 = 0 ?


3x^2 + 3y^2 - 21x + 6 y + 7 = 0


{{{3x^2 + 3y^2 - 21x + 6 y + 7 = 0}}}

{{{3x^2-21x+3y^2+6y+7=0}}}

{{{3(x^2-7y)+3(y^2+2y)+7=0}}}

Complete The Squares:
{{{3(x^2-7y+(7/2)^2)+3(y^2+2y+1)+7=3(49/4)+3}}}

{{{3(x-7/2)^2+3(y+1)^2+7=3(49/4)+3}}}

{{{(x-7/2)^2+(y+1)^2+7/3=49/4+1}}}

{{{(x-7/2)^2+(y+1)^2=49/4+1-7/3}}}

RHS is {{{(49*(3)+12-7(4))/12}}}
{{{131/12}}}
{{{10&11/12}}}


{{{highlight((x-7/2)^2+(y+1)^2=10&11/12)}}}


Center is (7/2,-1) and radius seems to be {{{sqrt(10&11/12)}}} ---- might be computational mistake