JEE Advanced | 2013Complex NumbersHard
Question
Let complex numbers α and 
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| =
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| = Options
A.
B.
C.
D.
Solution
Given : α satisfies |z - z0| = r ⇒ |α - z0 | = r ..........(1)
satisfies |z - z0| = 2r 
.........(2)squaring (1) and (2) we get
⇒
..........(3)&
⇒
⇒
⇒ 1+ 2 |z0|2 - 2 - |α|2 - |z0|2 + |z0|2 |α|2 = 8|z0|2 |α|2 - 8|α|2
⇒ -1 + |z0|2 - 7|z0|2 |α|2 + 7|α|2 = 0
⇒ (|z0|2 - 1) (7|α2|- 1) = 0
⇒ |z0| = 1 (rejected as r = 0)
&
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