Condition for formation of azeotropes

Consider a binary azeotropic solution of two volatile components $A, B$. I wanted to try and find out the required concentrations of $A, B$ in the solution in terms of their vapour pressures. I tried using Raoult's Law to do this, but the I got an unexpected answer:

Let the vapour pressure of $A, B$ at the current temperature be $P_A, P_B$, respectively. Let their concentrations be $\chi_A, \chi_B$ and their partial vapour pressures over the solution be $p_A, p_B$.

Then, by Raoult's Law,

$$p_A = P_A\chi_A$$

$$p_B = P_B\chi_B$$

By Dalton's Law of partial pressures, the mole fractions of $A, B$ in the vapor phase are given by $\frac{p_A}{p_A + p_B}$ and $\frac{p_B}{p_A + p_B}$ respectively. Since the solution is azeotropic, these two are equal to their mole fractions in solution.

That is:

$$\frac{P_A\chi_A}{P_A\chi_A + P_B\chi_B} = \chi_A \implies \frac{P_A}{P_B}(1-\chi_A) = \chi_B\tag1$$

Similarly,

$$\frac{P_B}{P_A}(1-\chi_B) = \chi_A \tag2$$

Substituting $(1)$ in $(2)$, yields

$$\frac{P_B}{P_A}\left(1-\frac{P_A}{P_B}(1-\chi_A)\right) = \chi_A$$

$$= \frac{P_B}{P_A}-1+\chi_A = \chi_A$$

$$\implies P_B = P_A$$

But there are numerous azeotropes where this condition is not satisfied. How come?