Intersection points on circle

Suppose the equation of the circle is $(x-h)^2+(y-k)^2=r^2$

Since it intersects with the graph $f(x)=x^2$ , the above equation can be rewritten as $(x-h)^2+(x^2-k)^2=r^2$

Note that the coefficient of $x^3$ of above quartic equation is 0.

Let the $x$ -coordinates of 4 intersection points be $\displaystyle \alpha_1, \alpha_2, \alpha_3$ and $\displaystyle\alpha_4$ . Then $\alpha_1+\alpha_2+\alpha_3+\alpha_4 = \color{#D61F06}0$ . On the other hand, from $g(x) = x^4 + ax^3 -2x^2+ bx +1$ , the sum of roots is $\displaystyle -\frac{a}{1}$ . So $\displaystyle a=0$ and hence $\lfloor{1000ab}\rfloor=0$ .