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

Given variable line is `(x)/(a) + (y)/(b) = 1 " " (1)`
Line perpendicular to (1) and passing through the origin is
`(x)/(b) - (y)/(a) = 0 " " (2)`
Now, foot of perpendicular P(h,k) from the origin upon the line (1) is the point of intersection of line (1) and (2).
` (h)/(a) + (k)/(b) =1 " " (3)`
`" and " (h)/(b) - (k)/(a) =0 " " (4)`
Squaring and adding (3) and (4), we get
`h^(2) ((1)/(a^(2))+ (1)/(b^(2))) +k^(2) ((1)/(b^(2))+ (1)/(a^(2))) = 1`
`"But, it is given that " (1)/(a^(2)) + (1)/(b^(2)) = (1)/(c^(2))`
`therefore (h^(2)+ k^(2))(1)/(c^(2)) = 1`
Hence, equation of locus of P is `x^(2) + y^(2) = c^(2)`.