11. B. Given the parabola $y=x^{2}$ and the moving line $y=(2 t-1) x-c$ have common points $\left(x_{1}, y_{1}\right) 、\left(x_{2}, y_{2}\right)$, and $x_{1}^{2}+x_{2}^{2}=t^{2}+2 t-3$.
(1) Find the range of real number $t$;
(2) For what value of $t$ does $c$ attain its minimum value, and find the minimum value of $c$.
11. B. (1) By combining $y=x^{2}$ and $y=(2 t-1) x- c$, eliminating $y$ yields the quadratic equation
$$
x^{2}-(2 t-1) x+c=0
$$

with real roots $x_{1}$ and $x_{2}$, then $x_{1}+x_{2}=2 t-1, x_{1} x_{2}=c$.
Thus, $c=x_{1} x_{2}=\frac{1}{2}\left[\left(x_{1}+x_{2}\right)^{2}-\left(x_{1}^{2}+x_{2}^{2}\right)\right]$
$$
\begin{array}{l}
=\frac{1}{2}\left[(2 t-1)^{2}-\left(t^{2}+2 t-3\right)\right] \\
=\frac{1}{2}\left(3 t^{2}-6 t+4\right) .
\end{array}
$$

Substituting equation (2) into equation (1) gives
$$
x^{2}-(2 t-1) x+\frac{1}{2}\left(3 t^{2}-6 t+4\right)=0 \text {. }
$$
The value of $t$ should satisfy
$$
t^{2}+2 t-3=x_{1}^{2}+x_{2}^{2} \geqslant 0 \text {, }
$$

and make equation (3) have real roots, i.e.,
$$
\begin{array}{l}
\Delta=(2 t-1)^{2}-2\left(3 t^{2}-6 t+4\right) \\
=-2 t^{2}+8 t-7 \geqslant 0 .
\end{array}
$$

Solving inequality (4) yields $t \leqslant-3$ or $t \geqslant 1$.
Solving inequality (5) yields $2-\frac{\sqrt{2}}{2} \leqslant t \leqslant 2+\frac{\sqrt{2}}{2}$.
Therefore, the range of $t$ is
$$
2-\frac{\sqrt{2}}{2} \leqslant t \leqslant 2+\frac{\sqrt{2}}{2} \text {. }
$$
(2) From equation (2), we know
$$
c=\frac{1}{2}\left(3 t^{2}-6 t+4\right)=\frac{3}{2}(t-1)^{2}+\frac{1}{2} \text {. }
$$

Since $c=\frac{3}{2}(t-1)^{2}+\frac{1}{2}$ is increasing in the interval $2-\frac{\sqrt{2}}{2} \leqslant t \leqslant 2+\frac{\sqrt{2}}{2}$, when $t=2-\frac{\sqrt{2}}{2}$,
$$
c_{\text {min }}=\frac{3}{2}\left(2-\frac{\sqrt{2}}{2}-1\right)^{2}+\frac{1}{2}=\frac{11-6 \sqrt{2}}{4} .
$$