= Solution
The first curve is a <circle> of <radius> $1$ centred at $(2,1)$. A line through the origin with equation $kx-y=0$ intersects it precisely when the perpendicular distance from that <centre> is at most the <radius>. The <distance from a point to a line> gives
$$
\frac{|2k-1|}{\sqrt{k^2+1}}\leq1
\quad\Longleftrightarrow\quad
(2k-1)^2\leq k^2+1
\quad\Longleftrightarrow\quad k(3k-4)\leq0.
$$
Hence the full parameter range is
$$
\boxed{0\leq k\leq\frac43.}
$$
The two endpoint lines are tangent, at $(2,0)$ and $(6/5,8/5)$ respectively; interior values give two real intersections. Both endpoints are therefore included.
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