Question

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

Answer 1

The given line is $\frac{x}{a}+\frac{y}{b}=\mathrm{1.....}\left(1\right)$
Any line through origin and perpendicular to (1) is $\frac{x}{b}-\frac{y}{a}=\mathrm{0.....}\left(2\right)$
Now foot of the perpenduclar is the point of intersection of lines (1) and (2) . In order to find its locus we have to eliminate the variables $a$ and $b$.
Squaring and adding (1) and (2) we get
$x^2(\frac{1}{a^2}+\frac{1}{b^2})+y^2(\frac{1}{b^2}+\frac{1}{a^2})=1$
or $x^2+y^2=c^2$
$\because \frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2}$ (given)