Complex Numbers 3 Question 12
12. Let $z _1=10+6 i$ and $z _2=4+6 i$. If $z$ is any complex number such that the argument of $\left(z-z _1\right) /\left(z-z _2\right)$ is $\pi / 4$, then prove that $|z-7-9 i|=3 \sqrt{2}$.
(1991, 4M)
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Solution:
- Since, $z _1=10+6 i, z _2=4+6 i$
and arg $\frac{z-z _1}{z-z _2}=\frac{\pi}{4}$ represents locus of $z$ is a circle shown as from the figure whose centre is $(7, y)$ and $\angle A O B=90^{\circ}$, clearly $O C=9 \Rightarrow O D=6+3=9$
$\therefore$ Centre $=(7,9)$ and radius $=\frac{6}{\sqrt{2}}=3 \sqrt{2}$
$\Rightarrow \quad$ Equation of circle is $|z-(7+9 i)|=3 \sqrt{2}$
