The number of values of $\mathrm{z} \in \mathrm{C}$, satisfying the equations $|\mathrm{z}-(4+8 \mathrm{i})|=\sqrt{10}$ and $|\mathrm{z}-(3+5 \mathrm{i})|+|\mathrm{z}-(5+11 \mathrm{i})|=4 \sqrt{5}$, is
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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.
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