Q.

Let the circle x2+y2=4 intersect x-axis at the points A(a,0), a>0 and B(b,0).  Let P(2cosα,2sinα) 0<α<π2 and Q(2cosβ,2sinβ) be two points such that (α-β)=π2.  Then the point of intersection of AQ and BP lies on :   [2026]

1 x2+y24x4=0  
2 x2+y24y4=0  
3 x2+y24x4y-4=0  
4 x2+y24x4y=0  

Ans.

(2)