By power of point, $$OE=\frac{OE\cdot OA}{OA}=\frac{OB\cdot OD}{OA}=\frac{\rho_1\rho_3}{1}=\frac{\rho_1\rho_2\rho_3}{\rho_2}=\frac{-1}{\rho_2}$$Therefore, by QMGM $$EC=\sqrt{\rho_2^2+\frac{1}{\rho_2^2}}\geq \sqrt{2}$$Equality can be achieved if $|\rho_2|=1$, and so the polynomial is of the form $(x^2+cx+1)(x+1)$ or $(x^2+dx-1)(x-1)$. If it is of the first form, it suffices for the discriminant of $x^2+cx+1$ to be strictly positive (if it were $0$ we would have two equal solutions both with absolute value $1$, and if it is positive, one must be of absolute value $>1$ and the other $<1$), i.e. $c^2\geq 4$. If it is the second form, the only case with nonnegative discriminant which isn't ok is $x^2+0x-1=(x+1)(x-1)$, so we must have $d^2+4>0$ which is always true (excluding $d=0$). Summing everything up, we must have $$(a,b)\in \{(c+1,c+1)|c\in(-\infty,-2)\cup (2,\infty)\}\cup \{(d-1,-1-d)|d\neq 0\}$$