Question

$y=mx$ is a chord of the circle of a radius a through the origin and whose diameter is along the axis of x. Find the equation of the circle whose diameter is the chord . Hence find the locus of its centre for all values of m.

Answer 1

Since the diameter is along x-axis and the chord y = mx passes through the origin

therefore centre is (a,0) where a is the given radius of the circle.
Hence its equation is
$S=(x-a{)}^2+y^2=a^2$
or $x^2+y^2-2ax=0$
P is y = mx (m is variable ). Circle through the intersection of
S = 0 and P = 0 is
$S+\lambda P=0or(x^2+y^2-2ax)+\lambda (y-mx)=0$
or $x^2+y^2-x(2a+m\lambda )+\lambda y=0$
Centre is $(\frac{2a+m\lambda }{2},-\frac{\lambda }{2})$
Since the chord y = mx is a diameter of the required circle therefore its centre lies on it $\therefore -\frac{\lambda }{2}=m.\frac{2a+m\lambda }{2}$
or $\lambda (1+m^2)=-2am$ $\therefore \lambda =\frac{-2am}{1+m^2}$
Putting for $\lambda$ , the required circle is
$(1+m^2)(x^2+y^2-2ax)-2am(y-mx)=0$
$(1+m^2)(x^2+y^2)-2a(x+my)=0$
If (h, k) be the coordinates of its centre , then $h=\frac{a}{1+m^2},k=\frac{am}{1+m^2}$
In order ti find its locus we have to eliminate m.
Dividing $\frac{K}{h}=m$ and putting in $h(1+m^2)=a$
We get $h(1+\frac{k^2}{h^2})=a$
or $h^2+k^2=ah$
Generalising , the required locus is
$x^2+y^2-ax=0$ which is a circle.