Question

The points in the $\{z\in C;Arg\left(\frac{z-2}{z-6i}\right)=\frac{\pi }{2}\}$ lie on the curve which is a (Where C denotes the set of all complex numbers).

• A. Circle
• B. Pair of straight lines
• C. Parabola
• D. Hyperbola

Answer 1

Answer: (A)

Circle

$Arg\left(\frac{z-2}{z-6i}\right)=\frac{\pi }{2}$

$Arg(z-2)-Arg(z-6i)=\frac{\pi }{2}$

$z=x+iy$

$Arg\left(\right(x-2)+iy)-Arg(x+(y-6\left)i\right)=\frac{\pi }{2}$

$={\tan }^{-1}\left(\frac{y}{x-2}\right)-{\tan }^{-1}\left(\frac{y-6}{x}\right)=\frac{\pi }{2}$

${\tan }^{-1}\left(\frac{\frac{y}{x-2}-\frac{y-6}{x}}{1+\frac{y(y-6)}{x(x-2)}}\right)=\frac{\pi }{2}$

${\tan }^{-1}\left(\frac{xy-(y-6)(x-2)}{x(x-2)+4(y-6)}\right)=\frac{\pi }{2}$

$\frac{xy-(x-2)(y-6)}{x(x-2)+y(y-2)}=\tan \frac{\pi }{2}$

$\frac{xy-(x-2)(y-6)}{x(x-2)+y(y-2)}=\frac{1}{0}$

$x(x-2)-y(y-2)=0$

$x^2-2x+1-4+7(y^2-2y+1-1)=0$

$(x-1{)}^2+(y-1{)}^2=0$

this is equation of point circle

Circle