Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Find the locus of the point of intersection of the tangents at the ends of a chord of the circle <img alt="x^2+y^2=a^2" src="https://entrancecorner.oncodecogs.com/gif.latex?x%5E2&plus;y%5E2%3Da

Find the locus of the point of intersection of the tangents at the ends of a chord of the circle x^2+y^2=a^2 which touches the circle x^2+y^2-2 a x=0.

Option: 1

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


Option: 2

\mathrm{y^2=2 a(a-2 x)}


Option: 3

\mathrm{y^2=2 a(a-x)}


Option: 4

\mathrm{y^2=a(a-2 x)}


Answers (1)

Let  \left(x_1, y_1\right)  be one such point. Chord of contact of tangents from \left(x_1, y_1\right) to x^2+y^2=a^2 is

x x_1+y y_1-a^2=0.This line is a tangent to x^2+y^2-2 a x=0 whose centre is (a, 0) and radius =a.

\therefore \frac{a x_1-a^2}{\sqrt{x_1^2+y_1^2}}= \pm a \quad ; \quad \therefore\left(x_1-a\right)^2=x_1^2+y_1^2 \quad \text { i.e, } y_1^2=a^2-2 a x_1

\therefore \text { Locus of }\left(x_1, y_1\right) \text { is } y^2=a(a-2 x) \text {. }

Posted by

Ramraj Saini

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 session 2 application correction begins on jeemain.nta.nic.in; edit details by tomorrow
anam-khan

JEE Main 2026 Registration LIVE: Session 2 exam from April 2 to 9; shift timings, admit card, exam pattern
anam-khan

JEE Main 2026 session 2 registration ends today for BTech, BArch; apply at jeemain.nta.nic.in

Latest Question