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  4. Let z1= 10 + 6i and z2 = 4 + 6i. If z is any complex number such that the argument of (z - z1) / (z - z2) is π/4 then prove that |z - 7 - 9i | = 3√ 2

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Q. 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 $|z - 7 - 9i | = 3\sqrt 2$

IIT JEEIIT JEE 1991Complex Numbers and Quadratic Equations

Solution:

Since, $z_1 = 10 + 6t, z_2 = 4 + 6i$
and $arg \bigg(\frac{z-z_1}{z-z_2}\bigg) =\frac{\pi}{4}$ 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 =$\frac{6}{\sqrt 2}=3\sqrt 2$
image
$\Rightarrow \, $ Equation of circle is $|z-(7+9i)|=3\sqrt 2$