33 The line $y=m x+b(|m|<1,|b|<1)$ intersects the circle $x^{2}+y^{2}=1$ at $P, Q$, and the hyperbola $x^{2}-y^{2}=1$ at $R, S$. If $P, Q$ trisect the segment $R S$, find the values of $m, b$.
33 Let the x-coordinates of $P$ and $Q$ be $x_{1}, x_{2}$. Then $x_{1}, x_{2}$ are the roots of the equation
$$
x^{2}+(m x+b)^{2}=1
$$

Thus,
$$
\left\{\begin{array}{l}
x_{1}+x_{2}=-\frac{2 m b}{1+m^{2}}, \\
x_{1} \cdot x_{2}=\frac{b^{2}-1}{1+m^{2}} .
\end{array}\right.
$$

Let the x-coordinates of $R$ and $S$ be $x_{1}^{\prime}, x_{2}^{\prime}$. Then they are the roots of the equation
$$
x^{2}-(m x+b)^{2}=1
$$

Thus,
$$
\left\{\begin{array}{l}
x_{1}^{\prime}+x_{2}^{\prime}=\frac{2 m b}{1-m^{2}} \\
x_{1}^{\prime} x_{2}^{\prime}=\frac{-b^{2}-1}{1-m^{2}} .
\end{array}\right.
$$

Notice that the midpoints of $P Q$ and $R S$ coincide. Therefore,
$$
x_{1}+x_{2}=x_{1}^{\prime}+x_{2}^{\prime} \text {. }
$$

Substituting (1) and (3) gives
$$
\frac{-2 m b}{1+m^{2}}=\frac{2 m b}{1-m^{2}},
$$

which implies $m b=0$.
When $m=0$, the line $y=m x+b$ becomes $y=b$. Substituting it into
$$
x^{2}+y^{2}=1 \quad \text { and } \quad x^{2}-y^{2}=1
$$

we get
$$
\begin{array}{l}
x_{1}=\sqrt{1-b^{2}}, x_{2}=-\sqrt{1-b^{2}} ; \quad\left(x_{1}>x_{2}\right) \\
x_{1}^{\prime}=\sqrt{1+b^{2}}, x_{2}^{\prime}=-\sqrt{1+b^{2}} . \quad\left(x_{1}^{\prime}>x_{2}^{\prime}\right)
\end{array}
$$

It is easy to see that
$$
x^{\prime}{ }_{2}<x_{2}<x_{1}<x_{1}^{\prime} .
$$

Since $P, Q$ trisect $R S$,
$$
\frac{1}{2}\left(x_{1}^{\prime}+x_{1}\right)=x_{2},
$$

which means
$$
\frac{1}{2}\left(-\sqrt{1+b^{2}}+\sqrt{1-b^{2}}\right)=-\sqrt{1-b^{2}} .
$$

Thus, $b= \pm \frac{2}{5} \sqrt{5}$.
When $b=0$, the line $y=m x+b$ becomes $y=m x$. Substituting it into
$$
x^{2}+y^{2}=1, x^{2}-y^{2}=1 \text {, }
$$

we get
$$
\begin{array}{l}
x_{1}=\sqrt{\frac{1}{1+m^{2}}}, x_{2}=-\sqrt{\frac{1}{1+m^{2}}} ; \\
x_{1}^{\prime}=\sqrt{\frac{1}{1-m^{2}}}, x_{2}^{\prime}=-\sqrt{\frac{1}{1-m^{2}}} .
\end{array}
$$

It is easy to see that
$$
x^{\prime}{ }_{2}<x_{2}<x_{1}<x_{1}^{\prime} .
$$

Thus,
$$
\frac{1}{2}\left(x_{2}^{\prime}+x_{1}\right)=x_{2},
$$

which means
$$
-\sqrt{\frac{1}{1-m^{2}}}+\sqrt{\frac{1}{1+m^{2}}}=-2 \sqrt{\frac{1}{1+m^{2}}} .
$$

Thus, $m= \pm \frac{2}{5} \sqrt{5}$.
In summary, we have
$$
\left\{\begin{array} { l } 
{ m = 0 } \\
{ b = \frac { 2 } { 5 } \sqrt { 5 } , }
\end{array} \left\{\begin{array} { l } 
{ m = 0 } \\
{ b = - \frac { 2 } { 5 } \sqrt { 5 } , }
\end{array} \left\{\begin{array} { l } 
{ m = \frac { 2 } { 5 } \sqrt { 5 } } \\
{ b = 0 , }
\end{array} \left\{\begin{array}{l}
m=-\frac{2}{5} \sqrt{5} \\
b=0 .
\end{array}\right.\right.\right.\right.
$$