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:

Identify points. The circle has radius $$2$$. Intersections with the x-axis are $$A(2, 0)$$ and $$B(-2, 0)$$.

Points $$P$$ and $$Q$$ are $$(2\cos \alpha, 2\sin \alpha)$$ and $$(2\cos \beta, 2\sin \beta)$$ with $$\alpha - \beta = \pi/2$$.

Equation of $$AQ$$ (through $$(2,0)$$ and $$(2\cos \beta, 2\sin \beta)$$):

$$y - 0 = \frac{2\sin \beta}{2\cos \beta - 2}(x - 2) \implies y = \frac{\sin \beta}{\cos \beta - 1}(x - 2)$$

Using half-angle formulas: $$y = -\cot(\beta/2)(x - 2)$$.

Equation of $$BP$$ (through $$(-2,0)$$ and $$(2\cos \alpha, 2\sin \alpha)$$):

$$y - 0 = \frac{2\sin \alpha}{2\cos \alpha + 2}(x + 2) \implies y = \tan(\alpha/2)(x + 2)$$

Relate $$\alpha$$ and $$\beta$$.

Since $$\alpha = \beta + \pi/2$$, then $$\tan(\alpha/2) = \tan(\beta/2 + \pi/4) = \frac{1 + \tan(\beta/2)}{1 - \tan(\beta/2)}$$.
 Solve for the intersection.

By substituting the trigonometric relations and eliminating the parameters, we find the point $$(x, y)$$ satisfies:

Correct Option: C ($$x^2 + y^2 - 4y - 4 = 0$$)