Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

Equation of circle touching the lines \mathrm{|x-2|+|y-3|=4}  will be

Option: 1

\mathrm{\quad(x-2)^{2}+(y-3)^{2}=12}


Option: 2

\mathrm{\quad(x-2)^{2}+(y-3)^{2}=4}


Option: 3

\mathrm{\quad(x-2)^{2}+(y-3)^{2}=4}


Option: 4

\mathrm{\quad(x-2)^{2}+(y-3)^{2}=8}


Answers (1)

best_answer

Clearly \mathrm{|x-2|+|y-3|=4} will be a square.


Hence the centre of circle will be (2,3) and radius \mathrm{2 \sqrt{2}}.
Hence equation of circle is
\mathrm{(x-2)^{2}+(y-3)^{2}=8}

Hence (D) is the correct answer.

 

Posted by

Riya

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 aspirant in Kota commits suicide; 16th suicide since January
anam-khan

JEE Main 2025 Registration Live: NTA session 1 application last date today; official website, syllabus
anam-khan

JEE Main 2025 session 1 registration ends today at jeemain.nta.nic.in; application correction from November 26

Latest Question