Q30 / 75 · JEE Main 2025 (2 Apr Shift 1) · Mathematics
Let $z$ be a complex number such that $|z|=1$. If $\frac{2+\mathrm{k}^2 \mathrm{z}}{\mathrm{k}+\overline{\mathrm{z}}}=\mathrm{kz}, \mathrm{k} \in \mathbf{R}$, then the maximum distance of $\mathrm{k}+\mathrm{ik}^2$ from the circle $|\mathrm{z}-(1+2 \mathrm{i})|=1$ is:
- Correct answerA$\sqrt{5}+1$
- B2
- C3
- D$\sqrt{3}+1$
Solution
$\begin{aligned} & \frac{2+\mathrm{k}^2 \mathrm{z}}{\mathrm{k}+\overline{\mathrm{z}}}=\mathrm{kz} \\ & |\mathrm{z}|^2 \mathrm{k}=2 \\ & \mathrm{k}=2\end{aligned}$
point $\mathrm{p}(2,4)$; center $(1,2)$
distance from circle
$(x-1)^2+(y-2)^2=1$ is max.
if $(\mathrm{OP}+\mathrm{r})=\sqrt{1+4}+1=\sqrt{5}+1$
point $\mathrm{p}(2,4)$; center $(1,2)$
distance from circle
$(x-1)^2+(y-2)^2=1$ is max.
if $(\mathrm{OP}+\mathrm{r})=\sqrt{1+4}+1=\sqrt{5}+1$
Hence, the correct answer is option: 'A'.