The equation for the tangent to the polynomial is y=m(x1,y1)*x+b(x1,y1) where m(x1,y1) and b(x1,y1) are a function of the tangent point (x1,y1) and can easily be determined from calculus. Therefore, the tangent point on the circle must satisfy the two equations,
y2=m(x1,y1)*x2+b(x1,y1)
(x2-0.9439)^2+(y2-0.1063)^2=0.0537^2
Also, (x1,y1) must satisfy the polynomial equations
P(x1,y1)=0
And you have a 4th equation to express the fact that the normal vector to the tangent line is perpendicular to the tangent line,
(x2-0.9439)-m(x1,y1)*(y2-0.1063)=0
Four nonlinear equations in four unknowns. I expect there will be two solutions.
