If the complex number z is such that $|\mathrm{z}-25|\le 25$ and $|\mathrm{z}-50|=25$ then the minimum value of $\left|\mathrm{z}\right|+25$ is 10k  then k is equal to

$|\mathrm{z}-25|\le 25$ represent interior and boundary of circle with 

centre (25, 0) and radius 25. Points z satisfying  $|\mathrm{z}-25|\le 25$

and  $|\mathrm{z}-50|=25$  lie on arc DAC hence min $\left\{\right|\mathrm{z}\left|\right\}=\mathrm{OA}=25$

$\therefore \left|\mathrm{z}\right|+25=50\cdot \text{ So, }\mathrm{k}=5$