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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 $m$.

Xander Henderson
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SUBHAM GHOSH
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  • Centre is already given as $(a,0).$ We take locus of those which are variable. When we take the trajectory of the path of a variable point then it is called its locus. If you take $a$ to be variable then the locus of the centre is the $X$-axis (if you allow $a$ to take $0$). Otherwise it will be $X$ axis $\setminus {0 }.$ – math maniac. Nov 30 '19 at 16:20
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    Hint : $$\dfrac hk=?$$ – lab bhattacharjee Nov 30 '19 at 16:45

1 Answers1

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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.

Christian Blatter
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