If the base of a triangle and the ratio of the lengths of the other two unequal sides are given, then the vertex lies on a/an

Let the base of the triangle be the line segment joining B (0, 0) and C (a, 0) and let the vertex be A (h, k), where a is fixed. Also, letABAC=λ,λ≠1⇒AB2=λ2AC2⇒h2+k2=λ2(h−a)2+k2⇒h21−λ2+k21−λ2+2aλ2h−a2λ2=0So, A (h, k) lies onx2+y2+2aλ21−λ2x−a2λ21−λ2=0 which is a circle