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Complex Numbers question

2021 · 18 Mar · Shift 1 · Q32
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  5. /2021 · 18 Mar · Shift 1 · Q32

Complex Numbers question

2021 · 18 Mar · Shift 1 · Q32

JEE MainMathematicsComplex NumbersMCQ+4 / −1
If the equation a∣z∣2+α‾z+αz‾‾+d=0a|z{|^2} + \overline {\overline \alpha z + \alpha \overline z } + d = 0a∣z∣2+αz+αz​+d=0 represents a circle where a, d are real constants then which of the following condition is correct?
  1. A
    |α\alphaα|2 −-− ad eee 0
  2. B
    |α\alphaα|2 −-− ad > 0 and a ∈\in∈ R −-− {0}
  3. C
    |α\alphaα|2 −-− ad ≥\ge≥ 0 and a ∈\in∈ R
  4. D
    α\alphaα = 0, a, d ∈\in∈ R+
View written solutionFree

Correct answer: B

  1. Interpret the given equation

    The equation is a∣z∣2+α‾z+αz‾‾+d=0,a|z|^2 + \overline{\overline{\alpha}z + \alpha \overline z} + d = 0,a∣z∣2+αz+αz​+d=0, where a,d∈Ra,d \in \mathbb{R}a,d∈R.

    First simplify the conjugate term:

    = \alpha \overline z + \overline{\alpha} z.$$ So the equation becomes $$a|z|^2 + \overline{\alpha}z + \alpha \overline z + d = 0.$$

  2. Compare with standard circle form

    The standard complex equation of a circle is a∣z∣2+β‾z+βz‾+d=0,a≠0,a|z|^2 + \overline{\beta}z + \beta \overline z + d = 0, \qquad a\neq 0,a∣z∣2+β​z+βz+d=0,a=0, with center −βa-\frac{\beta}{a}−aβ​ and radius satisfying r2=∣β∣2−ada2.r^2 = \frac{|\beta|^2-ad}{a^2}.r2=a2∣β∣2−ad​.

    Here, β=α\beta=\alphaβ=α. Hence r2=∣α∣2−ada2.r^2 = \frac{|\alpha|^2-ad}{a^2}.r2=a2∣α∣2−ad​.

  3. Condition for representing a circle

    For the equation to represent a real circle (non-imaginary radius), we need

    • a≠0a \neq 0a=0 (otherwise the quadratic term vanishes and it is not a circle),
    • r2>0r^2>0r2>0 for a proper circle.

    Thus, ∣α∣2−ada2>0.\frac{|\alpha|^2-ad}{a^2}>0.a2∣α∣2−ad​>0.

    Since a2>0a^2>0a2>0 whenever a≠0a\neq 0a=0, this gives ∣α∣2−ad>0.|\alpha|^2-ad>0.∣α∣2−ad>0.

  4. Check options

    • A: incomplete/incorrect.
    • B: ∣α∣2−ad>0|\alpha|^2-ad>0∣α∣2−ad>0 and a∈R∖{0}a\in \mathbb{R}\setminus\{0\}a∈R∖{0} ✅
    • C: ∣α∣2−ad≥0|\alpha|^2-ad\ge 0∣α∣2−ad≥0 allows the degenerate case r=0r=0r=0 (a point circle), not a proper circle.
    • D: unnecessary and incorrect.

  5. Conclusion

    The correct condition is ∣α∣2−ad>0anda≠0.|\alpha|^2-ad>0 \quad \text{and} \quad a\neq 0.∣α∣2−ad>0anda=0.

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