A right angle triangle ABC having right angle at C , CA = a, CB = b moves such that the angular points A and B slide along x-axis and y-axis respectively. Then the locus C is

Let the coordinates of A and B are respectively (p, 0) and (0, q). Let C be (h , k) , so that                                                                                                                          (h−p)2+k2=a2                            --------- (i)                                                                                 h2+(k−q)2=b2                          -------- (ii)                                                                              p2+q2=a2+b2                             ------- (iii)                                                                               From (i) , (ii)  and (iii), we get                                                                                                          [h±a2−k2]2+[k±b2−h2]2=a2+b2                                                       ⇒±2ha2−k2±2kb2−h2=0                                                                                           ⇒h2a2=k2b2                                                                                                                ∴  Locus  of  C  is  ax±by=0