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JEE Advanced 2013 Paper 1 Mathematics Q1
Mathematics
Problem
Let complex numbers
α
\alpha
α
and
1
/
α
‾
1/\overline{\alpha}
1/
α
lie on circles
(
x
−
x
0
)
2
+
(
y
−
y
0
)
2
=
r
2
(x-x_0)^2+(y-y_0)^2=r^2
(
x
−
x
0
)
2
+
(
y
−
y
0
)
2
=
r
2
and
(
x
−
x
0
)
2
+
(
y
−
y
0
)
2
=
4
r
2
(x-x_0)^2+(y-y_0)^2=4r^2
(
x
−
x
0
)
2
+
(
y
−
y
0
)
2
=
4
r
2
, respectively. If
z
0
=
x
0
+
i
y
0
z_0=x_0+iy_0
z
0
=
x
0
+
i
y
0
satisfies the equation
2
∣
z
0
∣
2
=
r
2
+
2
2|z_0|^2=r^2+2
2∣
z
0
∣
2
=
r
2
+
2
, then
∣
α
∣
|\alpha|
∣
α
∣
=
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