11. As shown in Figure 3, in the Cartesian coordinate system $x O y$, the graph of the quadratic function $y_{1}=-x^{2}+a x$ intersects the $x$-axis at points $O$ and $A$, and the graph of the quadratic function $y_{2}=x^{2}+b x$ intersects the $x$-axis at points $O$ and $B$. The graphs of $y_{1}=-x^{2}+a x$ and $y_{2}=x^{2}+b x$ intersect at points $O$ and $C$, and $O C \perp A C$, with $B$ being the midpoint of $O A$. Find
(1) $\angle O A C$;
(2) $S_{\triangle O A C}$.
Three, 11. (1) Let point $B(-b, 0), A(a, 0)$.
Since $B$ is the midpoint of $O A$, we have $-b=\frac{a}{2}$.
Thus, $y_{2}=x^{2}-\frac{a}{2} x$.
The x-coordinate of point $C$, $x_{C}$, satisfies
$-x^{2}+a x=x^{2}-\frac{a}{2} x$
$\Rightarrow x_{C}=\frac{3}{4} a=\frac{x_{A}+x_{B}}{2}$.
Therefore, point $C$ lies on the perpendicular bisector of $A B$, and $C B=C A$.
Since $O C \perp A C$ and $B$ is the midpoint of $O A$, we have
$B C=A B$.
Thus, $\triangle A B C$ is an equilateral triangle, and $\angle O A C=60^{\circ}$.
(2) From the problem, we know
$y_{c}=-x_{c}^{2}+a x_{c}=\frac{3}{16} a^{2}$.
The height of the equilateral $\triangle A B C$ is
$h=\left|y_{C}\right|=\frac{\sqrt{3}}{2} A B=\frac{\sqrt{3}}{4} a$,
Thus, $\frac{3}{16} a^{2}=\frac{\sqrt{3}}{4} a \Rightarrow a=\frac{4}{\sqrt{3}}$.
Therefore, $S_{\triangle O A C}=2 S_{\triangle A B C}=\frac{\sqrt{3}}{8} a^{2}=\frac{2 \sqrt{3}}{3}$.