Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Confused! kindly explain, - Co-ordinate geometry - JEE Main-2

Let C be the circle with centre (0, 0) and radius 3 units. The equation of the locus of the mid points of chord of the circle C        that subtend an angle of  2\pi /3  at its centre is

  • Option 1)

    x^{2}+y^{2}=\frac{3}{2}

  • Option 2)

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

  • Option 3)

    x^{2}+y^{2}=\frac{27}{4}

  • Option 4)

    x^{2}+y^{2}=\frac{9}{4}

 

Answers (1)

best_answer

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.

 In \Delta AOB

cos\frac{\pi}{3}=\frac{OB}{OA}

\sqrt{\frac{h^{2}+k^{2}}{3}}=\frac{1}{2}

h2 + k2 = 9/4


Option 1)

x^{2}+y^{2}=\frac{3}{2}

This is incorrect option

Option 2)

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

This is incorrect option

Option 3)

x^{2}+y^{2}=\frac{27}{4}

This is incorrect option

Option 4)

x^{2}+y^{2}=\frac{9}{4}

This is correct option

Posted by

Aadil

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 Advanced 2024: Chemical engineering cut-offs at IITs; know opening, closing ranks
anam-khan

IIT JEE Mains Result 2024 Session 2: How is overall merit list prepared?
anam-khan

JEE Mains Session 2 result 2024 soon; List of top NITs in India

Latest Question