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)

 

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

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