I1.2 If $\left\{\begin{array}{c}x+y=2 \\ x y-z^{2}=a \\ b=x+y+z\end{array}\right.$, find the value of $b$.
(2), $x y=1+z^{2}>0$; together with (1) we have $x>0$ and $y>0$ by A.M. $\geq$ G.M. in (1) $x+y \geq 2 \sqrt{x y} \Rightarrow 2 \geq 2 \sqrt{1+z^{2}}$

After simplification, $0 \geq z^{2} \Rightarrow z=0$
(3): $b=x+y+z=2+0=2$