Mathematics · Circles

JEE Main 2024 — 5 April, Shift 1 — Question 12

Let a circle C of radius 1 and closer to the origin be such that the lines passing through the point $(3,2)$ and parallel to the coordinate axes touch it. Then the shortest distance of the circle C from the point $(5,5)$ is :

Option A:
$2 \sqrt{2}$
Option B:
5
Option C:
$4 \sqrt{2}$
Option D:
4
Correct

Answer: D

Step-by-step solution

Coordinates of the centre will be $(2,1)$

Equation of circle will be

$(x-2)^{2}+(y-1)^{2}=1$

$\mathrm{QC}=\sqrt{(5-2)^{2}+(5-1)^{2}}$

$\mathrm{QC}=5$

shortest distance

$=\mathrm{RQ}=\mathrm{CQ}-\mathrm{CR}$

$=5-1$

$=4$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 5 April, Shift 1
Subject: Mathematics
Chapter: Circles
Topic: Considering a Line or a Point wrt a Circle