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

A

$$x^{2}+y^{2}-4x-4y=0$$

B

$$x^{2}+y^{2}-4x-4=0$$

C

$$x^{2}+y^{2}-4y-4=0$$

D

$$x^{2}+y^{2}-4x-4y-4=0$$