Animated Solution for Mathematics - Circles: The length of the diameter of the circle which touches the x-axis at the point $(1,0)$ and passes through the point $(2,3)$ is:

Imagine you are standing on the coordinate plane, looking at a circle that kisses the x-axis at the point $(1,0)$. This is a geometric puzzle where the fundamental truth is that the radius drawn to the point of tangency is perpendicular to the tangent line.

Since our tangent is the horizontal x-axis, the radius must be a perfectly vertical line. Because the circle touches the x-axis at $(1,0)$, the center must lie on the vertical line $x=1$.

We know the x-coordinate of the center is $1$. The distance from the center to the x-axis is, by definition, the radius $r$. Therefore, the center of our circle is $C(1,r)$.

Using the standard form of a circle equation, $(x-h{)}^2+(y-k{)}^2=r^2$, we substitute our center $(1,r)$ to obtain:

This equation is the heart of our solution, encapsulating everything we know about the circle's position and size.

The problem states that the circle passes through the point $(2,3)$. Since this point lies on the circle, it must satisfy our equation. Substituting $x=2$ and $y=3$ into the equation, we get:

Expanding this, we note that $(2-1{)}^2=1$. Expanding $(3-r{)}^2$ using the identity $(a-b{)}^2=a^2-2ab+b^2$, we obtain $9-6r+r^2$. Combining these, we have:

Observe the equation: $10-6r+r^2=r^2$. The $r^2$ terms on both sides cancel out, leaving us with a simple linear equation:

Solving for $r$, we move $6r$ to the other side to get $6r=10$, which simplifies to:

We have found the radius $r=\frac{5}{3}$. However, the question asks for the diameter, not the radius.

The diameter $d$ is defined as $2r$. Therefore:

The final answer is $\frac{10}{3}$.