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 α, 2 sin α), 0<α<(π)/2 and Q(2 cos β, 2 sin β)…

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