Four distinct points $(2k,3k)$, $(1,0)$, $(0,1)$ and $(0,0)$ lie on a circle for $k$ equal to:
Step-by-Step Solution
Key Concept: Since $(0,0)$, $(1,0)$, $(0,1)$ lie on a circle, use the circle through three points. Circle through these: $x^2+y^2-x-y=0$ (diameter endpoints $(1,0)$ and $(0,1)$: $(x-1)x+(y-1)y=0$). Substitute $(2k,3k)$: $4k^2+9k^2-2k-3k=0\Rightarrow13k^2-5k=0\Rightarrow k(13k-5)=0$, so $k=5/13$.
Substitute $(2k,3k)$ in $x^2+y^2-x-y=0$: $13k^2-5k=0\Rightarrow k=5/13$ (since $k\ne0$ for distinct points).
Correct Answer: 3