$\text{ If the line }\frac{x}{a}+\frac{y}{b}=1\text{ moves in such a way that }\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2}\text{ where }c\text{ is a constant, then the locus of }$the foot of perpendicular from the origin on the straight line is 

 Variable line is
$\frac{x}{a}+\frac{y}{b}=1-----\left(1\right)$

Any line perpendicular to (l) and passing through the origin will be 

$\frac{x}{b}-\frac{y}{a}=0-----\left(2\right)$

Now foot of the perpendicular from the origin to line (l) is the point of intersection (1) and (2). 

$\text{ Let it be }P(\alpha ,\beta ),\text{ then }$

$\frac{\alpha }{a}+\frac{\beta }{b}=1-----\left(3\right)$

$\text{ and }\frac{\alpha }{b}-\frac{\beta }{a}=0-----\left(4\right)$

Squaring and adding (3) and (4), we get
$\begin{array}{l}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\alpha }^2\left(\frac{1}{a^2}+\frac{1}{b^2}\right)+{\beta }^2\left(\frac{1}{b^2}+\frac{1}{a^2}\right)=1\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\left({\alpha }^2+{\beta }^2\right)\frac{1}{c^2}=1\end{array}$

$\text{ Hence, the locus of }P(\alpha ,\beta )\text{ is }{x}^2+y^2=c^2\text{ . }$