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<2π and Q(2cosβ,2sinβ) be two points such that (α−β)=2π. Then the point of intersection of AQ and BP lies on :
Held on 28 Jan 2026 · Verified 6 Jul 2026.
x2+y2−4y−4=0
x2+y2−4x−4y=0
x2+y2−4x−4=0
x2+y2−4x−4y−4=0
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