Let P=z C:|z+2-3i| 1 and Q=z C:z(1+i)+z(1-i) -8.

Question

Let $P={z\in C:|z+2-3i|\le 1}$ and $Q={z\in C:z(1+i)+\vec{z}(1-i)\le -8}$. Let in $P\cap Q$, $|z-3+2i|$ be maximum and minimum at ${z}_{1}$ and ${z}_{2}$ respectively. If ${|{z}_{1}|}^{2}+2{|{z}_{2}|}^{2}=\alpha +\beta \sqrt{2}$, where $\alpha ,\beta $ are integers, then $\alpha +\beta $ equals \_\_\_\_\_\_\_\_\_\_\_\_\_.

Answer: 36

Accepted Answer

Clearly for the shaded region ${z}_{1}$ is the intersection of the circle and the line passing through $P({L}_{1})$ and ${z}_{2}$ is intersection of line ${L}_{1}\&{L}_{2}$

Circle : $(x+2{)}^{2}+(y-3{)}^{2}=1$

$\begin{gathered} {L}_{1}:x+y-1=0 \ {L}_{2}:x-y+4=0 \end{gathered}$

On solving circle $\&{L}_{1}$ we get

${z}_{1}:(-2-\frac{1}{\sqrt{2}},3+\frac{1}{\sqrt{2}})$

On solving ${L}_{1}$ and ${z}_{2}$ is intersection of line ${L}_{1}\&{L}_{2}$ we get ${z}_{2}:(\frac{-3}{2},\frac{5}{2})$

$\begin{gathered} {|{z}_{1}|}^{2}+2{|{z}_{2}|}^{2}=14+5\sqrt{2}+17 \ =31+5\sqrt{2} \end{gathered}$

$\begin{gathered} \alpha =31 \ \beta =5 \ \alpha +\beta =36 \end{gathered}$

Solutions · 1