Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Can someone explain - Two-dimensional Coordinate Geometry - BITSAT

If y=2x is a chord of the circle x^{2}+y^{2}=10x,then the equation of the circle whose diameter is this chord is 

  • Option 1)

    x^{2}+y^{2}+2x+4y=0

  • Option 2)

    x^{2}+y^{2}+2x-4y=0

  • Option 3)

    x^{2}+y^{2}-2x-4y=0

  • Option 4)

    None of these 

 

Answers (1)

best_answer

 

Family of circle -

S+KL= 0

- wherein

Equation of the family of circles passing through point of intersection S=0 \, and\, line\, L=0.

 

 (x^{2}+y^{2}-10x)+\lambda (2x-y)=0 is a family of circle

Centre \equiv \left ( 5-\lambda , \frac{\lambda }{2} \right )

it lies on y=2x

\therefore 10-2\lambda -\frac{\lambda }{2}=0

\Rightarrow \lambda =4

x^{2}+y^{2}-2x-4y=0

 


Option 1)

x^{2}+y^{2}+2x+4y=0

This solution is incorrect

Option 2)

x^{2}+y^{2}+2x-4y=0

This solution is incorrect

Option 3)

x^{2}+y^{2}-2x-4y=0

This solution is correct

Option 4)

None of these 

This solution is incorrect

Posted by

prateek

View full answer

Similar Questions

Trending Articles/News

anam-khan

BITSAT 2024 slot booking begins for session 1 at bitsadmission.com; admit card on May 15
anam-khan

BITSAT city allotment 2024 issued at bitsadmission.com; here’s how to download
anam-khan

BITSAT 2024 city intimation slip tomorrow at bitsadmission.com; exam pattern

Latest Question