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 α , 2 sin α), 0<α<π/2 and Q (2 cos β, 2 sin β) be two points such that (α-β)=π/2 . Then the point of intersection of AQ and BP lies on :

Mathematics · Circles

JEE Main 2026 — 28 January, Evening Shift — Question 12

Let the circle $x^{2}+y^{2}=4$ intersect $x$ -axis at the points $\mathrm{A}(\mathrm{a}, 0), \mathrm{a}>0$ and $\mathrm{B}(\mathrm{b}, 0)$ . Let $\mathrm{P}(2 \cos \alpha$ , $2 \sin \alpha), 0<\alpha<\frac{\pi}{2}$ and $\mathrm{Q}(2 \cos \beta, 2 \sin \beta)$ be two points such that $(\alpha-\beta)=\frac{\pi}{2}$ . Then the point of intersection of AQ and BP lies on :

Option A:
$x^{2}+y^{2}-4 y-4=0$
Correct
Option B:
$x^{2}+y^{2}-4 x-4=0$
Option C:
$x^{2}+y^{2}-4 x-4 y=0$
Option D:
$x^{2}+y^{2}-4 x-4 y-4=0$

Answer: A

Step-by-step solution

Let point of intersection $\mathrm{R}(\mathrm{h}, \mathrm{k})$ $\begin{aligned} & m_{B R}=m_{B P} \Rightarrow \frac{k}{h+2}=\frac{2 \sin \alpha}{2 \cos \alpha+2} \Rightarrow \frac{k}{h+2}=\tan \frac{\alpha}{2} & \end{aligned}$

$m_{A R}=m_{A Q} \Rightarrow \frac{k}{h-2}=\frac{2 \sin \beta}{2 \cos \beta-2}=\frac{\sin \beta}{\cos \beta-1}=-\cot \frac{\beta}{2}$

$\frac{\alpha}{2}-\frac{\beta}{2}=\frac{\pi}{4}$

$\tan \left(\frac{\alpha}{2}-\frac{\beta}{2}\right)=\tan \frac{\pi}{4}=1$ $\frac{\tan \frac{\alpha}{2}-\tan \frac{\beta}{2}}{1+\tan \frac{\alpha}{2} \tan \frac{\beta}{2}}=1$

$\frac{\frac{k}{h+2}+\frac{h-2}{k}}{1+\left(\frac{k}{h+2}\right)\left(\frac{2-h}{k}\right)}=1$ $\Rightarrow \frac{k^{2}+h^{2}-4}{\frac{k(h+2)}{4}}=1$ $\frac{h^{2}+k^{2}-4}{4 k}=1$

$x^{2}+y^{2}-4 y-4=0$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 28 January, Evening Shift
Subject: Mathematics
Chapter: Circles
Topic: Angle of intersection, Orthogonal circles.