Let the circle ${x}^{2}+{y}^{2}=4$ intersect $x$-axis at the points $A(a,0),a0$ and $B(b,0)$. Let $P(2\cos \alpha ,2\sin \alpha ),$and $Q(2\cos \beta ,2\sin \beta )$ be two points such that $\left(\alpha −\beta \right)=\frac{\pi }{2}.$ Then the point of intersection of $AQ$ and $BP$ lies on: [JEE Main 2026, 28 Jan (Shift 2)]

Question

Let the circle ${x}^{2}+{y}^{2}=4$ intersect $x$-axis at the points $A(a,0),a>0$ and $B(b,0)$. Let $P(2\cos \alpha ,2\sin \alpha ),$and $Q(2\cos \beta ,2\sin \beta )$ be two points such that $\left(\alpha −\beta \right)=\frac{\pi }{2}.$ Then the point of intersection of $AQ$ and $BP$ lies on: [JEE Main 2026, 28 Jan (Shift 2)]

YESPYQ · Accepted Answer

Options are free to see. Unlock the correct answer and full explanation with PYQ Pass.

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 \al…

Practice more JEE Maths PYQs