Question

The number of complex number(s) $z$ satisfying $|z-3-i|=|z-9-i|$ and $|z-3+3i|=3$ is

• A. one
• B. two
• C. four
• D. infinitely many

Answer 1

The correct option is A one

Let $z=x+iy$. Then,
$|z-3-i|=|z-9-i|$
$\Rightarrow \sqrt{(x-3{)}^2+(y-1{)}^2}=\sqrt{(x-9{)}^2+(y-1{)}^2}$
$\Rightarrow (x-3{)}^2+(y-1{)}^2=(x-9{)}^2+(y-1{)}^2$
$\Rightarrow x=6$
Now, $|z-3+3i|=3$
$\Rightarrow \sqrt{(x-3{)}^2+(y+3{)}^2}=3$
$\Rightarrow (x-3{)}^2+(y+3{)}^2=9$
For $x=6,y=-3$.
$\therefore z=6-3i$