Example. Given the coordinates of points $A$ and $B$ are $(-2,4)$
and $(8,-2)$, find the circle with $AB$ as its diameter.
Solution 1: Let the center of the circle to be found be $O(x, y)$. Since $A(-2,4)$ and $B(8,-2)$ are the endpoints of the diameter of the circle, it is not difficult to find the coordinates of the center. Because the center is the midpoint of $A$ and $B$, we have
$$
x=\frac{-2+8}{2}=3, y=\frac{4-2}{2}=1 \text {. }
$$

Also, $|AB|=\sqrt{(8+2)^{2}+(-2-4)^{2}}=2 \sqrt{34}$,
so the radius of the circle $r=\sqrt{34}$, thus the equation of the circle is $(x-3)^{2}+$
$$
\begin{array}{l}
(y-1)^{2}=34 \text {. } \\
\text { Solution 2: From Solution 1, we know the coordinates of the center } O \text { are }(3,1) \text {, } \\
\text { Let the equation of the circle be } \\
x^{2}+y^{2}+D x+E y+F=0, \\
\text { which is }\left(x+\frac{D}{2}\right)^{2}+\left(y+\frac{E}{2}\right)^{2} \\
=\left(\frac{1}{2} \sqrt{D^{2}+E^{2}-4 F}\right)^{2} \text {, } \\
\end{array}
$$

Thus we have
$$
\begin{array}{l}
-\frac{D}{2}=3 \cdots \cdots(1) \\
-\frac{E}{2}=1 \cdots(2)
\end{array}
$$

Since point $(-2,4)$ is on the circle, substituting its coordinates into the equation of the circle and simplifying, we get:
$$
-2 D+4 E+F+20=0 \ldots \ldots . .(3)
$$
$$
\begin{array}{l}
\text { Solving the system of equations (1), (2), and (3), we get, } \\
D=-6, E=-2, F=-24 .
\end{array}
$$

Therefore, the equation of the circle is $x^{2}+y^{2}-6 x-2 y-24=0$.
Solution 3: Let $P(x, y)$ be a point on the circle, then $\angle APB=90^\circ$, so $\triangle PAB$ is a right triangle. From the Pythagorean theorem, we know: $PA^{2}+PB^{2}=AB^{2}$,

which is $\left[(x+2)^{2}+(y-4)^{2}\right]+\left[(x-8)^{2}+(y+2)^{2}\right]$.
$$
=(8+2)^{2}+(-2-4)^{2} .
$$

Simplifying, we get:
$\mathbf{x}^{2}+\mathbf{y}^{2}-6 \mathbf{x}-2 \mathbf{y}-24=0$.
Thus, the equation of the circle is: $\mathbf{x}^{2}+y^{2}-6 x-2 y-24=0$.
Solution 4: Let $P(x, y)$ be a point on the circle
$\because AP \perp PB$

which is $\frac{y-4}{x+2}-\frac{y+2}{x-8}=-1$.
Simplifying, we find the equation of the circle is $x^{2}+y^{2}-6 x-2 y-24=0$.
Solution 5: This solution has several steps, mainly combining some plane geometry knowledge, although the method is more complex, it can review some plane geometry (Figure 4 (1)).
( - ) (1) Find the equation of line $AB$ using the two-point form, and find the slope $k_1$ of $AB$;
(2) Let $P(x, y)$ be a point on the circle, draw a perpendicular from $P$ to $AB$ intersecting at
$D$, then $k_{PD}=\frac{1}{k_{AP}}$, from
$k_{PD}$ and $P(x, y)$, we can find the equation of $PD$.
(3) Solve the equations representing $AB$ and $PD$ to find the coordinates of $D$;
(4) Find $|PD|$ from the coordinates of $P$ and $D$;
(5) From the property of a right triangle: $|PD| \cdot|AB|$
$=|PA| \cdot|PB|$, we can find the equation of the circle.
(II) Similar to (I), after finding the coordinates of $D$, use the distance formula between two points to find $|AD|$, $|DB|$, $|PD|$, from the formula $|PD|^{2}=|AD| \cdot|DB|$, we can also find the equation of the circle.
(III) Similar to (I), after finding the coordinates of $D$, find $|AD|$, $|PA|$, from the formula $|PA|^{2}=|AD| \cdot|BA|$, we can also find the equation of the circle.
(IV) Similar to (I), after finding the coordinates of $D$, find $|AD|$ and $|BD|$, from the formula $|PA|^{2}:|PB|^{2}=$ $|AD|:|DB|$, find the equation of the circle.

Solution 6: This solution is based on the area relationship, and can be divided into two types:
(I) Let $P(x, y)$ be a point on the circumference of the circle, combining the coordinates of points $A$ and $B$, use a third-order determinant to find the area of $\triangle PAB$. From $\frac{1}{2}\left|\begin{array}{rrr}x & y & 1 \\ -2 & 4 & 1 \\ 8 & -2 & 1\end{array}\right|$ being equal to $\left.\frac{1}{2}|\mathrm{PA}| \cdot \right|\, \mathrm{PB}|$, we can find the equation of the circle.
(II) From Solution (I), we know the coordinates of the center $O$ are $(3,1)$. Connect $PO$, then the area of $\triangle PAB = 2 \cdot$ the area of $\triangle PAO$. The area of $\triangle PAB = \frac{1}{2}|PA| \cdot|PB| \cdot$ the absolute value of the determinant $\left|\begin{array}{rrr}x & y & 1 \\ -2 & 4 & 1 \\ 3 & 1 & 1\end{array}\right|$, we can find the equation of the circle (Figure 4 (2)).

Solution 7: Let $P(x, y)$ be any point on the circle, connect $PA$, $PB$, draw a line parallel to $PB$ through $O$ intersecting $AP$ at $F$, then $F$ is the midpoint of $PA$. From the coordinates of $P$ and $A$, we can find the coordinates of $F$. From the coordinates of $P$ and $B$, we can find the slope of $PB$. From the coordinates of $O$ and $F$, we can find the slope of $OF$, from $k_{OF}=k_{PB}$, we can find the equation of the circle (Figure 4 (3)).