A line is at a constant distance $c$ from the origin and meets the coordinates axes in $A$ and $B$. The locus of the centre of the circle passing through $O, A, B$ is

Let the equation of line be

$\displaystyle \frac{x}{a}+\frac{y}{b}=1$

where $a$ and $b$ are the x-intercept and y-intercept.

Then the coordinates of A and B are $(a,0)$ and $(0,b)$

Distance of origin to the line is $c$

$\Rightarrow c=\displaystyle \frac{|-1|}{\sqrt{(\dfrac{1}{a})^2+(\dfrac{1}{b})^2}}$

$\Rightarrow \displaystyle \frac{1}{c}=\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}$

$\Rightarrow \displaystyle \frac{1}{c^2}=\frac{1}{a^2}+\frac{1}{b^2}$ ....(1)

Let the center of circle through O, A, B be $(h,k)$

Points $O(0,0), A(a,0),(0,b)$ forms a right triangle.

So, the center of circle is the mid-point of AB i.e.$\displaystyle(\frac{a+0}{2},\frac{b+0}{2})$

$\Rightarrow h=\displaystyle \frac{a}{2}, k=\frac{b}{2}$

$\Rightarrow a=2h, b=2k$

Substitute this value in (1), we get

$\displaystyle \frac{1}{c^2}=\frac{1}{4h^2}+\frac{1}{4k^2}$

$\Rightarrow \displaystyle \frac{4}{c^2}=\frac{1}{h^2}+\frac{1}{k^2}$

$\Rightarrow 4c^{-2}=x^{-2}+y^{-2}$ (Replacing $h,k$ by $x,y$ )