KVPY SX SB Exam 05-Nov-2017 Question Paper

Case (i) :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, ${\left(\mathit{x}-\mathit{a}\right)}^2+{\mathit{y}}^2=2{\mathit{a}}^2$
Circle
Case (iii) :
∠A = ∠B
AC = BC
$\sqrt{{\mathit{h}}^2+{\left(\mathit{k}-\mathit{a}\right)}^2}=\sqrt{2{\mathit{a}}^2}$
${\mathit{x}}^2+{\left(\mathit{y}-\mathit{a}\right)}^2=2{\mathit{a}}^2$
also a circle
So union of two circle and a line.