Question

Let $C$ be the set of all complex numbers.
Let $S_1=\{z\in C:|z-2|\le 1\}$ and
$S_2=\{z\in C:z(1+i)+\bar{z}(1-i)\ge 4\}$
Then, the maximum value of ${\left|z-\frac{5}{2}\right|}^2$ for $z\in S_1\cap S_2$ is equal to :

• A. $\frac{3+2\sqrt{2}}{4}$
• B. $\frac{5+2\sqrt{2}}{2}$
• C. $\frac{3+2\sqrt{2}}{2}$
• D. $\frac{5+2\sqrt{2}}{4}$

Answer 1

Answer: (D)

$|t-2|\le 1$ Put $t=x+iy$

$(x-2{)}^2+y^2\le 1$
Also, $t(1+i)+\bar{t}(1-i)\ge 4$
Gives $x-y\ge 2$
Let point on circle be $A(2+\cos \theta ,\sin \theta )$
$\theta \in \left[-\frac{3\pi }{4},\frac{\pi }{4}\right]$
$(AP{)}^2={\left(2+\cos \theta -\frac{5}{2}\right)}^2+{\sin }^2\theta$
$={\cos }^2\theta -\cos \theta +\frac{1}{4}+{\sin }^2\theta$
$=\frac{5}{4}-\cos \theta$
For $(AP{)}^2$ maximum $\theta =-\frac{3\pi }{4}$
$(AP{)}^2=\frac{5}{4}+\frac{1}{\sqrt{2}}=\frac{5\sqrt{2}+4}{4\sqrt{2}}$