Complex Numbers – Geometry – JEE Main 29 Jan 2025 Shift 1

Solution:

The inequality $|z_{1}-(8+2i)| \le 1$ represents the interior and boundary of a circle $C_1$ with center $A(8, 2)$ and radius $r_1 = 1$.

The inequality $|z_{2}-(2-6i)| \le 2$ represents the interior and boundary of a circle $C_2$ with center $B(2, -6)$ and radius $r_2 = 2$.

We need to find the minimum value of $|z_1 – z_2|$, which corresponds to the shortest distance between these two circular regions.

First, find the distance between the centers $A$ and $B$:

$AB = \sqrt{(8-2)^2 + (2-(-6))^2} = \sqrt{6^2 + 8^2} = \sqrt{36+64} = \sqrt{100} = 10$.

Since $AB > r_1 + r_2$ ($10 > 1+2$), the circles are disjoint and separated.

The minimum distance is the distance along the line connecting the centers, subtracting the radii.

Min Distance $= AB – (r_1 + r_2)$.

$= 10 – (1 + 2) = 10 – 3 = 7$.

Ans. (2)