Complex Numbers
Circle / Half-plane Intersection — Max/Min of Modulus
nta_pyq_2024_jan
Grade 11
Question:
Let $P=\{z\in\mathbb{C}:|z+2-3i|\le 1\}$ and $Q=\{z\in\mathbb{C}: z(1+i)+\bar{z}(1-i)\le -8\}$. 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
Step-by-Step Solution
Key Concept: Identify $P$ as a disk of radius 1 centred at $(-2,3)$, and $Q$ as a half-plane $x-y+4\le 0$. The intersection $P\cap Q$ is the part of the disk on one side of the line. $|z-3+2i|$ is distance from $(3,-2)$; maximum and minimum occur at specific points on the arc/chord.
Circle: $(x+2)^2+(y-3)^2=1$, $L_1: x+y-1=0$, $L_2: x-y+4=0$.
$z_1$: intersection of circle and $L_1$ through $P$ = $\left(-2-\frac{1}{\sqrt{2}}, 3+\frac{1}{\sqrt{2}}\right)$.
$z_2$: intersection of $L_1$ and $L_2$ = $\left(-\frac{3}{2},\frac{5}{2}\right)$.
$|z_1|^2=\left(2+\frac{1}{\sqrt{2}}\right)^2+\left(3+\frac{1}{\sqrt{2}}\right)^2=14+5\sqrt{2}$, $2|z_2|^2=2\left(\frac{9}{4}+\frac{25}{4}\right)=17$.
$\alpha+\beta\sqrt{2}=31+5\sqrt{2}\Rightarrow\alpha=31,\beta=5$, $\alpha+\beta=36$.
Correct Answer: 36