Complex Numbers
Collinear Points on Circle — Minimum Modulus
nta_pyq_2023_apr
Grade 11
Question:
Let $C$ be the circle in the complex plane with centre $z_0=\dfrac{1}{2}(1+3i)$ and radius $r=1$. Let $z_1=1+i$ and the complex number $z_2$ be outside circle $C$ such that $|z_1-z_0||z_2-z_0|=1$. If $z_0,z_1$ and $z_2$ are collinear, then the smaller value of $|z_2|^2$ is equal to
$\dfrac{5}{2}$
$\dfrac{7}{2}$
$\dfrac{13}{2}$
$\dfrac{3}{2}$
Step-by-Step Solution
Key Concept: $|z_1-z_0|=\frac{1}{\sqrt{2}}$. Since $|z_1-z_0||z_2-z_0|=1$, $|z_2-z_0|=\sqrt{2}$. Collinearity: $z_2=z_0\pm 2(z_1-z_0)$.
$z_2=\frac{3}{2}+\frac{i}{2}$: $|z_2|^2=\frac{5}{2}$ (outside $C$). Smaller value $=\frac{5}{2}$.
Correct Answer: 1