If a circle of radius R passes through the origin O and intersects the coordinate axes at …

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

Accepted Answer

$\angle AOB=90^\circ$ so $AB$ is a diameter, $AB=2R$. Let $M(h,k)$ be the foot of perpendicular from $O$; slope of $OM$ is $k/h$, so $AB\perp OM$: line $AB$: $hx+ky=h^2+k^2$.

$$A=\left(\frac{h^2+k^2}h,0\right),\qquad B=\left(0,\frac{h^2+k^2}k\right).$$

$$AB^2=4R^2\Rightarrow\left(\frac{h^2+k^2}h\right)^2+\left(\frac{h^2+k^2}k\right)^2=4R^2\ \Longrightarrow\ (x^2+y^2)^3=4R^2x^2y^2.$$

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