4. As shown in Figure $4, \odot O$ is the circumcircle of $\triangle A B C$, $B C=$ $a, C A=b$, and $\angle A-\angle B$ $=90^{\circ}$. Then the radius of $\odot O$ is $\qquad$
4. $\frac{1}{2} \sqrt{a^{2}+b^{2}}$.

Draw the diameter $C D$ of $\odot O$, and connect $D B$.
Since $\angle A=\angle B+90^{\circ}=\angle B+\angle C B D=\angle A B D$,
we have $\overparen{C D B}=\overparen{A C D}$.
Thus, $\overparen{A C}=\overparen{B D}, B D=A C=b$.
At this point, $C D=\sqrt{a^{2}+b^{2}}$.
Therefore, the radius of $\odot O$ is $\frac{1}{2} \sqrt{a^{2}+b^{2}}$.