For $\alpha, \beta, z \in \mathbb{C}$ and $\lambda>1$, if $\sqrt{\lambda-1}$ is the radius of the circle $|z-\alpha|^{2}+|z-\beta|^{2}=2 \lambda$, then $|\alpha-\beta|$ is equal to ____.

JEE Main · 2023 · 06 Apr 2023 · Shift 2 · Complex Numbers

For

α, β, z \in 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 ____.

Answer: 2

The relation

∣ z - α ∣^{2} + ∣ z - β ∣^{2} = 2 λ

describes a circle centred at

m = \frac{α + β}{2}

. Using the identity

∣ z - α ∣^{2} + ∣ z - β ∣^{2} = 2 ∣ z - m ∣^{2} + \frac{∣ α - β ∣ ^{2}}{2}

, the equation becomes

∣ z - m ∣^{2} = λ - \frac{∣ α - β ∣ ^{2}}{4}

, so the radius satisfies

r^{2} = λ - \frac{∣ α - β ∣ ^{2}}{4}

. Since

r = λ - 1

, this gives

λ - \frac{∣ α - β ∣ ^{2}}{4} = λ - 1

, so

\frac{∣ α - β ∣ ^{2}}{4} = 1

. Therefore

∣ α - β ∣ = 2

JEE Main · 2023 · 06 Apr 2023 · Shift 2 · Integer answer