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¯z2−z2¯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 B2z1z2(¯z2−¯z1)z1¯z2−z2¯z1
As △OAC is a right-angled 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 (2) from (1), we get 2(¯z1−¯z2)−z3(¯z1z1−¯z2z2)=0 ⇒2(¯z2−¯z1)=z3(¯z2z2−¯z1z1) ⇒2(¯z2−¯z1)=z3(z1¯z2−z2¯z1z1z2) ⇒z3=2z1z2(¯z2−¯z1)z1¯z2−z2¯z1