Q36 / 75 · JEE Main 2026 (28 January Shift 2) · Mathematics

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 :

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

Solution

Let point of intersection $R(h,k)$
$m_{BR} = m_{BP} \Rightarrow \dfrac{k}{h+2} = \dfrac{2\sin\alpha}{2\cos\alpha + 2} \Rightarrow \dfrac{k}{h+2} = \tan\dfrac{\alpha}{2}$
$m_{AR} = m_{AQ} \Rightarrow \dfrac{k}{h-2} = \dfrac{2\sin\beta}{2\cos\beta - 2} = \dfrac{\sin\beta}{\cos\beta - 1} = -\cot\dfrac{\beta}{2}$
$\dfrac{\alpha}{2} - \dfrac{\beta}{2} = \dfrac{\pi}{4}$
$\tan\left(\dfrac{\alpha}{2} - \dfrac{\beta}{2}\right) = \tan\dfrac{\pi}{4} = 1$
$\dfrac{\tan\dfrac{\alpha}{2} - \tan\dfrac{\beta}{2}}{1 + \tan\dfrac{\alpha}{2} \tan\dfrac{\beta}{2}} = 1$
$\dfrac{\dfrac{k}{h+2} + \dfrac{h-2}{k}}{1 + \left(\dfrac{k}{h+2}\right)\left(\dfrac{2-h}{k}\right)} = 1 \Rightarrow \dfrac{k^2 + h^2 - 4}{\dfrac{k(h+2)}{\dfrac{4}{h+2}}} = 1$
$\dfrac{h^2 + k^2 - 4}{4k} = 1$
$x^2 + y^2 - 4y - 4 = 0$

Hence, the correct answer is option: 'A'.

Relevant Questions