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 :

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

JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Circle

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 < \frac{π}{2}

and

Q (2 cos β, 2 sin β)

be two points such that

(α - β) = \frac{π}{2}

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

Answer: (A)

The circle meets the

x

-axis at

A (2, 0)

and

B (- 2, 0)

. Let

R (h, k)

be the intersection of

A Q

and

B P

. The slope of

B P

\frac{2 sin α}{2 cos α + 2} = tan \frac{α}{2}

, so

\frac{k}{h + 2} = tan \frac{α}{2}

. The slope of

A Q

\frac{2 sin β}{2 cos β - 2} = - cot \frac{β}{2}

, so

\frac{k}{h - 2} = - cot \frac{β}{2}

, giving

tan \frac{β}{2} = - \frac{h - 2}{k}

. Because

\frac{α}{2} - \frac{β}{2} = \frac{π}{4}

tan (\frac{α}{2} - \frac{β}{2}) = 1

\frac{tan \frac{α}{2} - tan \frac{β}{2}}{1 + tan \frac{α}{2} tan \frac{β}{2}} = 1.

Substituting

tan \frac{α}{2} = \frac{k}{h + 2}

and

tan \frac{β}{2} = - \frac{h - 2}{k}

and clearing denominators gives

k^{2} + (h - 2) (h + 2) = 4 k

, that is

h^{2} + k^{2} - 4 k - 4 = 0

. The point of intersection lies on

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

JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Single correct option