The locus of the centre of the circle passing through the origin O and the points of intersection A and B of any line through (a, b)  and the coordinate axes is ax+by=λ,where λ

Let the coordinates of A and B be (p, 0) and ( 0, q)respectively.  Then, equation of AB is xp+yq=1Since it passes through (a, b)∴ ap+bq=1The triangle OAB is a right-angled triangle. So, it is a diameter of the circle passing through O, A and B. So, coordinates of the centre of the circle are (p/2, q/2)Let (h, k) be the centre of the circle. Thenh=p/2, k=q/2⇒p=2h, q=2kSubstituting values of p, q in (i), we get: a2h+b2k=1Hence, the locus of (h, k) is a2x+b2y=1 or ax+by=2