Question 1207523


 

a circle formula is

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


so, complete squares in given equation


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

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

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

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

for {{{x }}}part {{{b=(1/2)/2=1/4}}}

for {{{y}}} part {{{b=(2/3)/2=1/3}}}


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

{{{2(x-1/4)^2-2(1/16)+3(y+1/3)^2-3(1/9)=-1}}}

{{{2(x-1/4)^2-1/8+3(y+1/3)^2-1/3=-1}}}

{{{2(x-1/4)^2+3(y+1/3)^2=-1+1/8+1/3}}}

{{{2(x-1/4)^2+3(y+1/3)^2=-13/24}}}

as you can see, we cannot get a formula of a circle which means given equation is NOT a circle