Chọn phương án A.

Ta có \(\mathrm{T}_{\overrightarrow{v}}\colon\begin{cases}
x'=x-3\\ y'=y-2
\end{cases}\Leftrightarrow\begin{cases}
x=x'+3\\ y=y'+2.
\end{cases}\)
$$\begin{eqnarray*}
\left(\mathscr{C}\right)\colon&x^2+(y-1)^2&=1\\
\Leftrightarrow&\left(x'+3\right)^2+\left(y'+2-1\right)^2&=1\\
\Leftrightarrow&\left(x'+3\right)^2+\left(y'+1\right)^2&=1.
\end{eqnarray*}$$
Vậy \(\left(\mathscr{C}'\right)\colon(x+3)^2+(y+1)^2=1\).