Question

A circle of constant radius $r$ passes through origin $O$ and cuts the axes of co-ordinates in points $A$ and $B$. Prove that the locus of the foot of perpendicular from $O$ to $AB$ is
$(x^2+y^2{)}^2(\frac{1}{x^2}+\frac{1}{y^2})=4r^2$.

Answer 1

The circle $OAB$ is $x^2+y^2-ax-by=0$
where $\frac{a^2}{4}+\frac{b^2}{4}=r^2.....\left(1\right)$
Equation of $AB$ is $\frac{x}{a}+\frac{y}{b}=1....\left(2\right)$
Any line through $(0,0)$, perpendicular to it
$\frac{x}{b}-\frac{y}{a}=0.....\left(3\right)$
Both $\left(2\right)$ and $\left(3\right)$ given the foot of perpendicular
We have to eliminate the variables $a$ and $b$ from the three relations $\left(1\right),\left(2\right)$ and $\left(3\right)$.
Solving $\left(2\right)$ and $\left(3\right)$ for $a$ and $b$, we have $b=\frac{ax}{y}$.
Putting in $\left(1\right)$, we get
$\frac{x}{a}+\frac{y.y}{ax}=1\therefore x^2+y^2=ax$
$\therefore a=\frac{x^2+y^2}{x}\therefore b=\frac{x^2+y^2}{y}$.
Putting in $\left(1\right)$, we get
$(x^2+y^2{)}^2[\frac{1}{x^2}+\frac{1}{y^2}]=4r^2$ is the required locus.