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Question

z1 and z2 lies on the circle with centre at the origin. The point of intersection z3 of the tangents at z1 and z2 is given by

A
12(¯z1+¯z2)
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B
2z1z2(¯z2¯z1)z1¯z2z2¯z1
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C
12(1z1+1z2)
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D
z1+z2¯z1¯z2
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Solution

The correct option is B 2z1z2(¯z2¯z1)z1¯z2z2¯z1


As OAC is a right-angled triangle with right angle at A, so
|z1|2+|z3z1|2=|z3|2
2|z1|2¯z3z1¯z1z3=0
2¯z1¯z3¯z1z1z3=0 (1)
Similarly, 2¯z2¯z3¯z2z2z3=0 (2)

Substracting (2) from (1), we get
2(¯z1¯z2)z3(¯z1z1¯z2z2)=0
2(¯z2¯z1)=z3(¯z2z2¯z1z1)
2(¯z2¯z1)=z3(z1¯z2z2¯z1z1z2)
z3=2z1z2(¯z2¯z1)z1¯z2z2¯z1

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