Example 1 Given that a circle has intercepts $a, b$ on the $x$-axis and an intercept $c(c \neq 0)$ on the $y$-axis, find the equation of this circle.
Solution 1 Since the circle passes through points $A(a, 0), B(b, 0), C(0, c)$, the center $M$ lies on the perpendicular bisectors of $AB$ and $AC$, i.e., point $M$ is the intersection of the lines $x=\frac{1}{2}(a+b),(x-a)^{2}+y^{2}=x^{2}+(y-c)^{2}$. Solving these, we get $M\left(\frac{a+b}{2}, \frac{c^{2}+a b}{2 c}\right)$, and the radius $r=|M A|=\sqrt{\left(\frac{a+b}{2}-b\right)^{2}+\left(\frac{c^{2}+a b}{2 c}\right)^{2}}$. Therefore, by property 1, the equation of the circle is $\left(x-\frac{a+b}{2}\right)^{2}+\left(y-\frac{c^{2}+a b}{2}\right)^{2}=\left(\frac{b-a}{2}\right)^{2}+\left(\frac{c^{2}+a b}{2 c}\right)^{2}$.
Solution 2 Let the equation of the circle be $x^{2}+y^{2}+D x+E y+F=0$.
The distance between the circle and the $x$-axis is $a, b$, so $a^{2}+D a+F=0$, (1) $\quad b^{2}+D b+F=0$.
From (1) and (2), $a, b$ are the roots of the equation $x^{2}+D x+F=0$, thus by Vieta's formulas, we have $a+b=-D$, $a b=F$.
Since the circle intersects the $y$-axis at $c$, we have $c^{2}+E c+F=0$.
Thus, $c^{2}+E c+a b=0$, which gives $E=-\frac{c^{2}+a b}{c}$.
Clearly, $D^{2}+E^{2}-4 F>0$ is satisfied, so by property 2, the equation of the circle is
$$
x^{2}+y^{2}-(a+b) x-\left(c+\frac{a b}{c}\right) y+a b=0 .
$$