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