11. (20 points) Given that $P$ is a moving point on the parabola $y=x^{2}-1$, and let point $Q(-1,0)$. If the circle with diameter $P Q$ intersects the parabola at only two points, find the range of the area of the circle.

 untranslated text:
已知 $P$ 是抛物线 $y=x^{2}-1$上一动点, 设点 $Q(-1,0)$. 若以 $P Q$ 为直径的圆与抛物线只有两个公共点,求该圆面积的取值范围.

 translated text:
Given that $P$ is a moving point on the parabola $y=x^{2}-1$, and let point $Q(-1,0)$. If the circle with diameter $P Q$ intersects the parabola at only two points, find the range of the area of the circle.
11. Consider the opposite.

Let the circle with diameter $P Q$ intersect the parabola at a point $R\left(t, t^{2}-1\right)$ other than points $P$ and $Q$.

If $t=1$, then $k_{Q R}=0$. Thus, $P R$ is perpendicular to the $x$-axis, which is a contradiction. Therefore, $t \neq 1$.
Let $P(x, y)$.
Since $k_{Q R}=\frac{t^{2}-1}{t+1}=t-1$, we have
$$
k_{P R}=\frac{1}{1-t} \text {. }
$$

Then the line $l_{P R}$ is given by: $y-t^{2}+1=\frac{1}{1-t}(x-t)$.
Combining this with $y=x^{2}-1$, and eliminating $y$, we get
$$
(1-t)\left(x^{2}-t^{2}\right)=x-t \text {. }
$$

Since $x \neq t$, we have
$$
\begin{array}{l}
(1-t)(x+t)=1 \\
\Rightarrow t^{2}+(x-1) t-x+1=0 .
\end{array}
$$

From the quadratic equation in $t$ (1), we know
$$
\begin{array}{l}
\Delta=(x-1)^{2}-4(1-x) \geqslant 0 \\
\Rightarrow x \leqslant-3 \text { or } x \geqslant 1 .
\end{array}
$$

And $4 S_{\text {intersection }}=\pi|P Q|^{2}$
$$
\begin{array}{l}
=\pi\left[(x+1)^{2}+y^{2}\right] \\
=\pi\left[(x+1)^{2}+\left(x^{2}-1\right)^{2}\right] . \\
\text { Let } f(x)=(x+1)^{2}+\left(x^{2}-1\right)^{2} .
\end{array}
$$

When $x \geqslant 1$, $f(x)$ is an increasing function, so
$$
f(x) \geqslant f(1)=4 \text {; }
$$

When $x \leqslant-3$, $f(x)$ is a decreasing function, so
$$
f(x) \geqslant f(-3)=68 \text {. }
$$

Then $4 S_{\text {circle }}=\pi f(x) \geqslant 4 \pi$.
Thus, $S_{\text {circle }} \geqslant \pi$.
Therefore, if the circle with diameter $P Q$ intersects the parabola at only two points, then $0<S_{\text {intersection }}<\pi$.
Hence, the range of the area of the circle is $(0, \pi)$.