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 $\frac{a}{x}+\frac{b}{y}=\lambda$,where $\lambda$

Let the coordinates of $A$ and $B$ be $(p, 0)$ and $( 0, q)$

respectively.  Then, equation of $AB$is 

$\frac{x}{p}+\frac{y}{q}=1$

Since it passes through $(a, b)$

$\therefore \frac{a}{p}+\frac{b}{q}=1$

The 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. Then

$h=p/2, k=q/2\Rightarrow p=2h, q=2k$

Substituting values of $p$, $q$in $\left(i\right)$, we get: 

$\frac{a}{2h}+\frac{b}{2k}=1$

Hence, the locus of $(h, k)$ is $\frac{a}{2x}+\frac{b}{2y}=1$ or $\frac{a}{x}+\frac{b}{y}=2$