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 B2z1z2z1+z2 Let point of intersection of tangents be z3
As △OAC is a right -angle triangle with right angle at A , so |z1|2+|z3−z1|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¯¯¯¯¯z1z1−z2¯¯¯¯¯z2z2)=z3(z1¯¯¯¯¯z1z21−z2¯¯¯¯¯z2z22) ⇒2r2(z2−z1)z1z2=z3r2(z22−z21z21z22)[∵|z1|2=|z2|2=r2] ⇒z3=2z1z2z2+z1