The side AB of triangle ABC is fixed and is o | vedclass

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(B) Let the vertex $C$ be represented by the complex number $z = x + iy$. The coordinates of $A$ are $(0, 0)$ and $B$ are $(2a, 0)$.Since the angle $\

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$x^2+y^2-2ax-2ay \cot \alpha=0$

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(B) Let the vertex $C$ be represented by the complex number $z = x + iy$. The coordinates of $A$ are $(0, 0)$ and $B$ are $(2a, 0)$.Since the angle $\angle ACB = \alpha$, the argument of the ratio of the vectors $\vec{CA}$ and $\vec{CB}$ is $\alpha$.$\arg \left( \frac{0 - z}{2a - z} \right) = \alpha$$\arg \left( \frac{-z}{2a - z} \right) = \alpha$$\arg \left( \frac{z}{z - 2a} \right) = \alpha$Using the property $\arg \left( \frac{z - z_1}{z - z_2} \right) = \alpha$, the locus is a circle passing through $A(0,0)$ and $B(2a,0)$.The equation of the circle is $(x^2 + y^2) - 2ax - 2ay \cot \alpha = 0$.

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

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