Question

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 :

• A. $(x^2+y^2{)}^3=4R^2{x}^2{y}^2$
• B. $(x^2+y^2{)}^2=4R^2{x}^2{y}^2$
• C. $(x^2+y^2{)}^2=4R{x}^2{y}^2$
• D. $(x^2+y^2)(x+y)=R^{2}xy$

Answer 1

The correct option is A $(x^2+y^2{)}^3=4R^2{x}^2{y}^2$


Slope of $OP=\frac{k}{h}$
Slope of $AB=\frac{-h}{k}$
Equation of line $AB$ is :
$y-k=\frac{-h}{k}(x-h)$
$\Rightarrow hx+yk=h^2+k^2$

Therefore, coordinates of $A$ and $B$ are $(\frac{h^2+k^2}{h},0)$ and $(0,\frac{h^2+k^2}{k})$, respectively.

$\angle AOB={90}^{\circ }$
$\therefore AB$ is diameter of the circle.
$AB=2R$
$\Rightarrow {\left(\frac{h^2+k^2}{h}\right)}^2+{\left(\frac{h^2+k^2}{k}\right)}^2=4R^2$
$\Rightarrow {(h^2+k^2)}^3=4R^2{h}^2{k}^2$

$\therefore$ Locus is $(x^2+y^2{)}^3=4R^2{x}^2{y}^2$