JEE Mathematics 2026 Question with Solution

Given: The circle is $x^2+y^2=4$. Its intersections with the $x$-axis are $A(-2,0)$ and $B(2,0)$. Also, $P(2\cos\alpha,2\sin\alpha)$ and $Q(2\cos\beta,2\sin\beta)$ lie on the circle with $(\alpha-\beta)=\frac{\pi}{2}$.

Find: The locus on which the intersection point of lines $AQ$ and $BP$ lies.

From the circle $x^2+y^2=4$, the points on the $x$-axis are:

$$A(-2,0),\quad B(2,0)$$

Using

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

the solution gives

$$\cos\beta=\sin\alpha,\quad \sin\beta=-\cos\alpha$$

Hence,

$$Q=(2\sin\alpha,-2\cos\alpha)$$

Now form the equations of lines $AQ$ and $BP$ using the two-point form. On solving these two line equations simultaneously by elimination, the intersection point $(x,y)$ satisfies

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

This relation does not depend on $\alpha$, so it is the required locus.

Therefore, the point of intersection lies on $x^2+y^2-4x-4y-4=0$. Hence, the correct option is A.