Let $z_{1}=10+6 i$ and $z_{2}=4+6 i$. If $z$ is any complex number such that the argument of $\left(

Question

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 $|z-7-9 i|=$.

Professor PC Thomas Classes · Answer

$\arg \frac{z-z_{1}}{z-z_{2}}=\frac{\pi}{4}$\ $\Rightarrow \arg \left(z-z_{1}ight)-\arg \left(z-z_{2}ight)=\frac{\pi}{4}$\ $\Rightarrow \arg \{(x-10)+(y-6) i\}$\ $-\arg \{(x-4)+(y-6) i\}=\frac{\pi}{4}$\ $\Rightarrow 	an ^{-1} \frac{y-6}{x-10}-	an ^{-1} \frac{y-6}{x-4}=\frac{\pi}{4}$\ $\Rightarrow \frac{(y-6)(x-4-x+10)}{x^{2}-14 x+40+(y-6)^{2}}=	an \frac{\pi}{4}$\ $\Rightarrow 6(y-6)=y^{2}-12 y+36+x^{2}-14 x+40$\ $\Rightarrow y^{2}-18 y+112+x^{2}-14 x=0$\ $\Rightarrow \quad(y-9)^{2}+(x-7)^{2}=18$\ $\Rightarrow \quad \mid(x-7)+i(y-9)^{2}=18$\ $\Rightarrow|(x+i y)-7-9 i|=3 \sqrt{2}$\ $\Rightarrow|z-7-9 i|=3 \sqrt{2}$\