3. As shown in Figure 3, $\triangle A B C$ is inscribed in $\odot O, B C=a, C A=$ $b, \angle A-\angle B=90^{\circ}$. Then the area of $\odot O$ is
3. $\frac{1}{4} \pi\left(a^{2}+b^{2}\right)$

The auxiliary lines are shown in Figure 6.
It is easy to see that $\angle A=\angle A B D$,
i.e., $\overparen{C D B}=\overparen{A C D} \Rightarrow \overparen{A C}=\overparen{B D}$
$$
\Rightarrow B D=A C=b \text {. }
$$

At this point, the radius $R$ of $\odot O$ is
$$
\frac{1}{2} C D=\frac{1}{2} \sqrt{a^{2}+b^{2}} \text {. }
$$

Therefore, the area of $\odot O$ is $=\pi R^{2}=\frac{1}{4} \pi\left(a^{2}+b^{2}\right)$.