Question

A variable chord is drawn to the circle $x^2+y^2-2ax=0$ from the origin, then the locus of the centre of the circle which is made by taking the chord as the diameter, is

• A. $x^2+y^2+ax=0$
• B. $x^2+y^2-ax=0$
• C. $x^2+y^2+ay=0$
• D. $x^2+y^2-ay=0$

Answer 1

Answer: (B)

The given circle $x^2+y^2-2ax=0$ ...(i)
Let centre of the drawn circle be $(\alpha ,\beta )$ let variable chord $y=mx$ ...(ii) On solving Eqs. (i) and (ii), we get
$x=0,\frac{2a}{1+m^2}$ and $y=0,\frac{2am}{1+m^2}$
$\therefore$ Intersection points are
$(0,0)\left(\frac{2a}{1+m^2},\frac{2am}{1+m^2}\right)$
Let coordinate of the centre of circle be $(\alpha ,\beta )$
$\therefore$ $\alpha =\frac{0+\frac{2a}{1+m^2}}{2}=\frac{a}{1+m^2},$ and $\beta =\frac{0+\frac{2am}{1+m^2}}{2}=\frac{am}{1+m^2},$
Thus, $\beta =\alpha m$
$\Rightarrow$ $m=\frac{\beta }{\alpha }$
Then, $\alpha =\frac{a}{1+\frac{{\beta }^2}{{\alpha }^2}}$
$\Rightarrow$ $\alpha =\frac{a{\alpha }^2}{{\alpha }^2+{\beta }^2}$
$\Rightarrow$ $\alpha ({\alpha }^2+{\beta }^2)=a{\alpha }^2$
$\Rightarrow$ ${\alpha }^2+{\beta }^2-a\alpha =0$
Hence, locus of the centre of circle is $x^2+y^2-ax=0$