Let S 1 = z 1 ∈ C : z 1 - 3 = 1 2 and S 2 = z 2 ∈ C : z 2 - z 2 + 1 = z 2 + z 2 - 1 . Then, for…

Let S1=z1∈C:z1-3=12 and S2=z2∈C:z2-z2+1=z2+z2-1. Then, for z1∈S1 and z2∈S2, the least value of z2-z1 is
  1. 0
  2. 12
  3. 32
  4. 52

Solution

Here z1-3=12 represents a circle on argand plane with centre 3,0 and radius 12

Given z2+z2−12=z2−z2+12

⇒z2+z2-1z¯2+z2-1=z2-z2+1z¯2-z2+1

⇒z2z2-1+z2+1+z¯2z2-1+z2+1=z2+12-z2-12

⇒z2+z¯2z2+1+z2-1=2z2+z¯2

⇒ Either z2+z¯2=0 or z2+1+z2-1=2

i.e. z2 lies on imaginary axis or it lies on the line segment joining -1,0 and 1,0

So, the minimum distance between z1 & z2 will be the distance between the points 1,0 & 52,0

Hence, z1-z2min=32

Asked in: JEE Main 2022 (28 Jul Shift 1)

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