136 If a circle centered at the origin with radius 1 is tangent to the line $\frac{x}{a}+\frac{y}{b}=1(a>0, b>0)$, then the minimum value of the product $a b$ is $\qquad$.
136 2. From $\frac{x}{a}+\frac{y}{b}=1,(a>0, b>0)$, multiplying both sides by $a b$, we get
$$
b x+a y-a b=0 .
$$

Since the condition for the line (1) to be tangent to the circle $x^{2}+y^{2}=1$ is that the distance from the origin to the line (1) equals 1, we have
$$
\frac{a b}{\sqrt{a^{2}+b^{2}}}=1 \text {. }
$$

Thus,
$$
a^{2} b^{2}=a^{2}+b^{2} \geqslant 2 a b .
$$

That is, $a b \geqslant 2$, with equality holding if and only if $a=b=\sqrt{2}$. Therefore, the minimum value of $a b$ is 2.