Let a ,   b   ∈ R and a 2 + b 2 ≠ 0 . Suppose S = z ∈ C   : z = 1 a + i b t…

Let a, b ∈R and a2+b2≠0 . Suppose S=z∈C :z=1a+ibt, t ∈R, t ≠0, where i= -1. If z=x+iy and z∈S, then x, y lies on
  1. The circle with radius 12a and centre 12a, 0 for a>0, b ≠0
  2. The circle with radius -12a and centre -12a, 0 for a<0, b≠0
  3. The x - axis for a ≠0, b=0
  4. The y - axis for a=0, b ≠0

Solution

x+iy=1a+ibt (Multiply & divide by conjugate a+ibt)
x+iy=a-ibta2+b2 t2
Let a ≠0 and b ≠0
x=aa2+b2 t2 ......(i)
y=-bta2+b2 t2 ......(ii)
yx=-bta ⇒  t= -aybx
Put in (i)
xa2+b2.a2y2b2x2=a
⇒ a2 x2+y2=ax
x2+y2-1ax=0
x-12a2+y2=14a2 
Circle with centre ( 1 2a ,0 ) , radius = 1 2| a | { a>0 ;  1 2a a<0 ;  −1 2a
For a≠0, b=0
x+iy=1a
x=1a, y=0 ⇒z lies on x - axis  
For a=0,  b≠0
x+iy=1ibt
y= -1bt, x=0
⇒ z lies on y - axis.

Asked in: JEE Advanced 2016 (Paper 2)

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