The sum of the real parts of the complex numbers satisfying the equations $\left|\frac{z-4i}{z-2i}\right|=1$ and $\left|\frac{z-8+3i}{z+3i}\right|=\frac{3}{5}is\frac{k}{5},$ then ‘k’ satisfies 

$$\begin{gathered} \left|\frac{z-4i}{z-2i}\right|=1\Rightarrow z=x+3iu\sin gthisin\left|\frac{z-8+3i}{z+3i}\right|=\frac{3}{5}\Rightarrow 5|x-8+6i|=3|x+6i| \\ \Rightarrow x=8,17 \end{gathered}$$

$twocomplexnumbers8+3i,17+3i$         $sumofrealpart=8+17=25$