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JEE Main 201912 JanuaryEvening Shift
Exam
JEE Main
Year
2019
Date
12 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Complex Number

JEE Main 2019 · 12 January · Evening Shift · Mathematics · Complex Number

Let z _ 1 and z _ 2 be two complex numbers satisfying | z _ 1 | = 9...

Let z1andz2{z}_{1} and {z}_{2} be two complex numbers satisfying ∣z1∣=9and∣z2−3−4i∣=4| {z}_{1} | = 9 and | {z}_{2} - 3 - 4i | = 4. Then the minimum value of \left|{z}_{1} - {z}_{2}\right| is :Let z _ 1 and z _ 2 be two complex numbers satisfying | z _ 1 | = 9 and | z _ 2 - 3 - 4i | = 4 . Then the minimum value

Options

  1. A22
  2. B2\sqrt{2}
  3. C00
  4. D11

Correct answer

C. 00

Step-by-step solution

We have,

C1:∣z1∣=9{C}_{1}: \left|{z}_{1}\right| = 9

C2:∣z2−3−4i∣=4{C}_{2}: \left|{z}_{2} - 3 - 4i\right| = 4

Let z _ 1 and z _ 2 be two complex numbers satisfying | z _ 1 | = 9 and | z _ 2 - 3 - 4i | = 4 . Then the minimum value

Centres of the circles are c1≡{c}_{1} ≡\left(0, 0\right)andand{c}_{2} ≡ \left(3, 4\right).</p><p>Radiusofthecirclesare.</p><p>Radius of the circles are{r}_{1} = 9andand{r}_{2} = 4.</p><p>.</p><p>⇒ {c}_{1}{c}_{2} = \sqrt{{3}^{2} + {4}^{2}} = 5andand{r}_{1} - {r}_{2} = 5</p><p>Circlestoucheachotherinternally.<br/><br/></p><p>Circles touch each other internally.<br /><br />⇒ {\left|{z}_{1} - {z}_{2}\right|}_\min = 0 at the point of contact.

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