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

Correct answer: A. x^ 2 +y^ 2 -4 y-4=0 Let point of intersection R(h,k) m_ BR = m_ BP (k)/(h+2) = (2 α)/(2 α + 2) (k)/(h+2) = (α)/(2) m_ AR = m_ AQ (k)/(h-2) = (2 β)/(2 β - 2) = ( β)/( β - 1) = - (β)/(2) (α)/(2) - (β)/(2) = (π)/(4) ((α)/(2) - (β)/(2)) = (π)/(4) = 1 ( (α)/(2) - (β)/(2))/(1 + (α)/(2) (β)/(2)) = 1 ((k)/(h+2) + (h-2)/(k))/(1 + ((k)/(h+2))((2-h)/(k))) = 1 (k^2 + h^2 - 4)/((k(h+2))/((4)/(h+2))) = 1 (h^2 + k^2 - 4)/(4k) = 1 x^2 + y^2 - 4y - 4 = 0

JEE Main 202628 JanuaryEvening Shift

Exam: JEE Main
Year: 2026
Date: 28 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Circle

JEE Main 2026 · 28 January · Evening Shift · Mathematics · Circle

Let the circle x^ 2 +y^ 2 =4 intersect x -axis at the points A(a,...

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 :

Options

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

Correct answer

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

Step-by-step 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

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