If the complex number z is such that |z−25|≤25 and |z−50|=25 then the minimum value of |z|+25 is 10k then k is equal to
|z−25|≤25 represent interior and boundary of circle with
centre (25, 0) and radius 25. Points z satisfying |z−25|≤25
and |z−50|=25 lie on arc DAC hence min {|z|}=OA=25
∴|z|+25=50⋅ So, k=5