3. The minimum value of the area of a right-angled triangle circumscribed around a circle with radius 1 is ( ).
(A) $3-2 \sqrt{2}$
(B) $3+2 \sqrt{2}$
(C) $6-4 \sqrt{2}$
(D) $6+4 \sqrt{2}$
3. (B).

Let the lengths of the two legs of a right triangle be $a$ and $b$, and the area be $\mathrm{S}$. Then we have
$$
\frac{1}{2}\left(a+b+\sqrt{a^{2}+b^{2}}\right) \times 1=\frac{1}{2} a b .
$$
(The area of the triangle is equal to the product of the semi-perimeter and the inradius.)
Simplifying, we get $a b-2 a-2 b+2=0$.
$$
\begin{array}{l}
\because b=\frac{2 S}{a}, \\
\therefore 2 S-2 a-\frac{4 S}{a}+2=0,
\end{array}
$$

This gives $a^{2}-(S+1) a+2 S=0$.
$$
\therefore \Delta_{a}=(S+1)^{2}-4 \times 2 S \geqslant 0 \text {. }
$$

Solving this, we get $S \leqslant 3-2 \sqrt{2}$ or $S \geqslant 3+2 \sqrt{2}$.
$$
\because S>S_{\text {incircle }}=\pi, \therefore S \geqslant 3+2 \sqrt{2} \text {. }
$$

Thus, the minimum value of $S$ is $3+2 \sqrt{2}$.