Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • The equation of the circle of radius which touches the line <img alt="x+y=1 at \: (2,-1)" src="https://en

The equation of the circle of radius \sqrt{2} which touches the line x+y=1 at \: (2,-1) is

Option: 1

\mathrm{x^{2}+y^{2}-4 x+2 y+3=0}


Option: 2

\mathrm{x^{2}+y^{2}+6 x+7=0}


Option: 3

\mathrm{x^{2}+y^{2}-2 x+4 y+3=0}


Option: 4

none of these


Answers (1)

best_answer

Family of circle touches \mathrm{x+y-1=0} at \mathrm{(2,-1)} is

\mathrm{(x-2)^{2}+(y+1)^{2}+\lambda(x+y-1)=0}

Now radius of this circle is \mathrm{\sqrt{2}}

Hence \mathrm{I= \pm 2}

Hence (C) is the correct answer.

Posted by

chirag

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Mains 2025 Jan 29 Shift 2 Analysis: Overall paper moderate, maths lengthy; subject-wise difficulty level
anam-khan

JEE Main January 29 shift 1 exam paper analysis: Maths most ‘time-consuming’, physics easier
anam-khan

JEE Main 2025 paper 2 for BArch, BPlanning in single shift tomorrow; exam pattern

Latest Question