Question

If $z_1=6+i,z_2=4-3i$ and z be a complex number such that arg $\left(\frac{z-z_1}{z_2-z}\right)=\frac{\pi }{2},$ then z satisfies

• A. $|z-(5-i)|=5$
• B. $|z-(5-i)|=\sqrt{5}$
• C. $|z-(5+i)|=5$
• D. $|z-(5+i)|=\sqrt{5}$

Answer 1

Answer: (B)

$\because$ $\arg \left(\frac{z-z_1}{z_2-z}\right)=\frac{\pi }{2}$ $\Rightarrow$ $\mathrm{Re}\left(\frac{z-z_1}{z_2-z}\right)=0$ $\Rightarrow$ $\frac{z-z_1}{z_2-z}+\frac{\overline{z}-{\overline{z}}_1}{{\overline{z}}_2-\overline{z}}=0$ $\Rightarrow$ $(z-z_1)({\overline{z}}_2-\overline{z})+(z_2+-z)(\overline{z}-{\overline{z}}_1)=0$ $\Rightarrow$ $z({\overline{z}}_1+{\overline{z}}_2)+\overline{z}(z_1+z_2)2z\overline{z}$ $-(z_1{\overline{z}}_2+z_2\overline{z_1})=0$ $\Rightarrow$ $z\overline{z}-(5+i)z+(5-i)\overline{z}+21=0$ $\therefore$ $|z-(5-i)|=\sqrt{5}$