A rectangle of maximum area is inscribed in the circle |z−3−4i|=1. If one vertex of the rectangle is 4+4i, then another adjacent vertex of this rectangle can be
A
2+4i
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B
3+5i
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C
3+3i
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D
3−3i
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Solution
The correct options are B3+5i C3+3i
For area to be maximum the inscribed rectangle has to a square. Let the adjacent vertex be z. Then 3+4i−z(3+4i)−(4+4i)=e±iπ2 (By rotation about center) 3+4i−z=±i(−1) z=3+5ior3+3i