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If \mathrm{y=2 x}be the equation of a chord of the circle \mathrm{x^2+y^2=2 a x}, then the equation of the circle, of which this chord is a diameter, is
 

Option: 1

\mathrm{2\left(x^2+y^2\right)-5 a(x+2 y)=0}
 


Option: 2

\mathrm{x^2+y^2-2 a(x+2 y)=0}
 


Option: 3

\mathrm{5\left(x^2+y^2\right)-2 a(x+2 y)=0}

 


Option: 4

None of these.


Answers (1)

best_answer

The equation of any circle passing through the ends of the chord is given by

\mathrm{\mathrm{x}^2+\mathrm{y}^2-2 a \mathrm{x}+\lambda(\mathrm{y}-2 \mathrm{x})=0}
If \mathrm{ y=2 x} is a diameter then the coordinates of the centre of this circle, i.e.
\mathrm{\left(\frac{2 a+2 \lambda}{2},-\frac{\lambda}{2}\right)} should lie on this line i.e., \mathrm{-\frac{\lambda}{2}=2 \cdot \frac{(2 a+2 \lambda)}{2} \Rightarrow \lambda=-\frac{4}{5} a}
Putting \mathrm{\lambda=-\frac{4}{5} a} in (1), equation of the required circle is

\mathrm{ x^2+y^2-2 a x-\frac{4}{5} a(y-2 x)=0 }

\mathrm{ \Rightarrow 5\left(x^2+y^2\right)-2 a(x+2 y)=0}

Hence option 3 is correct.



 

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chirag

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