Let |z₁ − 8 − 2i| ≤ 1 and |z₂ − 2 + 6i| ≤ 2, z₁, z₂ ∈ C. Then the Minimum Value of |z₁ − z₂| Is

JEE Main 2025 — 29 January, Shift 1. Previous Year Question.

Problem

Let $|z_1-8-2i|\le1$ and $|z_2-2+6i|\le2$, $z_1,z_2\in\mathbb C$. Then the minimum value of $|z_1-z_2|$ is:

Key insight. Each inequality $|z-z_0|\le r$ describes every point inside and on a circle of radius $r$ centred at $z_0$ — recognising both conditions as solid discs turns “minimum distance between two complex numbers satisfying these” into “minimum distance between two circular regions,” a question with a standard geometric answer once the discs are confirmed not to overlap.

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Approach

Interpret each modulus inequality as a disc in the Argand plane — $z_1$ ranges over a disc of radius 1 centred at $(8,2)$, and $z_2$ over a disc of radius 2 centred at $(2,-6)$. First check whether the two discs overlap by comparing the distance between their centres to the sum of their radii; if they don’t overlap, the minimum distance between any point in one disc and any point in the other is simply the centre-to-centre distance minus both radii.

Solution

Step 1 — Identify the two discs

$|z_1-8-2i|\le1$ means $|z_1-(8+2i)|\le1$: $z_1$ lies in the closed disc of radius $1$ centred at $C_1=(8,2)$.

$|z_2-2+6i|\le2$ means $|z_2-(2-6i)|\le2$: $z_2$ lies in the closed disc of radius $2$ centred at $C_2=(2,-6)$.

Step 2 — Find the distance between the centres

$$C_1C_2 = \sqrt{(8-2)^2+(2-(-6))^2} = \sqrt{6^2+8^2} = \sqrt{36+64} = \sqrt{100} = 10$$

Step 3 — Check whether the discs overlap

The sum of the two radii is $1+2=3$. Since $10 > 3$, the discs are entirely separate — nowhere close to touching, let alone overlapping.

Step 4 — Find the minimum distance between the discs

When two discs don’t overlap, the shortest path from one to the other runs straight along the line joining their centres, and it’s shortened at each end by that disc’s own radius (moving from the centre out to the near edge). So:

$$\min|z_1-z_2| = C_1C_2-r_1-r_2 = 10-1-2 = 7$$

Answer

Common mistakes

• Skipping the overlap check. If the discs did overlap or intersect, the minimum distance would be $0$ (some point would lie in both discs) — assuming the “centre distance minus radii” formula always applies, without first confirming $C_1C_2>r_1+r_2$, can silently give a wrong answer for a different pair of discs.
• Getting a centre’s sign wrong when converting from the modulus expression. $|z_2-2+6i|$ means $|z_2-(2-6i)|$, so the centre is $(2,-6)$, not $(2,6)$ — the minus sign in front of $6i$ in the original expression flips to a plus once it’s pulled inside the subtraction, and it’s easy to lose track of that flip.
• Confusing minimum with maximum distance. The maximum value of $|z_1-z_2|$ for these same discs would instead be $C_1C_2+r_1+r_2=13$ — reaching for the sum instead of the difference gives the maximum, not the minimum, value.

Practise next

• Let $|z_1-5-12i|\le3$ and $|z_2|\le4$, $z_1,z_2\in\mathbb C$. Find the minimum value of $|z_1-z_2|$.