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

Option: 1

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


Option: 2

x^2+y^2-2 a(x+2 y)=0


Option: 3

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


Option: 4

none of these.


Answers (1)

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

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

Posted by

Ramraj Saini

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