Let A=z C:|z-2| 4 and B=z C:|z-2|+|z+2|=5.

Question

Let
$A={z\in C:|z-2|\le 4}$ and
$B={z\in C:|z-2|+|z+2|=5}$.
Then the max ${|{z}_{1}-{z}_{2}|:{z}_{1}\in A{z}_{2}\in B}$ is :

Accepted Answer

Analyze Set $A$ :

The set $A$ is defined as ${z\in C:|z-2|\le 4}$.

This represents a circle (and its interior) centered at $(2,0)$ with a radius $r=4$.

The circle extends from $x=-2$ to $x=6$ along the real axis.

Analyze Set $B$ :

The set $B$ is defined as ${z\in C:|z-2|+|z+2|=5}$.

By the definition of an ellipse, the sum of distances from two fixed points (foci) is constant.

Here, the foci are ${F}_{1}(2,0)$ and ${F}_{2}(-2,0)$, and the length of the major axis $2a=5$, so $a=2.5$.

By the definition of an ellipse, the sum of distances from two fixed points (foci) is constant.

Here, the foci are ${F}_{1}(2,0)$ and ${F}_{2}(-2,0)$, and the length of the major axis $2a=5$, so $a=2.5$.

The distance between foci is $2ae=4$, so $e=\frac{4}{5}=0.8$.

The semi-minor axis $b$ can be found using ${b}^{2}={a}^{2}(1-{e}^{2})=(2.5{)}^{2}(1-0.64)=6.25(0.36)=2.25$, so $b=1.5$.

The ellipse is centered at the origin $(0,0)$ and extends from $x=-2.5$ to $x=2.5$ along the real axis.

To find $max{|{z}_{1}-{z}_{2}|:{z}_{1}\in A$ and ${z}_{2}\in B}\,$ This means maximize the distance between a point ${z}_{1}$ in the circle and a point ${z}_{2}$ on the ellipse.

The furthest point in set $A$ (the circle) is at the rightmost edge, ${z}_{1}=(2+4,0)=(6,0)$.

The furthest point in set $B$ (the ellipse) from ${z}_{1}$ is at its leftmost edge, ${z}_{2}=(-2.5,0)$.

Maximum Distance $|{z}_{1}-{z}_{2}|=|6-(-2.5)|=|6+2.5|=8.5=\frac{17}{2}$

Solutions · 1