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:

Let the circle of fixed radius $$R$$ pass through the origin $$O(0,0)$$ and cut the coordinate axes at the points $$A(a,0)$$ on the $$x$$-axis and $$B(0,b)$$ on the $$y$$-axis. Because $$A$$ and $$B$$ lie on the axes, the straight line joining them has intercept form

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

We first rewrite this line in the form $$lx+my+n=0$$, because the formulas for the foot of a perpendicular from a point to a line are most convenient in that form. Multiplying by $$ab$$ gives

$$bx+ay-ab=0.$$

Thus $$l=b,\;m=a,\;n=-ab.$$

The foot of the perpendicular drawn from the origin to a line $$lx+my+n=0$$ is given by the standard formula

$$\bigl(h,k\bigr)=\left(\!-\frac{ln}{l^{2}+m^{2}},-\frac{mn}{l^{2}+m^{2}}\!\right).$$

Substituting $$l=b,\;m=a,\;n=-ab$$ we obtain

$$h=-\frac{b(-ab)}{b^{2}+a^{2}}=\frac{ab^{2}}{a^{2}+b^{2}},\qquad k=-\frac{a(-ab)}{b^{2}+a^{2}}=\frac{a^{2}b}{a^{2}+b^{2}}.$$

Hence, if $$P(h,k)$$ is the foot of the perpendicular from $$O$$ to $$AB$$, its coordinates are

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

Next we employ the fact that $$O,\,A,\,B$$ lie on the given circle of radius $$R$$. Since $$OA$$ is along the $$x$$-axis and $$OB$$ is along the $$y$$-axis, the angle at $$O$$ is a right angle. For a right-angled triangle, the circumradius equals half the hypotenuse. Therefore, using the hypotenuse $$AB,$$ we have

$$R=\frac{1}{2}\,AB=\frac{1}{2}\sqrt{a^{2}+b^{2}}\;,$$

so

$$a^{2}+b^{2}=(2R)^{2}=4R^{2}.$$

To obtain the required locus, we eliminate $$a$$ and $$b$$ from the expressions for $$x$$ and $$y$$. First compute $$x^{2}+y^{2}:$$

$$\begin{aligned} x^{2}+y^{2}&=\left(\frac{ab^{2}}{a^{2}+b^{2}}\right)^{\!2}+\left(\frac{a^{2}b}{a^{2}+b^{2}}\right)^{\!2} \\ &=\frac{a^{2}b^{4}+a^{4}b^{2}}{(a^{2}+b^{2})^{2}}\\ &=\frac{a^{2}b^{2}(b^{2}+a^{2})}{(a^{2}+b^{2})^{2}}\\ &=\frac{a^{2}b^{2}}{a^{2}+b^{2}}\\[4pt] &=\frac{a^{2}b^{2}}{4R^{2}}\quad\bigl(\text{because }a^{2}+b^{2}=4R^{2}\bigr). \end{aligned}$$

Now compute $$x^{2}y^{2}:$$

$$x^{2}y^{2}=\left(\frac{ab^{2}}{a^{2}+b^{2}}\right)^{\!2}\!\left(\frac{a^{2}b}{a^{2}+b^{2}}\right)^{\!2} =\frac{a^{6}b^{6}}{(a^{2}+b^{2})^{4}} =\frac{a^{6}b^{6}}{(4R^{2})^{4}} =\frac{a^{6}b^{6}}{256R^{8}}.$$

We are now ready to relate $$x^{2}+y^{2}$$ to $$x^{2}y^{2}$$. Cubing the earlier expression for $$x^{2}+y^{2}$$ gives

$$\bigl(x^{2}+y^{2}\bigr)^{3} =\left(\frac{a^{2}b^{2}}{4R^{2}}\right)^{\!3} =\frac{a^{6}b^{6}}{64R^{6}}.$$

On the other hand, multiplying $$x^{2}y^{2}$$ by $$4R^{2}$$ yields

$$4R^{2}x^{2}y^{2}=4R^{2}\cdot\frac{a^{6}b^{6}}{256R^{8}} =\frac{a^{6}b^{6}}{64R^{6}}.$$

Both expressions are identical, so we have the relation

$$\bigl(x^{2}+y^{2}\bigr)^{3}=4R^{2}x^{2}y^{2}.$$

This is precisely the equation listed in Option B.

Hence, the correct answer is Option 2.