If the line `(x/a)+(y/b)=1` moves in such a way that `(1/(a^2))+(1/(b^2))=(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^2dot`

Explanation

The equation of the line is `x/a​+y/b​=1 ...(1)`

where `1/(a^2)​+1/(b^2)​=1/(c^2)​ ...(2)`

We know that any line perpendicular to (1) is `x/a​−y/b​+k=0`

and If it passes through the origin then `k=0`
.
∴ Equation of the line through the origin and perpendicular to (1) is` x/b​−y/a​=0 ...(3)`


Squaring (1) and (3) and adding, we get

`(x^2)/(a^2)+(y^2)/(b^2)+(x^2)/(a^2)-(y^2)/(b^2)=1+0`

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