Vtyjkyuk

$Y=mx$ 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.

The centre of the given circle lies at the point
$$(a,0)$$
because it passes through the origin and its diameter is along the x-axis.
The equation of this circle can be written as:
$$(x-a)^2+(y-0)^2 = a^2
\x^2+y^2-2ax = 0$$
The equation of the family of circles passing through the points of intersection of a circle $$S=0$$ and a line $$L=0$$ is given by:
$$S+lambdaL = 0$$
The equation of the family of circles passing through $$x^2+y^2-2ax = 0$$ and $$y-mx = 0$$ is given by:
$$x^2+y^2-2ax+lambda(y-mx) = 0
\x^2+y^2-(2a+lambdam)x+lambday = 0$$
Coordinates of the centre of a circle are given by
$$(-g,-f)$$
where 2g and 2f are respectively the x-coefficient and y-coefficient in the equation of the circle.
Hence, the coordinates of the centre, C of the family of circles are given by:
$$left(frac2a+lambdam2,frac-lambda2right)$$
The centre of the required circle lies on the line $$y = mx$$ because it is the diameter of the circle. Hence the coordinates of point C must satisfy the equation of this line.
$$y = mx
\frac-lambda2 = mleft(frac2a+lambdam2right)$$
On solving, we get
$$lambda = frac-2am1+m^2$$
On putting $$lambda = frac-2am1+m^2$$ in the equation of the family of circles, we get the equation of the required circle as:
$$(1+m^2)(x^2+y^2)-2a(x+my) = 0$$