JEE Main 2021 · 18 March · morning · Single correct · +4 / −1
If the equation a|z|² + α̅ z + α z̅̅ + d = 0 represents a circle where a, d are real constants then which of the following condition is correct?
- (a) |α|² - ad ≠ 0
- (b) |α|² - ad > 0 and a∈R - 0
- (c) |α|² - ad ≥ 0 and a∈R
- (d) α = 0, a, d∈R⁺
Answer and solution
Answer (b) |α|² - ad > 0 and a∈R - 0
Solution
Maths sir · 4 steps
Divide by a, complete the square to get the form |z − centre|² = r², then insist that r² is positive.
- Remove the bar. ᾱz + αz̄ is a number plus its own conjugate, so it is real, and its conjugate is itself. The equation becomes a zz̄ + ᾱz + αz̄ + d = 0.
- Need a nonzero. If a = 0, the |z|² term disappears and what is left is a straight line, not a circle. So take a ≠ 0 and divide by it. Then compare with |z+c|² = zz̄ + c̄z + cz̄ + |c|², using c = α/a.
|z + α/a|² = |α|²/a² - d/a = (|α|² - ad)/a²
- Read the radius. This is a circle with centre -α/a and r² = (|α|² - ad)/a². A real circle needs r² > 0. Since a² > 0, this means |α|² - ad > 0.
- So. The conditions are |α|² - ad > 0 and a ∈ ℝ - {0}. That is option (b).
The idea to keep. For azz̄ + ᾱz + αz̄ + d = 0, the centre is -α/a and r² = (|α|² - ad)/a². It is a circle only when a ≠ 0 and |α|² > ad.
Answer from the NTA final answer key. Solution by PrepWiser.