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Solution
$|Z-5| \leq 3$ center $C(5,0), R=3$ Point $A \equiv$ greatest argument $\Delta O A C$ is right angled triangled $\sin \theta=\frac{3}{5}, \cos \theta=\frac{4}{5}$
\[ \begin{aligned} & \mathrm{OM}=4 \cos \theta=\frac{16}{5} ; \mathrm{AM}=4 \sin \theta=\frac{12}{5} \\ & \mathrm{~A}\left(\frac{16}{5}, \frac{12}{5}\right) \mathrm{Z}=\frac{16+12 \mathrm{i}}{5} \\ & 34\left|\frac{5 \mathrm{z}-12}{5 \mathrm{zi}+16}\right|^{2}=34\left|\frac{16+12 \mathrm{i}-12}{(16+12 \mathrm{i}) \mathrm{i}+16}\right|^{2}=34\left|\frac{4+12 \mathrm{i}}{4+16 \mathrm{i}}\right|^{2}=34\left|\frac{1+3 \mathrm{i}}{1+4 \mathrm{i}}\right|^{2}=34 \frac{(1+9)}{(1+16)}=2(10)=20 \end{aligned} \]
Solution
Set S: $\frac{3}{2} \leq\left|z-\frac{3}{2}(1+i)\right| \leq \frac{7}{2}$ First circle $\left|z-\frac{3}{2}(1+i)\right|=\frac{3}{2}$
$\mathrm{C}_{1}\left(\frac{3}{2}, \frac{3}{2}\right), \mathrm{r}_{1}=\frac{3}{2}$ Second circle $\left|\mathrm{z}-\frac{3}{2}(1+\mathrm{i})\right|=\frac{7}{2}$
\[ \begin{aligned} & \mathrm{C}_{2}\left(\frac{3}{2}, \frac{3}{2}\right), \mathrm{r}_{2}=\frac{7}{2} \\ & \underset{\mathrm{z} \in \mathrm{~S}}{\operatorname{Min}}\left|\mathrm{z}-\left(-\frac{5}{2}-\frac{3 \mathrm{i}}{2}\right)\right|=\mathrm{PB} \\ & =\mathrm{PC}-\mathrm{r}_{2}=\sqrt{4^{2}+3^{2}}-\frac{7}{2}=5-\frac{7}{2}=\frac{3}{2} \text { Ans. } \end{aligned} \]Solution
minimum \& maximum distance lie along their common normal (line joining centres)
\[ \begin{aligned} & |z|=r \text { and } O A=r \\ & \text { minimum }\left|z_{1}-z_{2}\right|=2=A B \\ & \Rightarrow O A-O B=A B \\ & \Rightarrow r-10=2 \Rightarrow r=12 \end{aligned} \]$\operatorname{Max}\left|\mathrm{z}_{1}-\mathrm{z}_{2}\right|=\mathrm{CB}=\mathrm{OC}+\mathrm{OB}=\mathrm{r}+10=12+10=22$
Solution
$|\mathrm{z}-(3+5 \mathrm{i})|+|\mathrm{z}-(5+1 \mathrm{li})|=4 \sqrt{5} \Rightarrow \mathrm{PS}_{1}+\mathrm{PS}_{2}=4 \sqrt{5}=2 \mathrm{a}$
So, $\mathrm{S}_{1}(3,5) \& \mathrm{~S}_{2}(5,11)$ are two foci of ellipse, where $2 \mathrm{ae}=\mathrm{S}_{1} \mathrm{~S}_{2}=2 \sqrt{10} \& 2 \mathrm{a}=4 \sqrt{5}$\[ \Rightarrow \mathrm{a}=2 \sqrt{5}, \mathrm{e}=\frac{1}{\sqrt{2}}=\sqrt{1-\frac{\mathrm{b}^{2}}{20}} \Rightarrow \mathrm{~b}^{2}=10 \]Centre of an ellipse is mid point of both focus $\mathrm{S}_{1} \& \mathrm{~S}_{2}$ ∴ center $(4,8)$
Ellipse $\frac{(\mathrm{x}-4)^{2}}{20}+\frac{(\mathrm{y}-8)^{2}}{10}=1$
Equation of circle is $|\mathrm{z}-(4+8 \mathrm{i})|=\sqrt{10}$
Circle $(\mathrm{x}-4)^{2}+(\mathrm{y}-8)^{2}=10$
∴ Ellipse \& given circle are concentric \& radius of given circle is equal to length of minor axis of ellipse
from diagram we can see that both touch each other at ends of minor axis.
