For α, β, z ∈ ℂ and λ > 1, if √(λ-1) is the radius of the circle |z-α|²+|z-β|²=2 λ, then |α-β| is equal to __________.

Solution

Maths sir · 4 steps

Shift the origin to the midpoint of α and β, and the equation turns into a plain circle.

1. Use the midpoint. Let M = (α+β)/2, u = z - M and v = (β-α)/2. Then z-α = u+v and z-β = u-v.
2. Expand both moduli. |u+v|² + |u-v|² = 2|u|² + 2|v|², because the cross terms uv̄ + ūv cancel. Here 2|v|² = (|α-β|²)/2.

   |z-α|²+|z-β|² = 2|z-M|² + (|α-β|²)/2 = 2λ ⇒ |z-M|² = λ - (|α-β|²)/4

3. Match the radius. This is a circle with centre M and radius squared λ - (|α-β|²)/4. Set it equal to (√(λ-1))² = λ - 1, which gives (|α-β|²)/4 = 1.
4. So. |α-β|² = 4, and so |α-β| = 2.

The common slip: 4. Stopping at |α-β|² = 4 gives 4. Take the square root to get the distance itself.

The idea to keep. For |z-α|²+|z-β|² = k, the locus is a circle centred at the midpoint of α and β, with r² = k/2 - (|α-β|²)/4.

Answer from the NTA final answer key. Solution by PrepWiser.