If points A(3,5) and B are equidistant from H(2,5) and B has rational coordinates, then AB=

Let B≡(α,β), where α,β are rational. Given HA2=HB2⇒(3−2)2+(5−5)2=(α−2)2+(β−5)2⇒41−62−105=α2+β2+7−22α−25β⇒α2+β2−34+2(3−α)2+2(5−β)5=0Since α  and  βare rational, 5−β=0⇒β=5and3−α=0⇒α=3∴AB=0