Let C be a circle with radius 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)

Correct answer: A. aligned & 3- 2 aligned aligned & C: x^2+y^2=10 & A N=(M N)/(2)=1 & In O A N arrow(O N)^2=(O A)^2+(A N)^2 & 10=(O A)^2+1 arrow O A=3 aligned Perpendicular distance of center from PQ= |0+0-2| 2 = 2 Perpendicular distance between MN and aligned & PQ=OA+ 2 or |OA- 2 | & =3+ 2 or 3- 2 aligned

JEE Main 20244 AprilEvening Shift

Exam: JEE Main
Year: 2024
Date: 4 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Circle

JEE Main 2024 · 4 April · Evening Shift · Mathematics · Circle

Let C be a circle with radius 10 units and centre at the origin....

Let

C

be a circle with radius

\sqrt{10}

units and centre at the origin. Let the line

x+y=2

intersects the circle

\mathrm{C}

at the points

\mathrm{P}

and

\mathrm{Q}

. Let

\mathrm{MN}

be a chord of

\mathrm{C}

of length 2 unit and slope -1. Then, a distance (in units) between the chord PQ and the chord

\mathrm{MN}

Options

A $\begin{aligned}& 3-\sqrt{2} \end{aligned}$
B $\begin{aligned} &\sqrt{2}+1 \end{aligned}$
C $\begin{aligned} &\sqrt{2}-1 \end{aligned}$
D $\begin{aligned}& 2-\sqrt{3} \end{aligned}$

Correct answer

A. $\begin{aligned}& 3-\sqrt{2} \end{aligned}$

Step-by-step solution

\begin{aligned} & C: x^2+y^2=10 \\ & A N=\frac{M N}{2}=1 \\ & \therefore \text{\text{ In } } \triangle O A N \rightarrow(O N)^2=(O A)^2+(A N)^2 \\ & 10=(O A)^2+1 \rightarrow O A=3 \end{aligned}

Perpendicular distance of center from

\mathrm{PQ}=\frac{|0+0-2|}{\sqrt{2}}=\sqrt{2}

Perpendicular distance between

\mathrm{MN}

and

\begin{aligned} & \mathrm{PQ}=\mathrm{OA}+\sqrt{2} \text{ or }|\mathrm{OA}-\sqrt{2}| \\ & =3+\sqrt{2} \text{ or } 3-\sqrt{2} \end{aligned}

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