The acidity of an acid is dependent on the stability of the conjugate base. The more resonance forms you can draw, the better. You've correctly determined that the negative charge is in no way benefiting from having the methoxy group in the para position, from a resonance point of view. However, it would be better to draw how the negative charge is delocalised into the ring (rather than the electrons in the methoxygroup) - this shows the same pattern as plain old phenol. Then you could add a bit about the methoxy group being electron donating as you've done above.