If the line xa+yb=1 moves in such a way that 1a2+1b2=1c2 where c is a constant, then the locus of the foot of perpendicular from the origin on the straight line is

Variable line isxa+yb=1-----(1)Any line perpendicular to (l) and passing through the origin will be xb−ya=0-----(2)Now foot of the perpendicular from the origin to line (l) is the point of intersection (1) and (2).  Let it be P(α,β), then αa+βb=1-----(3) and αb−βa=0-----(4)Squaring and adding (3) and (4), we get    α21a2+1b2+β21b2+1a2=1∴     α2+β21c2=1 Hence, the locus of P(α,β) is x2+y2=c2 .