If a circle of radius R passes through the origin O and intersects...

Since, circle passing through origin intersect the coordinate axes at $A \text{ and } B$, hence $AB$ must be diameter and $AB = 2R$.

Now, let foot of the perpendicular from origin upon $AB$ be $P\left(h, k\right)$.

Slope of line $OP = \dfrac{k - 0}{h - 0} = \dfrac{k}{h}$

Since, line $AB ⊥ OP ⇒$ slope of $AB = - \dfrac{h}{k}$

Thus, equation of line $AB$ is $y - k = \dfrac{- h}{k}\left(x - h\right)$

For co-ordinates of $A$, put $y = 0 ⇒ 0 - k = \dfrac{- h}{k}\left(x - h\right) ⇒ x = \dfrac{{h}^{2} + {k}^{2}}{h} ⇒ A\left(\dfrac{{h}^{2} + {k}^{2}}{h}, 0\right)$.

For co-ordinates of $B$, put $x = 0 ⇒ y - k = \dfrac{- h}{k}\left(0 - h\right) ⇒ y = \dfrac{{h}^{2} + {k}^{2}}{k} ⇒ B\left(0, \dfrac{{h}^{2} + {k}^{2}}{k}\right)$.

Now, given $AB = 2R$

Applying distance formula,

$⇒ \sqrt{{\left(\dfrac{{h}^{2} + {k}^{2}}{h} - 0\right)}^{2} + {\left(0 - \dfrac{{h}^{2} + {k}^{2}}{k}\right)}^{2}} = 2R$

$⇒ {\left({h}^{2} + {k}^{2}\right)}^{2}\left(\dfrac{1}{{h}^{2}} + \dfrac{1}{{k}^{2}}\right) = 4{R}^{2} ⇒ {\left({h}^{2} + {k}^{2}\right)}^{3} = 4{R}^{2}{h}^{2}{k}^{2}$

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