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                                                                                                                          ${(h-p)}^2+k^2=a^2$                            --------- (i)                                                                             

    $h^2+{(k-q)}^2=b^2$                          -------- (ii)                                                                            

  $p^2+q^2=a^2+b^2$                             ------- (iii)                                                                           

    From (i) , (ii)  and (iii), we get                                                                                                

          ${\left[h\pm \sqrt{a^2-k^2}\right]}^2+{\left[k\pm \sqrt{b^2-h^2}\right]}^2=a^2+b^2$                                                      

 $\Rightarrow \pm 2h\sqrt{a^2-k^2}\pm 2k\sqrt{b^2-h^2}=0$                                                                                   

        $\Rightarrow h^2{a}^2=k^2{b}^2$                                                                                                                

$\therefore LocusofCisax\pm by=0$