$解：\left( 1 \right) 作CD⊥AB，垂足为点D\ $
$在Rt△ABC中， ∵AB=6\,\,\text{cm,}AC=3\,\,\text{cm}\ $
$∴BC=\sqrt{AB^2-AC^2}=3\sqrt{3}\,\,\text{cm}\ $
$∵点C到直线AB的距离CD=\frac{3×3\sqrt{3}}{6}=\frac{3\sqrt{3}}{2}\,\,\text{cm}\ $
$∵2\ \text{cm}\lt \frac{3\sqrt{3}}{2}\,\,\text{cm}\lt 4\,\,\text{cm}\ $
$∴半径为2\,\,\text{cm}的圆与直线AB相离，半径为4\,\,\text{cm}的圆与直线AB相交.\ $
$\left( 2 \right) ∵圆心C到直线AB的距离为\frac{3\sqrt{3}}{2}\,\,\text{cm}\ $
$∴当半径\ r=\frac{3\sqrt{3}}{2}\,\,\text{cm}时，直线AB与\odot C相切.\ $
$\left( 3 \right) ∵AC=3\,\,\text{cm}，BC=3\sqrt{3}\,\,\text{cm,}\odot C与边AB有一个公共交点\ $
$∴3\,\,\text{cm}＜r≤3\sqrt{3}cm或r=\frac{3\sqrt{3}}{2}cm.$
