Mathematics · Circles

JEE Main 2024 — 4 April, Shift 2 — Question 3

Let C be a circle with radius $\sqrt{10}$ units and centre at the origin. Let the line $x+y=2$ intersects the circle C at the points P and Q . Let MN be a chord of C of length 2 unit and slope -1 . Then, a distance (in units) between the chord PQ and the chord MN is

Option A:
$2-\sqrt{3}$
Option B:
$3-\sqrt{2}$
Correct
Option C:
$\sqrt{2}-1$
Option D:
$\sqrt{2}+1$

Answer: B

Step-by-step solution

$C: x^{2}+y^{2}=10$

$\mathrm{AN}=\frac{\mathrm{MN}}{2}=1$

$\therefore$ In $\triangle \mathrm{OAN} \rightarrow(\mathrm{ON})^{2}=(\mathrm{OA})^{2}+(\mathrm{AN})^{2}$

$10=(\mathrm{OA})^{2}+1 \rightarrow \mathrm{OA}=3$

Perpendicular distance of center from

$P Q=\frac{|0+0-2|}{\sqrt{2}}=\sqrt{2}$

Perpendicular distance between MN and

$\mathrm{PQ}=\mathrm{OA}+\sqrt{2}$ or

$|\mathrm{OA}-\sqrt{2}|$

$=3+\sqrt{2}$

$3-\sqrt{2}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 4 April, Shift 2
Subject: Mathematics
Chapter: Circles
Topic: Chords connected with a Circle