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 a circle with origin as centre and passing through the vertices of an equilateral triangle whose median is of length <img alt="3 \mathrm{a}" src="https://entrancecorner.oncodecogs.c

The equation of a circle with origin as centre and passing through the vertices of an equilateral triangle whose median is of length 3 \mathrm{a} is

Option: 1

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


Option: 2

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


Option: 3

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


Option: 4

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


Answers (1)

best_answer

Since the triangle is equilateral, therefore centroid of the triangle is same as the circumcentre and radius of the circumcircle =2 / 3$ (median $)=2 / 3(3 a)=2 a.

Hence, the equation of the circumcircle whose centre is at (0,0) and radius 2 a is \mathrm{x^{2}+y^{2}=(2 a)^{2}}.

Posted by

Rishi

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

Amrita BTech Admission 2025: JEE Main-based registration deadline extended to August 30
anam-khan

DASA, CSAB Special round 2 seat allotment result 2025 declared on csab.nic.in

Latest Question