Coordinate Geometry – Circle – JEE Main 28 Jan 2026 Shift 2

Solution:

For circle $x^2+y^2=4$, radius $r=2$. The intersection points with the x-axis are $A(2,0)$ and $B(-2,0)$.

Equation of line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$ is $y-y_1 = m(x-x_1)$.
Slope of line $BP$ (Points $B(-2,0)$ and $P(2\cos\alpha, 2\sin\alpha)$):
$$m_1 = \frac{2\sin\alpha – 0}{2\cos\alpha – (-2)} = \frac{\sin\alpha}{\cos\alpha+1} = \frac{2\sin(\frac{\alpha}{2})\cos(\frac{\alpha}{2})}{2\cos^2(\frac{\alpha}{2})} = \tan\frac{\alpha}{2}$$
Line $BP$: $y = \tan(\frac{\alpha}{2})(x+2)$
$$\tan\frac{\alpha}{2} = \frac{y}{x+2} \quad …(i)$$

Slope of line $AQ$ (Points $A(2,0)$ and $Q(2\cos\beta, 2\sin\beta)$):
$$m_2 = \frac{2\sin\beta – 0}{2\cos\beta – 2} = \frac{\sin\beta}{\cos\beta-1} = \frac{2\sin(\frac{\beta}{2})\cos(\frac{\beta}{2})}{-2\sin^2(\frac{\beta}{2})} = -\cot\frac{\beta}{2}$$
Line $AQ$: $y = -\cot(\frac{\beta}{2})(x-2)$
$$\tan\frac{\beta}{2} = -\frac{x-2}{y} = \frac{2-x}{y} \quad …(ii)$$

Given $\alpha – \beta = \frac{\pi}{2} \Rightarrow \frac{\alpha}{2} – \frac{\beta}{2} = \frac{\pi}{4}$.
Using $\tan(A-B) = \frac{\tan A – \tan B}{1 + \tan A \tan B}$:
$$\frac{\tan\frac{\alpha}{2} – \tan\frac{\beta}{2}}{1 + \tan\frac{\alpha}{2}\tan\frac{\beta}{2}} = 1$$
Substitute values from $(i)$ and $(ii)$:
$$\frac{\frac{y}{x+2} – \frac{2-x}{y}}{1 + \left(\frac{y}{x+2}\right)\left(\frac{2-x}{y}\right)} = 1$$
$$\frac{y^2 – (2-x)(x+2)}{y(x+2) + y(2-x)} = 1$$
$$\frac{y^2 – (4-x^2)}{y(x+2+2-x)} = 1$$
$$\frac{x^2+y^2-4}{4y} = 1$$
$$x^2+y^2-4 = 4y$$
$$x^2+y^2-4y-4 = 0$$

Ans. (1)