Question

Let $z_1=10+6i$ and $z_2=4+6i,$ where $i=\sqrt{-1}.$ If $z$ is any complex number such that $\arg \left(\frac{z-z_1}{z-z_2}\right)=\frac{\pi }{4},$ then the value of $|z-7-9i|$ is

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

Answer 1

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

Let $z=x+iy$
Then, $\frac{z-z_1}{z-z_2}=\frac{(x-10)+i(y-6)}{(x-4)+i(y-6)}$
$=\frac{\left[\right(x-10)+i(y-6\left)\right]\left[\right(x-4)-i(y-6\left)\right]}{(x-4{)}^2+(y-6{)}^2}$
$=\frac{(x-10)(x-4)+(y-6{)}^2}{(x-4{)}^2+(y-6{)}^2}+i\frac{6(y-6)}{(x-4{)}^2+(y-6{)}^2}$

Now, $\arg \left(\frac{z-z_1}{z-z_2}\right)=\frac{\pi }{4}$
$\Rightarrow \frac{6(y-6)}{(x-10)(x-4)+(y-6{)}^2}=1$
$\Rightarrow x^2+y^2-14x-18y+112=0\cdots \left(1\right)$

Now, $|z-7-9i{|}^2=(x-7{)}^2+(y-9{)}^2$
$=x^2+y^2-14x-18y+130$
$=18\left[\text{From}\right(1\left)\right]$
$\therefore |z-7-4i|=\sqrt{18}=3\sqrt{2}$