Circle

Let the circle $x^2 + y^2 = 4$ intersect x-axis at the points A$(a, 0)$, $a > 0$ and B$(b, 0)$. Let $P(2 \cos \alpha, 2 \sin \alpha)$, $0 < \alpha < \frac{\pi}{2}$ and $Q(2 \cos \beta, 2 \sin \beta)$ be two points such that $(\alpha - \beta) = \frac{\pi}{2}$. Then the point of intersection of AQ and BP lies on :