Question

If $|z-1|+|z+3|\le 8$,then range of $|z-4|$ is

• A. $(0,5)$
• B. $(7,0)$
• C. $(1,9)$
• D. $(5,0)$

Answer 1

The correct option is C $(1,9)$

Given:

$|z+3|+|z-1|\le 8$

z lies on and inner region of the ellipse $|z+3|+|z-1|=8$
From the figure:
Equation of ellipse is $SP+S^{\prime }P=2a$ where, S & S' are focii and $2a$ is major axis.

$|z+3|+|z-1|=8$
Comparing above eq with $SP+S^{\prime }P=2a$ we get,

$2a=8$
$\Rightarrow a=4$

$min|z-4|$=distance between point $P(4,0)$ & point $A(3,0)=d\left(PA\right)$

$min|z-4|=d\left(PA\right)=1$

$max|z-4|=$distance between point $P(4,0)$ & point $B(-5,0)=d\left(PB\right)$

$max|z-4|=d\left(PB\right)=9$
$min|z-4|\le |z-4|\le max|z-4|$

$\Rightarrow 1\le |z-4|\le 9$

$\Rightarrow |z-4|\in [1,9]$

Hence, option 'C' is correct.