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Question

If z1 and z2 lie on a circle having centre at the origin, then point of intersection of the tangents at z1 and z2 is

A
12(¯z1+¯z2)
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B
2z1z2z1+z2
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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 2z1z2z1+z2
Let point of intersection of tangents be z3


As OAC is a right -angle 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 (1) from (2), we get
2(¯z1¯z2)=z3(¯z1z1¯z2z2)2(z1¯¯¯¯¯z1z1z2¯¯¯¯¯z2z2)=z3(z1¯¯¯¯¯z1z21z2¯¯¯¯¯z2z22)
2r2(z2z1)z1z2=z3r2(z22z21z21z22) [|z1|2=|z2|2=r2]
z3=2z1z2z2+z1

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