Mathematics · Circles

JEE Main 2026 — 4 April, Morning Shift — Question 36

Suppose the two chords, drawn from the point (1, 2) on the circle $\mathbf{x}^2 +\mathbf{y}^2 +\mathbf{x} - 3\mathbf{y} = 0$ are bisected by the y-axis. If the other ends of these chords are R and S, and the mid point of the line segment RS is $(\alpha ,\beta)$ , then $6(\alpha +\beta)$ is equal to:

Option A:
1
Option B:
3
Correct
Option C:
4
Option D:
6

Answer: B

Step-by-step solution

chord PS bisected by y -axis at point $\mathrm{M}(0, \lambda)$ $\therefore \mathrm{S}(-1,2 \lambda-2)$ Which satisfy equation of circle $\Rightarrow(-1)^{2}+(2 \lambda-2)^{2}-1-3(2 \lambda-2)=0$ $\Rightarrow \lambda=1$ and

$\lambda=\frac{5}{2}$ Therefore two such chord PS & PR will exist When $\lambda=1 \Rightarrow \mathrm{~S}(-1,0)$ When $\lambda=\frac{5}{2} \mathrm{R}(-1,3)$ ∴ Mid point of $R$ & $S$ will be $\left(-1, \frac{3}{2}\right)$ $\therefore \alpha=-1, \beta=\frac{3}{2}$ $\therefore 6(\alpha+\beta)=3$

Answer key and solution verified before publishing.

Practise Circles

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2026
Paper: 4 April, Morning Shift
Subject: Mathematics
Chapter: Circles
Topic: Introduction to Circles