Hello, lance!
This is a messy one . . .
Find the equation of the circle touching the $\displaystyle x$axis
and passing through the points $\displaystyle A(7,2)$ and $\displaystyle B(0,9).$ Code:

 T
+*o*
 Ao (h,0) *
* *
Bo *

* C *
* * *
* (h,k) *

* *
* *
 * *
 * * *

$\displaystyle \text{The center is }C(h,k)\text{, directly below the point of tangency, }T(h,0)$
$\displaystyle CA, CB\text{ and }CT\text{ are radii. }\;\text{Hence: }\,CA \,=\,CB \,=\,k$
We have: . $\displaystyle \begin{array}{ccccc}
CA^2 & =& (h7)^2 + (k+2)^2 &=& k^2\\
CB^2 &=& h^2 + (k+9)^2 &=& k^2\end{array} $
Simplify: . $\displaystyle \begin{array}{ccccc}
h^2  14h + 4k + 53 &=& 0 & {\color{blue}[1]} \\
h^2 \qquad + 18k + 81 &=& 0 & {\color{blue}[2]} \end{array}$
Subtract [2]  [1]: .$\displaystyle 14h + 14k + 28 \:=\:0 \quad\Rightarrow\quad k \:=\:(h+2)\;\;{\color{blue}[3]}$
Substitute into [2]: .$\displaystyle h^2  18(h+2) + 81 \:=\:0 \quad\Rightarrow\quad h^2  18h + 45 \:=\:0$
Hence: .$\displaystyle (h3)(h15) \:=\:0 \quad\Rightarrow\quad h \:=\:3,\:15$
Substitute into [3]: .$\displaystyle k \:=\:5,\:17$
We have two solutions: . $\displaystyle \begin{array}{ccccc} \hline
\text{Center} & \text{Radius} & \text{Equation} &&\\ \hline \\[5mm]
(3,5) & 5 & (x3)^2 + (y+5)^2 &=& 25 \\ \\[5mm]
(15,17) & 17 & (x15)^2 + (y+17)^2 &=& 289 \\ \hline\end{array}$