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A(z1),B(z2),C(z3) are the vertices of a triangle ABC inscribed in the circle |z|=2 . Internal angle bisector of the angle A, meet the circum circle again at D(z4) and 2z42−kz2z3=0 then K =

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answer is 2.

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Detailed Solution

BOD=A=CODFrom rotation about the point ‘O’ we getz4z2=eiA , z3z4=eiA⇒z42=z2z3

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A(z1),B(z2),C(z3) are the vertices of a triangle ABC inscribed in the circle |z|=2 . Internal angle bisector of the angle A, meet the circum circle again at D(z4) and 2z42−kz2z3=0 then K =