<s>[INST] Locus of $z$ satisfying $\arg \left(\frac{z-8}{z-2}\right)=\frac{\pi}{2}$ 
For a given complex number, $z$, find the locus of points on the Argand diagram such that
$$\arg \left(\frac{z-8}{z-2}\right)=\frac{\pi}{2}$$

This is my approach:
$$\arg(z-8)-\arg(z-2)=\frac{\pi}{2}$$
Suppose $z=x+iy$:
$$\arctan\frac{y}{x-8}-\arctan\frac{y}{x-2}=\frac{\pi}{2}$$
This is where I'm unsure I made a 'legal' move. I reasoned that if I take the tangent of both sides of the equation, I would end up with a fraction that must be undefined as $\tan\frac{\pi}{2}$ is undefined; hence the denominator of the fraction I obtain must be $0$:
$$\tan\left(\arctan\frac{y}{x-8}-\arctan\frac{y}{x-2}\right)=\frac{\frac{y}{x-8}-\frac{y}{x-2}}{1+\frac{y^2}{(x-8)(x-2)}}=\frac{y(x-2)-y(x-8)}{(x-8)(x-2)+y^2}$$
But this is undefined; hence
$$(x-8)(x-2)+y^2=0\implies (x-5)^2+y^2=9$$
so it appears that the required locus is a circle with centre at $(5,0)$ on the Argand diagram with radius $3$.
Is my reasoning fully acceptable throughout my solution?
 [/INST] If $A(z_A),B(z_B),M(z_M)$ are distinct points in Argand plane, then $\arg \left(\frac{z_M-z_B}{z_M-z_A}\right)$ is an oriented angle between vectors $(\vec{AM},\vec{BM}).$
In the present case $z_M=z,z_A=2,z_B=8.$
$\arg \left(\frac{z-8}{z-2}\right)=\frac{\pi}{2}$ says that the vectors are orthogonal, and $M$ lies on an open semi-circle with diameter $AB.$
(It is the upper half of the circle with diameter $AB,$ without $A,B.$)
</s>