If a circle of radius R passes through the origin


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 :

  1. (1) $\left(x^{2}+y^{2}\right)^{2}=4 R^{2} x^{2} y^{2}$

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

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

  4. (4) $\left(x^{2}+y^{2}\right)(x+y)=R^{2} x y$

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 $A B$


$\Rightarrow \quad h x+k y=h^{2}+k^{2}$

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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