Question 815906
Try {{{y=-3/x}}} and substitute into the circle equation:

{{{x^2+(-3/x)^2=12}}}
{{{x^2+9/(x^2)=12}}}
{{{x^4+9=12x^2}}}
{{{x^4-12x^2+9=0}}} not factorable.


First solve for x^2 according to solution to quadratic formula;
{{{x^2=(12+- sqrt(144-4*9))/(2)}}}
{{{x^2=(12+- sqrt(108))/2}}}
{{{x^2=(12+- sqrt(4*9*3))/2}}}
{{{x^2=6+-3*sqrt(3)}}}


and then
x=-(6-3*sqrt(3))  or x=(6-3*sqrt(3))  or x=-(6+3*sqrt(3))   or   x=(6+3*sqrt(3))
'
FINAL ANSWER FOR JUST x:   {{{ x=-6+3*sqrt(3)}}} or {{{x=6-3*sqrt(3)}}}  or {{{x=-6-3*sqrt(3)}}}  or  {{{x=6+3*sqrt(3)}}}