If the equation of the locus point equidistant from the points a1,b1 and a2,b2 is a1−a2x+b1−b2y+c=0, then the value of c is

Let h,k be the points on the locus. Then by the given conditions,h−a2+k−b12=h−a22+k−b22or 2ha1−a2+2kb1−b2+a22+b22−a12−b12=0or 2ha1−a2+kb1−b2+12a22+b22−a12−b12=0         …(1)Also since h,k lies on the given locus, we have a1−a2h+bb1−b2k+c=0                                         …(2)Comparing (1) and (2) , we get c=12a22+b22−a12−b12