From the origin, a chord is drawn to the circle $x^2 +y^2-2ax=0$. Find the locus of the centre of the circle taking that chord as diameter.
Taking the equation of the chord as $y=mx$, I have found the point of intersections of the chord and the circle as $$(0,0) \qquad\text{and}\qquad \left(\frac{2a}{1+m^2} , \frac{2am}{1+m^2}\right).$$
If I take the center as $(h,k)$, then $h=\frac{a}{1+m^2}$ , $k= \frac{am}{1+m^2}$
I am having some trouble with elimination of $m$.
A hint:
Look at the problem geometrically. For each point ${\bf z}\ne(0,0)$ on the given circle you get the point ${1\over2}{\bf z}$ as a point of your locus.