1. Let the circle intersect the axes at $A(a,0), \qquad B(0,b)$ and since it passes through the origin, $O(0,0)$ is also on the circle.

2. Equation of the circle through $O, A, B$: A general circle through the origin is $x^2+y^2+2gx+2fy=0$ Since it passes through $(a,0)$, $a^2+2ga=0 \implies 2g=-a$ Since it passes through $(0,b)$, $b^2+2fb=0 \implies 2f=-b$ Hence the circle is $x^2+y^2-ax-by=0$

3. Use the radius condition: For the circle $x^2+y^2+2gx+2fy+c=0,$ radius is $\sqrt{g^2+f^2-c}$ Here $c=0$, $g=-\frac a2$, $f=-\frac b2$, so $R^2=\frac{a^2+b^2}{4}$ Therefore, $a^2+b^2=4R^2$

4. Equation of chord $AB$: Since $A(a,0)$ and $B(0,b)$ lie on the axes, $\frac{x}{a}+\frac{y}{b}=1$ or $bx+ay-ab=0$

5. Foot of perpendicular from $O$ to $AB$: Let the foot be $P(x,y)$.

   Since the line $AB$ is $bx+ay-ab=0,$ the perpendicular from origin to this line lies along the normal vector $(b,a)$. Hence $P=(\lambda b,\lambda a)$ for some parameter $\lambda$.

   Because $P$ lies on $AB$, $b(\lambda b)+a(\lambda a)-ab=0$ $\lambda(a^2+b^2)=ab$ $\lambda=\frac{ab}{a^2+b^2}$

   Therefore, $x=\frac{ab^2}{a^2+b^2}, \qquad y=\frac{a^2b}{a^2+b^2}$

6. Relate $x,y$ to $a,b$: From the above, $\frac{x}{y}=\frac{b}{a} \implies ay=bx$

   Also,

   =\frac{a^2b^2(a^2+b^2)}{(a^2+b^2)^2} =\frac{a^2b^2}{a^2+b^2}$$ Using $a^2+b^2=4R^2$, $$x^2+y^2=\frac{a^2b^2}{4R^2}$$ Now, $$xy=\frac{ab^2}{a^2+b^2}\cdot \frac{a^2b}{a^2+b^2} =\frac{a^3b^3}{(a^2+b^2)^2}$$ So, $$x^2y^2=\frac{a^6b^6}{(a^2+b^2)^4}$$ and $$(x^2+y^2)^3=\left(\frac{a^2b^2}{a^2+b^2}\right)^3 =\frac{a^6b^6}{(a^2+b^2)^3}$$ Therefore, $$\frac{(x^2+y^2)^3}{x^2y^2}=a^2+b^2=4R^2$$ Hence the locus is $$\boxed{(x^2+y^2)^3=4R^2x^2y^2}$$

7. Compare with options: This matches Option D.

Final Answer: $\boxed{(x^2+y^2)^3=4R^2x^2y^2}$