Question

If $z=x+iy,x,y\in R$ and if the point $P$ in the argand plane represents $z$, then the locus of $P$ satisfying the condition arg $\left(\frac{z-1}{z-3i}\right)=\frac{\pi }{2}$, is

• A. $\left\{z\in C/\left|z-\frac{1+3i}{2}\right|=\frac{\sqrt{10}}{2}\right\}$
• B. $\{z\in C/(3-i)z+(3+i)\bar{z}-6=0\}$
• C. $\{z\in C/(3-i)z+(3+i)\bar{z}-6>0$ $\left|z-\frac{1+3i}{2}\right|=\frac{\sqrt{10}}{2}\}$
• D. $\{z\in C/(3-i)z+(3+i)\bar{z}-6<0$ $\left|z-\frac{1+3i}{2}\right|=\frac{\sqrt{10}}{2}\}$

Answer 1

Answer: (C)

We have,
$\arg \left(\frac{z-1}{z-3i}\right)=\frac{\pi }{2}$
$\Rightarrow \arg (z-1)-\arg (z-3i)=\frac{\pi }{2}$
$\Rightarrow \arg [(x-1)+iy]-\arg [x+(y-3)i]=\frac{\pi }{2}$
$\Rightarrow {\tan }^{-1}\frac{y}{x-1}-{\tan }^{-1}\frac{y-3}{x}=\frac{\pi }{2}$
$\Rightarrow {\tan }^{-1}\left[\frac{\frac{y}{x-1}-\frac{y-3}{x}}{1+\frac{y}{x-1}\cdot \frac{y-3}{x}}\right]=\frac{\pi }{2}$
$\Rightarrow \frac{xy-(x-1)(x-3)}{x(x-1)+y(x-3)}=\tan \frac{\pi }{2}$
$\Rightarrow \frac{xy-(x-1)(y-3)}{x(x-1)+y(y-3)}=\frac{1}{0}$
$\Rightarrow x(x-1)+y(y-3)=0$
$\Rightarrow x^2+y^2-x-3y=0$
$\Rightarrow {\left(x-\frac{1}{2}\right)}^2+{\left(y-\frac{3}{2}\right)}^2=\frac{1}{4}+\frac{9}{4}$
$\Rightarrow {\left(x-\frac{1}{2}\right)}^2+{\left(y-\frac{3}{2}\right)}^2={\left(\frac{\sqrt{10}}{2}\right)}^2$
Which is a circle with centre
$\left(\frac{1}{2},\frac{3}{2}\right)$ and radius
$\frac{\sqrt{10}}{2}$.
$\therefore z\in C:(3-i)z+(3+i)\bar{z}-6>0,\left|z-\frac{1+3i}{2}\right|=\frac{\sqrt{10}}{2}$