The points $(1, 0)$, $(0, 1)$, $(0, 0)$, and $(2k, 3k)$, $k \ne 0$, are concyclic if $k = \dots$
$-\frac{5}{13}$
\(\frac{5}{13}\)
$\frac{1}{5}$
$-\frac{1}{5}$
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