Q.

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

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By Expert Faculty of Sri Chaitanya

answer is 5.

(Detailed Solution Below)

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Detailed Solution

|z−25|≤25 represent interior and boundary of circle with centre (25, 0) and radius 25. Points z satisfying |z−25|≤25and |z−50|=25 lie on arc DAC hence min {|z|}=OA=25∴|z|+25=50⋅ So, k=5

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