Question

The complex number $z$ which simultaneously satisfies the equations $\left|\frac{z-12}{z-8i}\right|=\frac{5}{3}$ and $\left|\frac{z-4}{z-8}\right|=1$ is

• A. $6-8i$
• B. $6+8i$
• C. $6+17i$
• D. $6-17i$

Answer 1

Answer: (B), (C)

$6+17i$

Let $z=x+iy$
Then
$\left|\frac{z-12}{z-8i}\right|=\frac{5}{3}\Rightarrow \left|\frac{x+iy-12}{x+iy-8i}\right|=\frac{5}{3}\Rightarrow \left|\frac{(x-12)+iy}{x+(y-8)i}\right|=\frac{5}{3}\Rightarrow 9(x-12{)}^2+9y^2=25x^2+25(y-8{)}^2\Rightarrow 9(x^2+144-24x)+9y^2=25x^2+25(y^2-16y+64)\Rightarrow 16x^2+16y^2+216x-400y+304=0\Rightarrow 2x^2+2y^2+27x-50y+38=0\cdots \left(1\right)$

And
$\left|\frac{z-4}{z-8}\right|=1\Rightarrow \left|\frac{x+iy-4}{x+iy-8}\right|=1\Rightarrow (x-4{)}^2+y^2=(x-8{)}^2+y^2\Rightarrow -8x+16=-16x+64\Rightarrow 8x=48\Rightarrow x=6\cdots \left(2\right)$

From equation $\left(1\right)$ and $\left(2\right)$, we get
$72+2y^2+162-50y+38=0\Rightarrow y^2-25y+136=0\therefore y=8,17$

Hence, the values of $z$ are $6+8i,6+17i$


Alternate Solution:
Check all the options

When $z=6-8i$
$\left|\frac{z-12}{z-8i}\right|=\frac{10}{2\sqrt{73}}=\frac{5}{\sqrt{73}}\ne \frac{5}{3}$

When $z=6+8i$
$\left|\frac{z-12}{z-8i}\right|=\frac{10}{6}=\frac{5}{3}$
$\left|\frac{z-4}{z-8}\right|=\frac{\sqrt{68}}{\sqrt{68}}=1$
Both the conditions are satisfied

When $z=6+17i$
$\left|\frac{z-12}{z-8i}\right|=\frac{5\sqrt{13}}{3\sqrt{13}}=\frac{5}{3}$
$\left|\frac{z-4}{z-8}\right|=\frac{\sqrt{293}}{\sqrt{293}}=1$
Both the conditions are satisfied

When $z=6-17i$
$\left|\frac{z-12}{z-8i}\right|=\frac{5\sqrt{13}}{\sqrt{661}}\ne \frac{5}{3}$