Example 1. Solve the equation $\mathrm{x}^{2}-2 \mathrm{x}-3=0$ geometrically.
We note that for any given equation (I), the fixed point $\mathrm{M}(0,1)$ lies on the circle (I) corresponding to the equation (I).
$$
\begin{array}{l}
\text { Let } p=-2, q=-3 \text { be substituted into (*), we get } O^{\prime} \\
(1,-1) . \text { Taking } O^{\prime} \text { as the }
\end{array}
$$

center, and the line segment connecting $\mathrm{O}^{\prime}$ and
$\mathrm{M}(0,1)$ as the
radius, draw a circle on the Cartesian coordinate paper
(see Figure 6). It can
be seen that this circle intersects the $x$-axis
at $\mathrm{A}(-1$,
$0)$ and $\mathrm{~B}(3,0)$. Therefore, the solutions to the original equation are $\mathrm{x}_{1}=$ $-1, x_{2}=3$