Problem 1349

11. (1) (Method 1) Note that the tangent line equation through point $E$ is $x+y-2=0$, and the tangent line equation through point $F$ is $x-y-2=0$. Their intersection point is $A(2,0)$.
Let $P\left(x_{0}, y_{0}\right)$, then the equation of the circumcircle is
$$\left(x-x_{0}\right)^{2}+\left(y-y_{0}\right)^{2}=\left(2-x_{0}\right)^{2}+\left(0-y_{0}\right)^{2},$$

which simplifies to
$$x^{2}+y^{2}-2 x_{0} x-2 y_{0} y+4 x_{0}-4=0 .$$

By combining the above equation with $x+y-2=0$, we can find the intersection point of the circumcircle and the tangent line through point $E$ as $C\left(x_{0}-y_{0}, y_{0}-x_{0}+2\right)$. Similarly, we can find the intersection point of the circumcircle and the tangent line through point $F$ as $B\left(x_{0}+y_{0}, x_{0}+y_{0}-2\right)$. Therefore, the equation of line $B C$ is
$$\left(x-x_{0}-y_{0}\right) \cdot\left(2 x_{0}-4\right)-\left(y-x_{0}-y_{0}+2\right) \cdot\left(2 y_{0}\right)=0,$$

which simplifies to
$$\left(x_{0}-2\right) x-y_{0} y+y_{0}^{2}-x_{0}^{2}+2 x_{0}=0 .$$

Since circle $\omega$ is tangent to line $B C$, we have
$$\left|y_{0}^{2}-x_{0}^{2}+2 x_{0}\right|=\sqrt{2} \cdot \sqrt{\left(x_{0}-2\right)^{2}+y_{0}^{2}},$$

Squaring the above equation, it can be seen as a quadratic equation in $y_{0}^{2}$, from which we can solve $y_{0}^{2}=\left(x_{0}-2\right)^{2}$ (discard) or $y_{0}^{2}=x_{0}^{2}-2$. Therefore, the trajectory equation of point $P$ is $x^{2}-y^{2}=2$.
(Method 2) Let $D(\sqrt{2} \cos t, \sqrt{2} \sin t)$, then the tangent line equation at point $D$ is $x \cos t+y \sin t=\sqrt{2}$. It intersects the tangent line through point $C$ at $C\left(\frac{\sqrt{2}-2 \sin t}{\cos t-\sin t}, \frac{-\sqrt{2}+2 \cos t}{\cos t-\sin t}\right)$, and intersects the tangent line through point $D$ at $B\left(\frac{\sqrt{2}+2 \sin t}{\cos t+\sin t}, \frac{\sqrt{2}-2 \cos t}{\cos t+\sin t}\right)$

Notice that $\triangle A B C$ is a right triangle (with $A$ as the right angle vertex), so the circumcenter $P$ of $\triangle A B C$ is the midpoint of side $B C$, i.e.,
$$P\left(\frac{\sqrt{2} \cos t-2 \sin ^{2} t}{\cos ^{2} t-\sin ^{2} t}, \frac{\sqrt{2} \sin t-2 \sin t \cos t}{\cos ^{2} t-\sin ^{2} t}\right) .$$

Eliminating $t$, we find that the trajectory equation of point $P$ is $x^{2}-y^{2}=2$.
Note When $\left(x_{0}, y_{0}\right)=\left(\frac{3}{2}, \frac{1}{2}\right)$ or $\left(\frac{3}{2},-\frac{1}{2}\right)$, the equation of line $B C$ becomes $x+y-2=0$ or $x-y-2=0$, i.e., $B$ or $C$ coincides with $A$, in which case the "circumcircle" does not exist. Therefore, strictly speaking, the trajectory equation should be $x^{2}-y^{2}=2\left(x \neq \frac{3}{2}\right)$. However, the degenerate case $x \neq \frac{3}{2}$ is easily explained geometrically, so not writing $x \neq \frac{3}{2}$ is not penalized.
(2) Take $G(-2,0)$, and construct a circle $\gamma$ with $G$ as the center and $2 \sqrt{2}$ as the radius. We prove that the circumcircle of $\triangle A B C$ is tangent to circle $\gamma$. For this, we only need to prove $|P A|-|P G|= \pm 2 \sqrt{2}$.

In fact, if $P\left(x_{0}, y_{0}\right)$, then $|P A|^{2}=\left(x_{0}-2\right)^{2}+y_{0}^{2},|P G|^{2}=\left(x_{0}+2\right)^{2}+y_{0}^{2}$. Substituting $y_{0}^{2}=x_{0}^{2}-2$, we get
$$|P A|^{2}=2\left(x_{0}-1\right)^{2}, \quad|P G|^{2}=2\left(x_{0}+1\right)^{2} \text {. }$$

Noting that $\left|x_{0}\right| \geqslant \sqrt{2}$, when $x_{0} \geqslant \sqrt{2}$, $|P A|=\sqrt{2}\left(x_{0}-1\right),|P G|=\sqrt{2}\left(x_{0}+1\right)$, in which case $|P A|-|P G|=$ $-2 \sqrt{2}$. When $x_{0} \leqslant-\sqrt{2}$, $|P A|=-\sqrt{2}\left(x_{0}-1\right),|P G|=-\sqrt{2}\left(x_{0}+1\right)$, in which case $|P A|-|P G|=$ $2 \sqrt{2}$. Therefore, the conclusion is proved.