Question

If z is a complex number such that arg $\left(\frac{z-6}{z+6}\right)=\frac{\pi }{3}$ then the locus of $z$ is

• A. $|z-2\sqrt{3i}|=4\sqrt{3}$, $Re\left(z\right)>0$
• B. $|z-2\sqrt{3i}|=4\sqrt{3}$, $Re\left(z\right)<0$
• C. $|z-2\sqrt{3i}|=4\sqrt{3}$, $Im\left(z\right)<0$
• D. $|z-2\sqrt{3i}|=4\sqrt{3}$, $Im\left(z\right)>0$

Answer 1

Answer: (D)

$|z-2\sqrt{3i}|=4\sqrt{3}$, $Im\left(z\right)>0$

$P\left(z\right)$ lies on a segment of a circle on $AB$ subtendlng an angle $\frac{\pi }{3}$ at $P$, where $A$ is $(-6,0)$, $B$ is $(6,0)$. Mid-point of $AB$ is $O(0,0).C$ is ths centre of the clrcle.
Let a be the radius , from $\Delta BOC,$ of the circle.
$sin\frac{\pi }{3}=\frac{6}{a}$
$\Rightarrow \frac{\sqrt{3}}{2}=\frac{6}{a}$
$\Rightarrow a=4\sqrt{3}$
Also $\tan \frac{\pi }{3}=\frac{6}{x}i.e.,\sqrt{3}=\frac{6}{x}$
$\Rightarrow x=2\sqrt{3}$
But OC is along the y axis
$\therefore$ $C$ is $(0,2\sqrt{3})$
The circle should lie above the $x$-axis (real axis)
$\therefore$ locus of z is $|z-2\sqrt{3i}|=4\sqrt{3}$
where Img $z>0$