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$Y=mx$ is a chord of circle of radius $a$ through the origin whose diameter is along the $x$-axis. Find the equation of the circle whose diameter is the chord.

We also need to find the locus of its centre. I got a relation $h=m^2 h+(a^2+c)^.5$. Where $h$ is abscissa of the centre, $c$ is the constant term in the circle's equation.

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    $\begingroup$ Would you please format mathematics as per the rules? $\endgroup$ – dbanet Aug 13 '16 at 9:13
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As the diameter of the circle is on the x axis.so it's center doesn't have any $y$ co-ordinate.and as it's radius is $a$,we get the equation of the circle is- $$ (x-a)^2+y^2=a^2...........(1)$$ the equation of the chord is $$y=mx.........(2)$$ solving equation (1) and (2) we get, $$(x,y)=(0,0) and \bigg(\dfrac{2a}{m^2+1},\dfrac{2am}{m^2+1}\bigg)$$ So ,the locus of the center of our required circle is $\bigg(\dfrac{a}{m^2+1},\dfrac{am}{m^2+1}\bigg)$ and the radius of the circle is $$\dfrac{\sqrt{\bigg(\dfrac{2a}{m^2+1}-0\bigg)^2+\bigg(\dfrac{2am}{m^2+1}-0\bigg)^2}}{2}=\dfrac{a}{\sqrt{m^2+1}}$$ Hence We can now make our required equation and that is- $$\bigg(x-\dfrac{a}{m^2+1}\bigg)^2+\bigg(y-\dfrac{am}{m^2+1}\bigg)^2=\dfrac{a^2}{m^2+1}$$ After some simplification you will find that the equation become, $$2a(my+x)=(m^2+1)(x^2+y^2)$$ graph of the circle

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