Question

Let 'z' be a complex number satisfying $|z-2-i|\le 5,$ Then |z-14-6i| lies in

• A. {8,18}
• B. {2,8}
• C. {0,2}
• D. {3,7}

Answer 1

The correct option is A {8,18}

Here, $|z-2-i|\le 5$ ………..$\left(1\right)$

Now, $|z-14-6i|=|z-2-i-12-5i|=|(z-2-i)-(12+5i)|$

$\le |z-2-i|+|12+5i|$

$\le 5+|12+5i|$

$=5+\sqrt{(12{)}^2+(5{)}^2}$

$=5+\sqrt{144+25}$

$=5+\sqrt{169}$

$=5+13$

$=18$.

$\therefore |z-14-6i|\le 18$

and $|z-14-6i|=|z-2-i-12-5i|$

$=|-(12+5i)+(z-2-i)|$

$\ge ||-(12+5i)|-|z-2-i||$

$\ge |12+5i|-5$ $[\because |z-2-i|\le 5$ $\Rightarrow -|z-2-i|\ge -5]$

$=13-5$

$=8$

$\therefore |z-14-6i|\ge 8$

$\therefore |z-14-6i|$ lies in $\{8,18\}$.