Complex Numbers
Locus of Complex Number — Circle Radius
nta_pyq_2023_apr
Grade 11
Question:
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 _____.
Step-by-Step Solution
Key Concept: The equation $|z-\alpha|^2+|z-\beta|^2=k$ represents a circle with centre $\frac{\alpha+\beta}{2}$ and radius $\sqrt{\frac{k}{2}-\frac{|\alpha-\beta|^2}{4}}$.
Radius $=\sqrt{\lambda-\frac{|\alpha-\beta|^2}{4}}=\sqrt{\lambda-1}\Rightarrow|\alpha-\beta|^2=4\Rightarrow|\alpha-\beta|=2$.
Correct Answer: 2