Question

Let, $z_1=10+6iand{z}_2=4+6i$. If $z$ is any complex number such that the argument of $\frac{(z-z_1)}{(z-z_2)}$ is $\frac{\pi }{4},$ then $|z-7-9i|$ is

• A. $2\sqrt{3}$
• B. $10$
• C. $63$
• D. $3\sqrt{2}$

Answer 1

The correct option is D $3\sqrt{2}$

$z_1=10+6i$ $;$ $z_2=4+6i$

$\Rightarrow arg\left(\frac{z-z_1}{z-z_2}\right)=\frac{\pi }{4}$

$\therefore |z-7-9i|=arg\left[\frac{(x+iy)-(10+60)}{(x+iy)-(4+6i)}\right]=\frac{\pi }{4}$

$\Rightarrow arg\left[\frac{(x-10)+i(y-6)}{(x-4)+i(y-6)}\right]=\frac{\pi }{4}$

$arg[(x-10)+i(y-6)]-arg[(x-4)+i(y-6)]=\frac{\pi }{4}$

${\theta }_1-{\theta }_2=\frac{\pi }{4}$

$\Rightarrow {\tan }^{-1}\left(\frac{y_1}{x_1}\right)-{\tan }^{-1}\left(\frac{y_2}{x_2}\right)={\tan }^{-1}1$

${\tan }^{-1}\left(\frac{y-6}{x-10}\right)-{\tan }^{-1}\left(\frac{y-6}{x-4}\right)={\tan }^{-1}1$

$\therefore \left(\frac{\frac{(y-6)}{(x-10)}-\frac{(y-6)}{(x-4)}}{1+\frac{(y-6)(y-6)}{(x-10)(x-4)}}\right)={\tan }^{-1}1$

$\Rightarrow \frac{6(y-6)}{x^2-14x+40+y^2-12y+36}=1$

$\Rightarrow x^2-14x+y^2-18y+112=0$

$(x^2-14x+49-49)+(y^2-18y+81-81)+112=0$

${(x-7)}^2-49+{(y-9)}^2-81+112=0$

${(x-7)}^2+{(y-9)}^2=18$

$\therefore |z-7-9i|=|x+iy-7-9i|$

$=|(x-7)+i(y-9)|$

$\Rightarrow \sqrt{{(x-7)}^2+{(y-9)}^2}=\sqrt{18}=3\sqrt{2}$

Hence, the answer is $3\sqrt{2}.$