2. Given vectors $\boldsymbol{a}, \boldsymbol{b}$ satisfy
$$
|a|=|b|=a \cdot b=2 \text {, }
$$

and $(a-c) \cdot(b-c)=0$.
Then the minimum value of $|2 b-c|$ is
2. $\sqrt{7}-1$.

Notice,
$$
\cos \langle\boldsymbol{a}, \boldsymbol{b}\rangle=\frac{\boldsymbol{a} \cdot \boldsymbol{b}}{|\boldsymbol{a}||\boldsymbol{b}|}=\frac{1}{2} \Rightarrow\langle\boldsymbol{a}, \boldsymbol{b}\rangle=\frac{\pi}{3} \text {. }
$$

From this, we can set $\boldsymbol{b}=(2,0), \boldsymbol{a}=(1, \sqrt{3})$.
Let $\boldsymbol{c}=(m, n)$.
$$
\begin{array}{l}
\text { By }(\boldsymbol{a}-\boldsymbol{c}) \cdot(\boldsymbol{b}-\boldsymbol{c})=0 \\
\Rightarrow(1-m)(2-m)+(0-n)(\sqrt{3}-n)=0 \\
\Rightarrow\left(m-\frac{3}{2}\right)^{2}+\left(n-\frac{\sqrt{3}}{2}\right)^{2}=1 .
\end{array}
$$

Let $m=\frac{3}{2}+\cos \alpha, n=\frac{\sqrt{3}}{2}+\sin \alpha$.
Also, $2 \boldsymbol{b}-\boldsymbol{c}=(4-m,-n)$, then
$$
\begin{array}{l}
|2 b-c|=\sqrt{(4-m)^{2}+n^{2}} \\
=\sqrt{8-(5 \cos \alpha-\sqrt{3} \sin \alpha)} \\
=\sqrt{8-2 \sqrt{7} \sin \left(-\alpha+\arctan \frac{5}{\sqrt{3}}\right)} \\
\geqslant \sqrt{8-2 \sqrt{7}}=\sqrt{7}-1 .
\end{array}
$$

Therefore, $|2 b-c|_{\min }=\sqrt{7}-1$.