Four distinct points (2k,3k),(1,0),(0,1) and (0,0) lie on a circle for k equal to:

Question

Four distinct points $(2k,3k),(1,0),(0,1)$ and $(0,0)$ lie on a circle for $k$ equal to:

Accepted Answer

Circle through $(0,0),(1,0),(0,1)$: $x^2+y^2-x-y=0$.

$(2k,3k)$ on it: $4k^2+9k^2-2k-3k=0 \Rightarrow 13k^2-5k=0 \Rightarrow k=\dfrac{5}{13}$ (for $k
e0$).

Solutions · 1