Question

Let $z_1=10+6i$ and $z_2=4+6i$. If $z$ is any complex number such that the argument of $(z-z_1)/(z-z_2)$ is $\pi /4$ then prove that $|z-7-9i|=3\sqrt{2}$

Answer 1

Since, $z_1=10+6t,z_2=4+6i$
and $arg\left(\frac{z-z_1}{z-z_2}\right)=\frac{\pi }{4}$ represents locus of $z$ is a circle
shown as from the figure whose centre is $(7,y)$ and
$\angle AOB=90^{\circ },$ clearly $OC=9$ .
$\Rightarrow OD=6+3=9$
$\therefore$ Centre $=(7,9)$ and radius =$\frac{6}{\sqrt{2}}=3\sqrt{2}$

$\Rightarrow$ Equation of circle is $|z-(7+9i)|=3\sqrt{2}$