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 : [28 Jan 2026 Shift 2]

Correct Answer is : x2+y2−4y−4=0

Circles Part 2

Section: Mathematics

Question : 1 of 78

Marks: +1, -0

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<‌

and Q(2‌cos‌β,2sin‌β) be two points such that (α−β)=‌

. Then the point of intersection of AQ and BP lies on :

[28 Jan 2026 Shift 2]

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