The number of solutions of the equations $\left|z-(4+8i)\right|=\sqrt{10}and\left|z-(3+5i)\right|+\left|z-(5+11i)\right|=4\sqrt{5,}$Where $i=\sqrt{-1}$

Here, $\left|z-(4+8i)\right|=\sqrt{10}$

Represents  a circle with centre(4,8) and radius  $\sqrt{10}$

 Also, $\left|z-(3+5i)\right|+\left|z-(5+11i)\right|=4\sqrt{5}$

Represents an ellipse.

$\therefore \left|(3+5i)-(5+11i)\right|=\sqrt{4+36}=\sqrt{40}<4\sqrt{5}$

With foci  $s_1(3,5)and{s}_2(5,11)$

Distance between foci $=s_1{s}_2=\sqrt{40}=2\sqrt{10}=Diameterofcircle$

i.e..                                  $2ae=2\sqrt{10}$

$\Rightarrow ae=\sqrt{10}and\text{\hspace{0.17em}}2a=4\sqrt{5}\Rightarrow a=2\sqrt{5}$

$\therefore e=\frac{ae}{a}=\frac{1}{\sqrt{2}}$

Now, $b=a\sqrt{(1-e^2)}=2\sqrt{5}\sqrt{(1-\frac{1}{2})}=\sqrt{10}=$ Radius of circle

$\therefore$ Centre of the ellipse=mid-point of $s_{1}and{s}_2$

$=\frac{3+5i+5+11i}{2}=4+8iie.,(4,8)$

Which coincides with the centre of the circle and length of minor –axis is equal to the radius of the circle. Hence , there are only(2) two solution of the given equations.