JEE Main 2025 Jan 29 Shift 1, Mathematics Q19: Introduction to Complex Numbers And Their Conjugates

JEE Main2025Jan 29, Shift 1Mathematics

Let $∣ z_{1} - 8 - 2 i ∣ \leq 1$ and $∣ z_{2} - 2 + 6 i ∣ \leq 2, z_{1}, z_{2} \in C .$ Then the minimum value of $∣ z_{1} - z_{2} ∣$ is :

A
$3$
B
$7$
C
$13$
D
$10$

Solution

$z_{1}$ lies inside or on the closed disc of radius $1$ centred at $A = 8 + 2 i$ ; $z_{2}$ lies inside or on the closed disc of radius $2$ centred at $B = 2 - 6 i .$

Distance between centres: $∣ A - B ∣ = ∣6 + 8 i ∣ = 36 + 64 = 10.$

Since $10 > 1 + 2 = 3$ , the discs are disjoint. The minimum separation between a point of one disc and a point of the other equals the centre-to-centre distance minus the two radii:

$∣ z_{1} - z_{2} ∣_{m i n} = 10 - 1 - 2 = 7.$

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JEE Main 2025 Jan 29 Shift 1, Mathematics Q19: Introduction to Complex Numbers And Their Conjugates

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