Since the centre of the circle is on the \(x\)-axis, let this centre be \((a . b)\). On the \(x\)-axis. \(y=0\). so \(h=0\) \(\Rightarrow(a . b)=(a \cdot o)\)
\(|O A|=|O B|\) radii
by distance between two points
\begin{array}{l}
|O A|=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \\
|O A|=\sqrt{(1-a)^{2}+(2-0)^{2}} \\
|O A|=\sqrt{(1-a)^{2}+4}
\end{array}
Similarly \(/ O B /\)
\begin{array}{l}
=\sqrt{(3-a)^{2}+(2-0)^{2}} \\
=\sqrt{(3-a)^{2}+4} \\
\Rightarrow|O A|=|O B| \\
\Rightarrow \sqrt{(1-a)^{2}+4}=\sqrt{(3-a)^{2}+4}
\end{array}
Square both sides
\begin{array}{l}
(1-a)^{2}+A=(3-a)^{2}+A \\
1-2 a+
ot h^{2}=9-6 a+
ot f^{2} \\
-2 a+6 a=9-1 \\
4 a=8 \\
a=\frac{8}{4}=2
\end{array}
so. the centre is \((2.0)\) and the radius \(r\)
\begin{array}{l}
r=|\mathrm{OA}|=\sqrt{(1-2)^{2}+4} \\
r=\sqrt{(-1)^{2}+4}=\sqrt{5} \text { and } r^{2}=5
\end{array}
the equation of a circle with centre \((a, b)=(2.0)\) and radius \(r\) is given by
\begin{array}{l}
(x-a)^{2}+(y-b)^{2}=r^{2} \\
(x-2) 2+(y-0)^{2}=5 \\
x^{2}-4 x+4+y^{2}=5 \\
x^{2}+y^{2}-4 x+4-5=0 \\
x^{2}+y^{2}-4 x-1=0
\end{array}