Chọn phương án D.

Ta có \((\mathscr{C})\colon x^2+y^2+x-\dfrac{1}{2}y-\dfrac{11}{16}=0\).

Khi đó \(\begin{cases}
-2a&=1\\ -2b&=-\dfrac{1}{2}\\ c&=-\dfrac{11}{16}
\end{cases}\Leftrightarrow\begin{cases}
a=-\dfrac{1}{2}\\ b=\dfrac{1}{4}\\ c=-\dfrac{11}{16}.
\end{cases}\)

Vậy \((\mathscr{C})\) có:

• Tâm \(I\left(-\dfrac{1}{2};\dfrac{1}{4}\right)\)
• Bán kính \(R=\sqrt{\left(-\dfrac{1}{2}\right)^2+\left(\dfrac{1}{4}\right)^2-\left(-\dfrac{11}{16}\right)}=1\).