Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Let AB be a chord of the circle  subtending a right ang

Let AB be a chord of the circle \mathrm{x^2+y^2=a^2} subtending a right angle at the centre. Then, the

locus of the centroid of the triangle PAB as P moves on the circle is

Option: 1

a parabola 


Option: 2

a circle


Option: 3

an ellipse


Option: 4

a pair of straight lines


Answers (1)

best_answer

The given circle is \mathrm{x^2+y^2=a^2} and AB is a chord of the circle.
As angle at the centre = double the angle at the circumference.

\mathrm{\begin{aligned} & \therefore<A O B=2<A P B \\ & \Rightarrow<A O M=<B O M=45^{\circ} \\ & \Rightarrow O M=A M=M B \\ & \Rightarrow A\left(-\frac{a}{\sqrt{2}},-\frac{a}{\sqrt{2}}\right), B\left(\frac{a}{\sqrt{2}},-\frac{a}{\sqrt{2}}\right) \end{aligned}}

Let P be \mathrm{(a \cos \theta, a \sin \theta)}

If centroid G is at (h, k) then

\mathrm{ h=\frac{a \cos \theta}{3}, k=\frac{-\frac{2 a}{\sqrt{2}}+a \sin \theta}{3}}

\mathrm{ Eliminate\: \theta:\left(\frac{3 h}{a}\right)^2+\left(\frac{3 k+\sqrt{2} a}{a}\right)^2=1}

\mathrm{ The \: locus\: of (h, k) is\: x^2+\left(y+\frac{a \sqrt{2}}{3}\right)^2=\frac{a^2}{9} }

which is the equation of a circle.

Posted by

Divya Prakash Singh

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 2026: JCECEB invites fresh applications for Jharkhand state merit list correction

Latest Question