Let $\mathrm{BC}$ be a fixed line segment in the plane. The locus of a point $A$ such that the triangle $\mathrm{ABC}$ is isosceles, is (with finitely many possible exceptional points)

• A

  a line
• B

  a circle
• C

  the union of a circle and a line
• Correct answer
  D

  the union of two circles and a line

Solution

Case (i):
If $\angle \mathrm{B}=\angle \mathrm{C}$ locus of $\mathrm{A}$ is $\perp$ bisector of $\mathrm{BC}$ So it is straight line Case (ii):
If $\angle \mathrm{A}=\angle \mathrm{C}$ BC fixed $\mathrm{B}(\mathrm{a}, 0), \mathrm{C}(0, \mathrm{a})$ $\mathrm{BC}=\mathrm{AB}$ So, $(x-a)^{2}+y^{2}=2 a^{2}$ Circle Case (iii): $\begin{array}{l} \angle \mathrm{A}=\angle \mathrm{B} \\ \mathrm{AC}=\mathrm{BC} \\ \sqrt{\mathrm{h}^{2}+(\mathrm{k}-\mathrm{a})^{2}}=\sqrt{2 \mathrm{a}^{2}} \\ \mathrm{x}^{2}+(\mathrm{y}-\mathrm{a})^{2}=2 \mathrm{a}^{2} \end{array}$ also a circle So union of two circle and a line.

Hence, the correct answer is option: 'D'.