Let $z_1$ and $z_2$ be two complex numbers such that $|z_1-8-2i|\leq1$ and $|z_2-2+6i|\leq2$. Then the minimum value of $|z_1-z_2|$ is
Step-by-Step Solution
Key Concept: The minimum distance between two closed discs equals the distance between their centres minus the sum of their radii (when the discs do not overlap).
Centre $A=(8,2)$, radius $r_1=1$; centre $B=(2,-6)$, radius $r_2=2$.
$$|AB|=\sqrt{(8-2)^2+(2-(-6))^2}=\sqrt{36+64}=10.$$
Since $|AB|=10>r_1+r_2=3$, the discs are disjoint.
Minimum $|z_1-z_2| = |AB|-r_1-r_2 = 10-1-2 = 7$.
Correct Answer: 4