Solve this following


Let $\mathbb{C}$ be the set of all complex numbers. Let

$\mathrm{S}_{1}=\{\mathrm{z} \in \mathbb{C}:|\mathrm{z}-2| \leq 1\}$ and

$\mathrm{S}_{2}=\{\mathrm{z} \in \mathbb{C}: \mathrm{z}(1+\mathrm{i})+\overline{\mathrm{Z}}(1-\mathrm{i}) \geq 4\}$

Then, the maximum value of $\left|\mathrm{z}-\frac{5}{2}\right|^{2}$ for

$\mathrm{z} \in \mathrm{S}_{1} \cap \mathrm{S}_{2}$ is equal to :


  1. $\frac{3+2 \sqrt{2}}{4}$

  2. $\frac{5+2 \sqrt{2}}{2}$

  3. $\frac{3+2 \sqrt{2}}{2}$

  4. $\frac{5+2 \sqrt{2}}{4}$

Correct Option: , 4


$|\mathrm{t}-2| \leq 1 \quad$ Put $\mathrm{t}=\mathrm{x}+\mathrm{iy}$

$(x-2)^{2}+y^{2} \leq 1$

Also, $\mathrm{t}(1+\mathrm{i})+\overline{\mathrm{t}}(1-\mathrm{i}) \geq 4$

Gives $x-y \geq 2$

Let point on circle be $\mathrm{A}(2+\cos \theta, \sin \theta)$

$\theta \in\left[-\frac{3 \pi}{4}, \frac{\pi}{4}\right]$

$(\mathrm{AP})^{2}=\left(2+\cos \theta-\frac{5}{2}\right)^{2}+\sin ^{2} \theta$

$=\cos ^{2} \theta-\cos \theta+\frac{1}{4}+\sin ^{2} \theta$

$=\frac{5}{4}-\cos \theta$

For $(\mathrm{AP})^{2}$ maximum $\theta=-\frac{3 \pi}{4}$

$(\mathrm{AP})^{2}=\frac{5}{4}+\frac{1}{\sqrt{2}}=\frac{5 \sqrt{2}+4}{4 \sqrt{2}}$


Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now