Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
JEE Main202118 Mar 2021Morning ShiftMathematicsComplex NumberActual

If the equation a | z | 2 + α ¯ z + α z ¯ ¯ + d = 0 represents a circle where a , d are real constants then which of the following condition is correct?

Options

  1. A| α | 2 - a d ≠ 0
  2. B| α | 2 - a d > 0 and a ∈ R - 0
  3. C| α | 2 - a d ≥ 0 and a ∈ R
  4. Dα = 0 , a , d ∈ R +

Correct answer

B. | α | 2 - a d > 0 and a ∈ R - 0

Step-by-step solution

Given a | z | 2 + α ¯ z + α z ¯ ¯ + d = 0 ∵ z 2 = z z ¯   &   α ¯ z + α z ¯ ¯ = α ¯ z ¯ + α z ¯ ¯ = α z ¯ + α ¯ z Then, a z z ¯ + α z ¯ + α ¯ z + d = 0   or z z ¯ + α a z ¯ + α ¯ a z + d a = 0 is the equation of circle. Centre = - α a ,   radius,  r = α α ¯ a 2 - d a = α α ¯ - a d a 2 α a

Practice Complex Number on Quantrex Academy →

More from Complex Number

The number of values of z C , satisfying the equations |z-(4+8i)|= 10 and |z-(3+5i)|+|z-(5+11i)|=4 5 , is: 2026Let S = z C : z^2 + 6 ,iz - 3 = 0 . Then _ z S z^8 is equal to : 2026Let the set of all values of k R such that the equation z( z + 2 + i) + k(2 + 3i) = 0 , z C , has at least one solution, be the interval [ , ] . Then 9( + ) is equal to: 2026Let z₁, z₂ C be the distinct solutions of the equation z^2 + 4z - (1 + 12i) = 0 . Then |z₁|^2 + |z₂|^2 is equal to : 2026Let S= z C : z^2+4z+16=0 . Then _ z S |z+ 3 i|^2 is equal to: 2026Let z be a complex number such that |z+2| = |z-2| and ( z+3 z-i ) = 4 . Then |z|^2 is equal to: 2026Let the circles C₁ : |z| = r and C₂ : |z - 3 - 4i| = 5 , z C , be such that C₂ lies within C₁ . If z₁ moves on C₁ , z₂ moves on C₂ and |z₁ - z₂| = 2 , then |z₁ - z₂| is equal to: 2026Let x and y be real numbers such that 50 ( 2x 1+3i - y 1-2i ) = 31 + 17i , i = -1 . Then the value of 10(x - 3y) is : 2026 Full Complex Number list All JEE Main PYQs