let agiven line L1 intersect the x and y axis at P &Q respectively.Let another line L2 perpendicular L1 cut the x and y axis at R & S respectively.Show that the locus of the point of intersection of the lines PS & QR is a circle passing through the origin.

let given line is L1: y=mx+c

(P,Q)=(-c/m,0) , (0,c)

now L2 is perpendicular to line L1

slope of L2*slope of L1=-1            (m1m2=-1)

 eq of L2: y=-x/m + k                 (k is any variable)

(Q,R)=(km,0),(0,k)

eq of PS is y=mkx/c + k and.................1

eq of QR is  y=-x/km + c ...................2

let the point if intersection be (X,Y)

 after solving 1 and 2

X= (c-k)/(mk/c + 1/mk )...............3

Y= (km+1/m) /(mk/c + 1/mk)...........4

 eliminate k by solving eq 3 and 4 you will get the equation in terms of X,Y and c .......this will give the required locus which is circle passing through origin...

Approved