The side $A B$ of $\triangle A B C$ is fixed and is of length $2 a$ unit. The vertex moves in the plane such that the vertical angle is always constant and is $\alpha$. Let $x$-axis be along $A B$ and the origin be at $A$. Then the locus of the vertex is

• A

  $x^2+y^2+2 a x \sin \alpha+a^2 \cos \alpha=0$
• Correct answer
  B

  $x^2+y^2-2 a x-2 a y \cot \alpha=0$
• C

  $x^2+y^2-2 a x \cos \alpha-a^2=0$
• D

  $x^2+y^2-a x \sin \alpha-$ ay $\cos \alpha=0$

Solution

Let, $c=z=x+i y$ $\begin{aligned} &\arg \left(\frac{2 \mathrm{a}-\mathrm{z}}{0-\mathrm{z}}\right)=\alpha \\ &\Rightarrow \arg \left(\frac{(\mathrm{x}-2 \mathrm{a})+\mathrm{iy}}{\mathrm{x}+\mathrm{iy}}\right)=\alpha \\ &\Rightarrow \mathrm{x}^2+\mathrm{y}^2-2 \mathrm{ax}-2 \mathrm{ay} \cot \alpha=0 \end{aligned}$

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