Let $z$ be a complex number such that $|z|=1$. If $\frac{2+k^{2}z}{k+\overline{z}}=kz$, $k \in \mathbb{R}$, then the maximum distance of $k+ik^{2}$ from the circle $|z-(1+2i)|=1$ is:

JEE Main · 2025 · 02 Apr 2025 · Shift 1 · Complex Numbers

Let

z

be a complex number such that

∣ z ∣ = 1

. If

\frac{2 + k ^{2} z}{k + z} = k z

k \in R

, then the maximum distance of

k + i k^{2}

from the circle

∣ z - (1 + 2 i) ∣ = 1

is:

Answer: (A)

Given

\frac{2 + k ^{2} z}{k + z} = k z

and

∣ z ∣ = 1

, so

z \overline{z} = 1

.

Cross-multiplying:

2 + k^{2} z = k z (k + \overline{z}) = k^{2} z + k z \overline{z} = k^{2} z + k

This gives

2 = k

, so

k = 2

.

The point is

k + i k^{2} = 2 + 4 i

, i.e.,

P (2, 4)

.

The circle has center

C (1, 2)

and radius

r = 1

.

Distance from

P

to center:

P C = (2 - 1)^{2} + (4 - 2)^{2} = 1 + 4 = 5

Maximum distance from

P

to the circle =

P C + r = 5 + 1

JEE Main · 2025 · 02 Apr 2025 · Shift 1 · Single correct option