Aspire Faculty ID #17934 · Topic: JEE Main 2026 (21 January Evening Shift) · Just now
JEE Main 2026 (21 January Evening Shift)

If $P$ is a point on the circle $x^2 + y^2 = 4$, $Q$ is a point on the straight line $5x + y + 2 = 0$ and $x - y + 1 = 0$ is the perpendicular bisector of $PQ$, then $13$ times the sum of abscissa of all such point $P$ is ______.

Solution




Let $P(2\cos\theta,; 2\sin\theta)$

$Q(\alpha,; -5\alpha - 2)$

Mid point of $PQ$ lies on $x - y + 1 = 0$

$\Rightarrow \frac{2\cos\theta + \alpha}{2} - \frac{2\sin\theta - 5\alpha - 2}{2} + 1 = 0$

$\Rightarrow 2\cos\theta + \alpha - 2\sin\theta + 5\alpha + 2 + 2 = 0$

$\Rightarrow \cos\theta - \sin\theta + 3\alpha + 2 = 0 \quad ...(1)$

$\because$ slope of $PQ = -1$

$\Rightarrow \frac{2\sin\theta + 5\alpha + 2}{2\cos\theta - \alpha} = -1$

$\Rightarrow 2\sin\theta + 5\alpha + 2 = -2\cos\theta + \alpha$

$\Rightarrow \sin\theta + \cos\theta + 2\alpha + 1 = 0 \quad ...(2)$

Eliminate $\alpha$ from (1) and (2)

$\Rightarrow \cos\theta + 5\sin\theta = 1,; \theta \in [0, 2\pi]$

$\Rightarrow 5\times 2\sin\frac{\theta}{2}\cos\frac{\theta}{2} = 2\sin^2\frac{\theta}{2}$

$\therefore \sin\frac{\theta}{2} = 0 \Rightarrow \cos\theta = 1$

or

$\sin\frac{\theta}{2} = \frac{5}{2}\cos\frac{\theta}{2} \Rightarrow \cos\theta = -\frac{12}{13}$

Sum of all possible values of abscissa of point $P$ is

$= 2\cdot 1 + 2\left(-\frac{12}{13}\right) = \frac{2}{13}$

$\therefore 13$ times sum of all values of abscissa of point $P$ is $2$
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