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Question:

Let complex numbers α and 1/α lie on circles (x−x₀)²+(y−y₀)²=r² and (x−x₀)²+(y−y₀)²=4r², respectively. If z₀=x₀+iy₀ satisfies the equation 2|z₀|²=r²+2, then |α|=

1√2

12

1√7

13

Solution:

|α−z₀|=r. (1)
|1/α−z₀|=2r (2)
2|z₀|²=r²+2. (3)
α.1/α=|α|² ⇒1/α=α/|α|²
Putting 1/α=α/|α|² in (2), we get
|α/|α|²−z₀|=2r (4)
Squaring (1), we get, |α|²+|z₀|²−(αz₀*+αz₀)=r² ⇒|α|²+(r²/2+1)−(αz₀z₀)=r² (5)
Squaring (4), we get, |α|⁴+|z₀|²−(αz₀
/|α|²+αz₀/|α|²)=4r² ⇒1/|α|²(1−(αz₀z₀))=4r²−|z₀|²=4r²−(r²/2+1)=7r²/2
⇒1−(αz₀
z₀)=|α|²(7r²/2)
⇒αz₀
+α*z₀=1−|α|²(7r²/2)
Substituting in (5), we get
|α|²+(r²/2+1)−(1−|α|²(7r²/2))=r²
⇒r²/2+|α|²(7r²/2)=r²
⇒|α|²(7r²/2)=r²/2
⇒|α|²=1/7
∴ |α|=1/√7