The points in the set $\left\{z \in C: \arg \left(\frac{z-2}{z-6 i}\right)=\frac{\pi}{2}\right\}$ (where $C$…
The points in the set $\left\{z \in C: \arg \left(\frac{z-2}{z-6 i}\right)=\frac{\pi}{2}\right\}$ (where $C$ denotes the set of all complex numbers) lie on the curve which is a
circle
pair of lines
parabola
hyperbola
Solution
Given that, $\arg \left(\frac{z-2}{z-6 i}\right)=\frac{\pi}{2}$
$
\therefore \quad \arg (z-2)-\arg (z-6 i)=\frac{\pi}{2}
$
Let $z=x+i y$
$
\begin{aligned}
& \Rightarrow \arg [(x-2)+i y]-\arg [x+i(y-6)]=\frac{\pi}{2} \\
& \Rightarrow \quad \tan ^{-1} \frac{y}{x-2}-\tan ^{-1} \frac{y-6}{x}=\frac{\pi}{2} \\
& \Rightarrow \quad\left(\frac{\frac{y}{x-2}-\frac{y-6}{x}}{1+\frac{y}{x-2} \cdot \frac{y-6}{x}}\right)=\tan \frac{\pi}{2} \\
& \Rightarrow \quad 1+\frac{y}{x-2} \cdot \frac{y-6}{x}=0 \\
& \Rightarrow \quad x(x-2)+y(y-6)=0
\end{aligned}
$
This is an equation of circle in diametric form