6. Given the circle $C: x^{2}+y^{2}=r^{2}$, points $P$ and $P^{\prime}$ lie on a ray starting from $O$, and satisfy $|O P|\left|O P^{\prime}\right|=r^{2}$, then points $P$ and $P^{\prime}$ are said to be symmetric with respect to the circumference of circle $C$. Then, the equation satisfied by the point $P^{\prime}$, which is symmetric to the point $P(x, y)$ on the hyperbola $x^{2}-y^{2}=1$ with respect to the unit circle $C$: $x^{2}+y^{2}=1$, is ( ).
(A) $x^{2}-y^{2}=x^{4}+y^{4}$
(B) $x^{2}-y^{2}=\left(x^{2}+y^{2}\right)^{2}$
(C) $x^{2}-y^{2}=2\left(x^{4}+y^{4}\right)$
(D) $x^{2}-y^{2}=2\left(x^{2}+y^{2}\right)^{2}$
6. B.

For any point $\left(x_{0}, y_{0}\right)$ on the hyperbola $x^{2}-y^{2}=1$, let its symmetric point with respect to the circle $C$ be $\left(x_{0} t, y_{0} t\right)$. Then
$$
\begin{array}{l}
\sqrt{x_{0}^{2}+y_{0}^{2}} \sqrt{x_{0}^{2} t^{2}+y_{0}^{2} t^{2}}=1 \\
\Rightarrow|t|=\frac{1}{x_{0}^{2}+y_{0}^{2}} .
\end{array}
$$

Let $x=x_{0} t, y=y_{0} t$. Then
$$
\begin{array}{c}
x^{2}=x_{0}^{2} t^{2}=\frac{x_{0}^{2}}{\left(x_{0}^{2}+y_{0}^{2}\right)^{2}}, \\
y^{2}=y_{0}^{2} t^{2}=\frac{y_{0}^{2}}{\left(x_{0}^{2}+y_{0}^{2}\right)^{2}} .
\end{array}
$$

Thus, $x^{2}-y^{2}=\frac{x_{0}^{2}-y_{0}^{2}}{\left(x_{0}^{2}+y_{0}^{2}\right)^{2}}=\frac{1}{\left(x_{0}^{2}+y_{0}^{2}\right)^{2}}$,
$$
x^{2}+y^{2}=\frac{x_{0}^{2}+y_{0}^{2}}{\left(x_{0}^{2}+y_{0}^{2}\right)^{2}}=\frac{1}{x_{0}^{2}+y_{0}^{2}} \text {. }
$$

Combining the above two equations, the equation satisfied by point $P^{\prime}$ is $x^{2}-y^{2}=\left(x^{2}+y^{2}\right)^{2}$.