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Question

If b1+b2+b3+b4=0 where biR such that the sum of no two values being zero and b1z1+b2z2+b3z3+b4z4=0 where z1,z2,z3,z4 are arbitary complex numbers such that no three of them are collinear, so the required condition for complex numbers to be concylic is

A
|b1b4||z1z4|2=|b2b3||z2z3|2
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B
|b1b3||z1z3|2=|b2b4||z2z4|2
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C
|b1b2||z1z2|2=|b3b4||z3z4|2
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D
None of these
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Solution

The correct option is C |b1b2||z1z2|2=|b3b4||z3z4|2

b1+b2=(b3+b4),b1z1+b2z2=(b3z3+b4z4)
b1z1+b2z2b1+b2=b3z3+b4z4b3+b4
Hence, the line joining the complex numbers A(z1),B(z2) and C(z3),D(z4) meet.
Let P(z) be the point of intersection. These points will be concyclic, if PA×PB=PC×PD. Now,
PA=b2b1+b2|z1z2|,PB=b1b1+b2|z1z2|
PC=b4b3+b4|z3z4|,PD=b3b3+b4|z3z4|
|b1b2||z1z2|2=|b3b4||z3z4|2(|b1+b2|=|b3+b4|)

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