Step 1: Check the options directly:
Options (A) and (D) give $y = \sqrt3 x + \ldots$ lines with slope $\pm\sqrt3$. Convert (C): $x - \sqrt3 y + 3 = 0$, slope $\frac{1}{\sqrt3}$.
Step 2: Verify both tangencies:
Parabola: with $m = \frac{1}{\sqrt3}$, the intercept should be $\frac{1}{m} = \sqrt3$. The line has intercept $\frac{3}{\sqrt3} = \sqrt3$. Matches.
Circle: distance of $(3, 0)$ from $x - \sqrt3 y + 3 = 0$ is $\frac{|3 + 3|}{\sqrt{1 + 3}} = \frac{6}{2} = 3$, equal to the radius. So (C) is a common tangent.
Final Answer:
$\sqrt3\,y = x + 3$.
\[ \boxed{\sqrt{3}\,y = x + 3} \]