If a circle of radius $\mathrm{R}$ passes through the origin $\mathrm{O}$ and intersects the coordinate axes at $\mathrm{A}$ and $\mathrm{B}$, then the locus of the foot of perpendicular from $\mathrm{O}$ on $\mathrm{AB}$ is :
Correct Option: , 2
As $\angle A O B=90^{\circ}$
Let $A B$ diameter and $M(h, k)$ be foot of perpendicular, then
$M_{A B}=-$
Then, equation of $A B$
$(y-k)=\frac{-h}{k}(x-h)$
$\Rightarrow \quad h x+k y=h^{2}+k^{2}$
Then, $A\left(\frac{h^{2}+k^{2}}{h}, 0\right)$ and $B\left(0, \frac{h^{2}+k^{2}}{k}\right)$
$\because A B$ is the diameter, then
$A B=2 R$
$\Rightarrow A B^{2}=4 R^{2}$
$\Rightarrow \quad\left(\frac{h^{2}+k^{2}}{h}\right)^{2}+\left(\frac{h^{2}+k^{2}}{k}\right)=4 R^{2}$
Hence, required locus is $\left(x^{2}+y^{2}\right)^{3}=4 R^{2} x^{2} y^{2}$
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