4. In $\triangle A B C$, $A B=1, B C=2, \angle B-\angle C$ $=90^{\circ}$. Then the radius of the circumcircle of $\triangle A B C$ is $\qquad$
4. $\frac{\sqrt{3}+1}{2}$.

As shown in Figure 5, let $O$ be the center of the circumcircle of $\triangle ABC$, and construct the diameter $AD$, connecting $DB$ and $DC$. Then
$$
\begin{array}{l}
\angle ABD=90^{\circ}. \\
\text{By } \angle ABC \\
=90^{\circ}+\angle ACB, \\
\angle ABC=90^{\circ}+\angle DBC, \\
\text{then } \angle ACB=\angle DBC \Rightarrow \overparen{DC}=\overparen{AB} \Rightarrow \overparen{BD}=\overparen{AC} \\
\Rightarrow DC=AB=1, BD=AC.
\end{array}
$$

Draw a perpendicular from point $C$ to $AD$, with the foot of the perpendicular at $F$.
Let $AD=2R, BD=AC=x$.
By the Pythagorean theorem, we have
$$
\begin{array}{l}
AC^{2}-CD^{2}=AF^{2}-FD^{2} \\
=(AF+FD)(AF-FD)=AD \cdot BC.
\end{array}
$$

Substituting in, we get
$$
x^{2}=1+4R \text{. }
$$

In the right triangle $\triangle ABD$, by the Pythagorean theorem, we have
$$
BD^{2}=AD^{2}-AB^{2} \text{, }
$$

which is $x^{2}=4R^{2}-1$.
From equations (1) and (2), we get $4R^{2}-1=1+4R$, which simplifies to
$$
2R^{2}-2R-1=0 \text{. }
$$

Solving this, we get $R=\frac{\sqrt{3}+1}{2}$ (the negative value is discarded).