If ∠B = ∠ C locus of A is ⊥ bisector of BC So it is straight line Case (ii) :

If ∠A = ∠C BC fixed B(a, 0), C(0, a) BC = AB So, (x−a)2+y2=2a2 Circle Case (iii) : ∠A = ∠B AC = BC √h2+(k−a)2=√2a2 x2+(y−a)2=2a2 also a circle So union of two circle and a line.