[ $m$ of given line $ = \frac{{ - b}}{a}$, $\therefore m$ of perpendicular $\left. { = \frac{a}{b}} \right]$
==> $by - ax = 0$
==> $\frac{x}{b} - \frac{y}{a} = 0$
Now, the locus of foot of perpendicular is the intersection point of line $\frac{x}{a} + \frac{y}{b} = 1$ .....$(i)$
and $\frac{x}{b} - \frac{y}{a} = 0$......$(ii)$
To find locus, squaring and adding $(i)$ and $(ii)$
${\left( {\frac{x}{a} + \frac{y}{b}} \right)^2} + {\left( {\frac{x}{b} - \frac{y}{a}} \right)^2} = 1$
==> ${x^2}\left( {\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}}} \right) + {y^2}\left( {\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}}} \right) = 1$
==> ${x^2}\left( {\frac{1}{{{c^2}}}} \right) + {y^2}\left( {\frac{1}{{{c^2}}}} \right) = 1$ ,
==> ${x^2} + {y^2} = {c^2}$.
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