19. If $a \circ b=\frac{\sqrt{a^{2}+3 a b+b^{2}-2 a-2 b+4}}{a b+4}$, find $((\cdots((2010 \circ 2009) \circ 2008) \circ \cdots \circ 2) \circ 1)$. (2 marks)若 $a \circ b=\frac{\sqrt{a^{2}+3 a b+b^{2}-2 a-2 b+4}}{a b+4}$, 求 $((\cdots((2010 \circ 2009) \circ 2008) \circ \cdots \circ 2) \circ 1)$ 。（2 分）
19. $\frac{\sqrt{15}}{9}$
19. Since $a^{2}+3 a b+b^{2}-2 a-2 b+4=(a-1)^{2}+(b-1)^{2}+3 a b+2>0$ for all $a$, $b$, we see that $a \circ b$ is always well-defined and positive when $a, b$ are positive. When $a$ is positive, we have
$$
a \circ 2=\frac{\sqrt{a^{2}+6 a+4-2 a-4+4}}{2 a+4}=\frac{\sqrt{(a+2)^{2}}}{2(a+2)}=\frac{1}{2} \text {. }
$$

Hence $((\cdots((2010 \circ 2009) \circ 2008)$
$\circ 2) \circ 1)=\frac{1}{2} \circ 1=\frac{\sqrt{15}}{9}$.

Remark. Clearly, the crucial observation of this question is that $a \circ 2$ is a constant. What leads us to consider this? One way is to write the expression in the square root as a function of a, i.e. $a^{2}+(3 b-2) a+\left(b^{2}-2 b+4\right)$, with discriminant $(3 b-2)^{2}-4\left(b^{2}-2 b+4\right)=5 b^{2}-4 b-12$. By some trial and error we see that this is a perfect square when $b=2$.