1. Use the condition of touching the $x$-axis at $(1,0)$

   If a circle touches the $x$-axis at $(1,0)$, then the radius at the point of contact is perpendicular to the $x$-axis.

   Hence the center must lie vertically above or below $(1,0)$.

   So let the center be $C=(1,r)$ where $r$ is the radius. Since the circle touches the $x$-axis, its radius is exactly the distance from the center to the $x$-axis, which is $|r|$. Taking the center above the axis, $C=(1,r), \quad \text{radius}=r.$

2. Use the fact that the circle passes through $(2,3)$

   The distance from the center $(1,r)$ to the point $(2,3)$ must be equal to the radius $r$: $\sqrt{(2-1)^2+(3-r)^2}=r.$

   Squaring both sides: $1+(3-r)^2=r^2.$

3. Simplify

   $1+9-6r+r^2=r^2$ $10-6r=0$ $6r=10$ $r=\frac{5}{3}.$

4. Find the diameter

   Diameter $=2r$: $2\cdot \frac{5}{3}=\frac{10}{3}.$

5. Check options

   • A: $\frac{10}{3}$ ✅
   • B: $\frac{3}{5}$
   • C: $\frac{6}{5}$
   • D: $\frac{5}{3}$

Therefore, the correct answer is A.