Let the circles C_1:|z|=r and C_2:|z-3-4i|=5, z C, be such that C_2 lies within C_1.

Question

Let the circles ${C}_{1}:|z|=r$ and ${C}_{2}:|z-3-4i|=5$, $z\in C$, be such that ${C}_{2}$ lies within ${C}_{1}$.
If ${z}_{1}$ moves on ${C}_{1}$, ${z}_{2}$ moves on ${C}_{2}$ and $min|{z}_{1}-{z}_{2}|=2$, then $max|{z}_{1}-{z}_{2}|$ is equal to :

Accepted Answer

$\begin{gathered} |z|=r \ OA=r \end{gathered}$

Circle ${C}_{1}:|z|=r$ has centre $O(0,0)$ and radius $r$ (so $OA=r$): 1: Circle ${C}_{1}:|z|=r$ has centre $O(0,0)$ and radius $r$ (so $OA=r$).

Circle ${C}_{2}:|z-3-4i|=5$ has centre at the point $(3,4)$ and radius $5$: 2: Circle ${C}_{2}:|z-3-4i|=5$ has centre at the point $(3,4)$ and radius $5$.

Distance between the centres i: 3: Distance between the centres is

$OC=\sqrt{{3}^{2}+{4}^{2}}=5$

Since ${C}_{2}$ lies completely inside ${C}_{1}$, the minimum distance between a point ${z}_{1}$ on ${C}_{1}$ and a point ${z}_{2}$ on ${C}_{2}$ happens when the two points are on the same line through the centres, on the nearest sides: 4: Since ${C}_{2}$ lies completely inside ${C}_{1}$, the minimum distance between a point ${z}_{1}$ on ${C}_{1}$ and a point ${z}_{2}$ on ${C}_{2}$ happens when the two points are on the same line through the centres, on the nearest sides.

So,

minimum $|{z}_{1}-{z}_{2}|=r-(OC+5)=r-(5+5)=r-10$

Given minimum $|{z}_{1}-{z}_{2}|=2$, we get

$\Rightarrow r-10=2\Rightarrow r=12$

The maximum distance $|{z}_{1}-{z}_{2}|$ happens when ${z}_{1}$ and ${z}_{2}$ are on opposite sides along the same line through the centres: 5: The maximum distance $|{z}_{1}-{z}_{2}|$ happens when ${z}_{1}$ and ${z}_{2}$ are on opposite sides along the same line through the centres.

Then the farthest distance is

$|{z}_{1}-{z}_{2}|=r+(OC+5)=r+10$

Substituting $r=12$,

$|{z}_{1}-{z}_{2}|=12+10=22$

Solutions · 1