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Complex Numbers question

2002 · Shift 0 · Q99
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Complex Numbers question

2002 · Shift 0 · Q99

JEE MainMathematicsComplex NumbersMCQ+4 / −1
The locus of the centre of a circle which touches the circle ∣z−z1∣=a\left| {z - {z_1}} \right| = a∣z−z1​∣=a and ∣z−z2∣=b \left| {z - {z_2}} \right| = b\,∣z−z2​∣=b externally (z, z1 & z2 z,\,{z_1}\,\& \,{z_2}\,z,z1​&z2​ are complex numbers) will be :
  1. A
    an ellipse
  2. B
    a hyperbola
  3. C
    a circle
  4. D
    none of these
View written solutionFree

Correct answer: B

  1. Let the required circle have centre at the complex point zzz and radius rrr.

  2. Since it touches the circle ∣z−z1∣=a|z-z_1|=a∣z−z1​∣=a externally, the distance between their centres equals the sum of radii: ∣z−z1∣=r+a.|z-z_1|=r+a.∣z−z1​∣=r+a. Similarly, since it touches the circle ∣z−z2∣=b|z-z_2|=b∣z−z2​∣=b externally, ∣z−z2∣=r+b.|z-z_2|=r+b.∣z−z2​∣=r+b.

  3. Subtract the two equations: ∣z−z1∣−∣z−z2∣=(r+a)−(r+b)=a−b.|z-z_1|-|z-z_2|=(r+a)-(r+b)=a-b.∣z−z1​∣−∣z−z2​∣=(r+a)−(r+b)=a−b. Hence, ∣z−z1∣−∣z−z2∣=a−b.|z-z_1|-|z-z_2|=a-b.∣z−z1​∣−∣z−z2​∣=a−b. Taking modulus if needed, ∣∣z−z1∣−∣z−z2∣∣=∣a−b∣.\big||z-z_1|-|z-z_2|\big|=|a-b|.​∣z−z1​∣−∣z−z2​∣​=∣a−b∣.

  4. This is the standard locus of a point whose difference of distances from two fixed points z1z_1z1​ and z2z_2z2​ is constant. The fixed points z1z_1z1​ and z2z_2z2​ act as the foci.

  5. Therefore, the locus is a hyperbola.

  6. Option check:

  • A: ellipse →\to→ false, because ellipse corresponds to constant sum of distances.
  • B: hyperbola →\to→ true, because hyperbola corresponds to constant difference of distances.
  • C: circle →\to→ false.
  • D: none of these →\to→ false.

Therefore, the correct answer is B: a hyperbola.

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