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: