Complex Numbers
Minimum distance between loci in complex plane
MJAT_TS3_P2
Grade 12
Question:
If complex number $z_1$ satisfies $\arg z = \dfrac{\pi}{4}$ and $|z-2-i|=1$, and $z_2$ lies on the curve $|z-25i|=15$ with the least positive argument, then the minimum value of $|z_1-z_2|$ is:
A) $\sqrt{346}$
B) $3\sqrt{74}$
C) $2\sqrt{74}$
D) $2\sqrt{76}$
Step-by-Step Solution
Key Concept: $z_1$ satisfies $z_1=\alpha(1+i)$, $\alpha>0$, and $|z_1-2-i|=1$. Solving: $\alpha=1$ or $\alpha=2$, giving $z_1=1+i$ or $z_1=2+2i$. $z_2$ on circle with centre $25i$, radius $15$, with least positive argument: $z_2=12+16i$ (tangent from origin direction).
$\min|z_1-z_2|=2\sqrt{74}$. Answer: **C**.
Correct Answer: C