Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • The locus of the mid-point of a chord of the circle    w

The locus of the mid-point of a chord of the circle  \mathrm{x^2+y^2=4}  which subtends a right angle at the origin is
 

Option: 1

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


Option: 2

\mathrm{ x+y=2}


Option: 3

\mathrm{ x^2+y^2=1}


Option: 4

\mathrm{ x+y=1}


Answers (1)

best_answer

As, we have to find locus of mid-point of chord and we know perpendicular from centre bisectS the chord.

\angle \mathrm{OAC}=45^{\circ}

or  \frac{\mathrm{OC}}{\mathrm{OA}}=\sin 45^{\circ} \Rightarrow \mathrm{OC}=\frac{2}{\sqrt{2}}

\therefore \sqrt{\mathrm{h}^2+\mathrm{k}^2}=\mathrm{OC}^2

or \mathrm{x^2+y^2=2}  is required equation of locus of mid-point of chord subtending right angle at centre.

Posted by

shivangi.shekhar

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: IIT Delhi’s CSE, EE most opted; last 2 year’s closing ranks for all categories
anam-khan

Must provide CBSE 11th registration numbers to JEE Mains aspirants, board tells schools
anam-khan

JEE Main 2026: Engineering at NIT Surathkal tougher for insiders, show BTech cut-offs

Latest Question