10. (20 points) In the Cartesian coordinate system $x O y$, the vertices of $\triangle A B C$ are $B(-1,0)$, $C(1,0)$, and $A$ is a moving point. $H$ and $G$ are the orthocenter and centroid of $\triangle A B C$, respectively. If the midpoint $K$ of segment $H G$ lies on the line $B C$:
(1) Find the equation of the trajectory $E$ of point $A$;
(2) A line passing through point $C$ with a slope of 1 intersects the curve $E$ at point $D$. If there exist two points $P$ and $Q$ on the curve $E$ such that points $P$, $C$, $Q$, and $D$ are concyclic, find the minimum value of the area of the circumcircle of quadrilateral $P C Q D$.
10. (1) Let point $A(x, y)(y \neq 0)$.

Then the orthocenter $H\left(x_{H}, y_{H}\right)$ of $\triangle A B C$ satisfies the system of equations
\[
\left\{\begin{array}{l}
x_{H}=x, \\
\left(x_{H}+1\right)(1-x)-y_{H} y=0
\end{array}\right.
\]
\[
\Rightarrow x_{H}=x, y_{H}=\frac{1-x^{2}}{y}.
\]
The centroid $G\left(\frac{x}{3}, \frac{y}{3}\right)$, thus the midpoint $K\left(\frac{2 x}{3}, \frac{3-3 x^{2}+y^{2}}{6 y}\right)$ of segment $H G$.

Since point $K$ lies on the $x$-axis, we have $3-3 x^{2}+y^{2}=0$, which means the locus equation of point $A$ is $x^{2}-\frac{y^{2}}{3}=1(y \neq 0)$.
(2) As shown in Figure 2, let $C D$ and $P Q$ intersect at point $M\left(x_{0}, y_{0}\right)$. Then the parametric equation of line $C D$ is
\[
\left\{\begin{array}{l}
x=x_{0}+\frac{\sqrt{2}}{2} t, \\
y=y_{0}+\frac{\sqrt{2}}{2} t
\end{array} \text { ( } t \text { is a parameter). }\right.
\]

Substituting into the equation of curve $E$ gives
\[
t^{2}+\left(3 x_{0}-2 y_{0}\right) t+3 x_{0}^{2}-y_{0}^{2}-3=0.
\]

Thus, $|M C||M D|=-t_{1} t_{2}=-\left(3 x_{0}^{2}-y_{0}^{2}-3\right)$.
Let the inclination angle of line $P Q$ be $\alpha$, and the parametric equation of line $P Q$ be
\[
\left\{\begin{array}{l}
x=x_{0}+t \cos \alpha, \\
y=y_{0}+t \sin \alpha
\end{array} \text { ( } t \text { is a parameter). }\right.
\]

Substituting into the equation of curve $E$ gives
\[
\left(3-4 \sin ^{2} \alpha\right) t^{2}+2\left(3 x_{0} \cos \alpha-y_{0} \sin \alpha\right) t+3 x_{0}^{2}-y_{0}^{2}-3=0.
\]
Then $|M P||M Q|=-t_{3} t_{4}=-\frac{3 x_{0}^{2}-y_{0}^{2}-3}{3-4 \sin ^{2} \alpha}$.
Since points $P, C, Q, D$ are concyclic,
\[
\Rightarrow|M C||M D|=|M P||M Q|
\]
\[
\Rightarrow-\left(3 x_{0}^{2}-y_{0}^{2}-3\right)=-\frac{3 x_{0}^{2}-y_{0}^{2}-3}{3-4 \sin ^{2} \alpha}
\]
\[
\Rightarrow 3-4 \sin ^{2} \alpha=1 \Rightarrow \sin ^{2} \alpha=\frac{1}{2}.
\]
Since $\alpha \in[0, \pi)$ and $C D$ intersects $P Q$, we have $\alpha=\frac{3 \pi}{4}, k_{P Q}=-1$.
By the problem, we can set $l_{P Q}: y=-x+m$, and combining $l_{c D}$: $y=x-1$, the equation of the quadratic curve system passing through points $P, C, Q, D$ is
\[
\begin{array}{l}
(x-y-1)(x+y-m)+\lambda\left(3 x^{2}-y^{2}-3\right)=0 \\
\Rightarrow(1+3 \lambda) x^{2}-(1+\lambda) y^{2}-(1+m) x+ \\
\quad(m-1) y+m-3 \lambda=0 .
\end{array}
\]

Since equation (1) represents a circle, we have
\[
1+3 \lambda=-(1+\lambda) \Rightarrow \lambda=-\frac{1}{2}.
\]

At this point, equation (1) becomes
\[
x^{2}+y^{2}+2(m+1) x-2(m-1) y-2 m-3=0.
\]

Let the radius of the corresponding circle be $R$, then
\[
R^{2}=2 m^{2}+2 m+5 \geqslant \frac{9}{2}.
\]

When $m=-\frac{1}{2}$, i.e., $l_{P Q}: y=-x-\frac{1}{2}$, the area of the circle reaches its minimum value $\frac{9 \pi}{2}$.