Exam

JEE Main

Year

2026

Date

21 January

Shift

Evening Shift

Subject

Mathematics

Chapter

Circle

JEE Main 2026 · 21 January · Evening Shift · Mathematics · Circle

If P is a point on the circle x^ 2 +y^ 2 =4, Q is a point on the...

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

Step-by-step solution

Mid point of PQ lies on $x - y + 1 = 0$
$\dfrac{2\cos\theta + \alpha}{2} - \dfrac{2\sin\theta - 5\alpha - 2}{2} + 1 = 0$
$2\cos\theta + \alpha - 2\sin\theta + 5\alpha + 2 + 2 = 0$
$\cos\theta - \sin\theta + 3\alpha + 2 = 0$ ...(1)
$\because$ Slope of PQ is $-1$
$\dfrac{2\sin\theta + 5\alpha + 2}{2\cos\theta - \alpha} = -1$
$2\sin\theta + 5\alpha + 2 = -2\cos\theta + \alpha$
$\sin\theta + \cos\theta + 2\alpha + 1 = 0$ ...(2)
eliminate $\alpha$ from (1) and (2)
$\Rightarrow \cos\theta + 5\sin\theta = 1$ , $\theta \in [0, 2\pi]$
$\Rightarrow 5 \times 2\sin\dfrac{\theta}{2}\cos\dfrac{\theta}{2} = 2\sin^2\dfrac{\theta}{2}$
$\therefore \sin\dfrac{\theta}{2} = 0 \Rightarrow \cos\theta = 1$
or
$\sin\dfrac{\theta}{2} = 5 \Rightarrow \cos\theta = -\dfrac{12}{13}$
Sum of all possible values of abscissa of point P is
$= 2 \times 1 + 2\left(\dfrac{-12}{13}\right) = \dfrac{2}{13}$
$\therefore$ 13 times sum of all possible values of abscissa of point P is 2.