Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

The equation of a circle with origin as a centre and passing through equilateral triangle whose median is of length 3a is

  • Option 1)

    x^{2}+y^{2}=9a^{2}

  • Option 2)

    x^{2}+y^{2}=16a^{2}

  • Option 3)

    x^{2}+y^{2}=4a^{2}

  • Option 4)

    x^{2}+y^{2}=a^{2}

 

Answers (1)

As we learnt in

Equation of a circle -

x^{2}+y^{2}=r^{2}

- wherein

Circle with centre \left ( O,O \right ) and radius r.

 AD=3a

AO =2a

Centroid divides median in 2 : 1

Hence equation is 

\\ x^{2}+y^{2} =(2a)^{2} \\ x^{2}+y^{2} =4a^{2}

 


Option 1)

x^{2}+y^{2}=9a^{2}

This option is incorrect.

Option 2)

x^{2}+y^{2}=16a^{2}

This option is incorrect.

Option 3)

x^{2}+y^{2}=4a^{2}

This option is correct.

Option 4)

x^{2}+y^{2}=a^{2}

This option is incorrect.

Posted by

Vakul

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 Main 2025 registration ends tomorrow; changes introduced in exam this year
anam-khan

BTech Admission 2025 Live: JEE Mains registrations closing tomorrow; VITEEE, MET exam dates out
anam-khan

JEE Main 2025 Live: NTA to close JEE registration on Friday; apply at jeemain.nta.nic.in, correction window

Latest Question