Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
JEE MainMathematicsComplex Number

Let A = z C : |z - 3| + |z + 3| = 10 and B = z C : |z - i| = r , where r > 0 . If the set A B contains exactly 3 elements, then the value of r is

Options

  1. A3
  2. B26
  3. C5
  4. D6

Correct answer

C. 5

Step-by-step solution

The set A represents an ellipse with foci at ( 3, 0) . The sum of focal distances is 2a = 10 , so the semi-major axis is a = 5 . The distance from the center to a focus is c = 3 . The semi-minor axis b is given by b^2 = a^2 - c^2 = 25 - 9 = 16 , so b = 4 . The Cartesian equation of the ellipse is x^2 25 + y^2 16 = 1 . The set B represents a circle centered at (0, 1) with radius r . Its equation is x^2 + (y - 1)^2 = r^2 . For the circle and ellipse to have exactly 3 points of intersection, the circle must be tangent

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