Let coordinates of the points A and B are (5 , 0) and (0 , 7) respectively. P and Q are the variable points lying on the x-and y-axis respectively so that PQ is always perpendicular to the line AB. Then locus of the point of intersection of BP and AQ is

Let the equation of PQ be xa+yb=1 ---------- (i)                                                                            Then equation of BP is xa+y7=1        ----------- (ii)                                                                          and equation of AQ is x5+yb=1          --------- (iii)                                                                               Since AB⊥PQ , therefore −ba×−75=−1                                                                                            ⇒5a+7b=0                                                                                                                    Now, slope of BP is −7a  and slope of AQ is −b5                                                                                So, product of these slopes is 7b5a  = -1   [from (iv)]                                                                  ∴  BP and AQ intersect each other at right angle. So, the locus of the point of intersection of BP and AQ is a circle with AB as diameter. Thus equation of locus is                                                                  x(x−5)+y(y−7)=0⇒x2+y2−5x−7y=0