If a circle of radius R passes through the origin O and intersects the coordinate axes at A and B, then the locus of the foot of perpendicular from O on AB is:

Consider the figure 

As $\mathrm{\angle }\mathrm{AOB}={90}^{\circ }$ ( angle between the axes)

SInce the angle in semi circle is right anlge, the line AB is diameter and suppose that  M (h, k) be foot of perpendicular, t

The slope of $AB$ is $m=-\frac{1}{slope of OM}=-\frac{h}{k}$

Then, equation of AB

$(\mathrm{y}-\mathrm{k})=-\frac{\mathrm{h}}{\mathrm{k}}(\mathrm{x}-\mathrm{h})\Rightarrow \mathrm{hx}+\mathrm{ky}={\mathrm{h}}^2+{\mathrm{k}}^2$

Coordinates of points $A,B$ are $A\left(\frac{h^2+k^2}{h},0\right)$ and $B\left(0,\frac{h^2+k^2}{k}\right)$

The length of the diameter is $AB=2R$

Then, ${\left(\frac{{\mathrm{h}}^2+{\mathrm{k}}^2}{\mathrm{h}}\right)}^2+{\left(\frac{{\mathrm{h}}^2+{\mathrm{k}}^2}{\mathrm{k}}\right)}^2=4{\mathrm{R}}^2$

Equation of locus is ${\left(x^2+y^2\right)}^3=4R^2{x}^2{y}^2$

Hence, required locus is $\boxed{{\left({\mathrm{x}}^2+{\mathrm{y}}^2\right)}^3=4{\mathrm{R}}^2{\mathrm{x}}^2{\mathrm{y}}^2}$