3. In a right triangle $\triangle A O B$, the right-angle vertex $O$ is on the unit circle, and the hypotenuse is tangent to the unit circle. Extend $A O, B O$ to intersect the unit circle at $C, D$ respectively. Then the minimum value of the area of quadrilateral $A B C D$ is $\qquad$
3. $\frac{1}{2}(3+2 \sqrt{2})$.

Let the point of tangency be $E$,
then $O E=1$, thus
$A B$
$$
\begin{array}{l}
=O E \cdot A B \\
=A O \cdot B O \\
=\frac{A O^{2}+B O^{2}}{2} \\
-\frac{(A O-B O)^{2}}{2} \\
\leqslant \frac{A O^{2}+B O^{2}}{2} \\
=\frac{A B^{2}}{2}
\end{array}
$$

Thus, $A B \geqslant 2$.
When $A O=B O$, $A B$ has a minimum value of 2, hence
$$
\begin{array}{l}
S_{A B C D}=\frac{1}{2} A C \cdot B D \\
=\frac{1}{2}(1+A O)(1+B O) \\
=\frac{1}{2}(1+A O+B O+A O \cdot B O) \\
\geqslant \frac{1}{2}(1+2 \sqrt{A O \cdot B O}+A O \cdot B O) \\
=\frac{1}{2}(1+\sqrt{A O} \cdot \overline{B O})^{2} \\
=\frac{1}{2}(1+\sqrt{O E \cdot A B})^{2} \\
=\frac{1}{2}(1+\sqrt{A B}) \\
\geqslant \frac{1}{2}(1+\sqrt{2})^{2} \\
=\frac{1}{2}(3+2 \sqrt{2}) .
\end{array}
$$

When $A O=B O$, the area has a minimum value of $\frac{1}{2}(3+2 \sqrt{2})$.