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:

Let (h,k) be the centre.
The equation of the circle is (x-h)² + (y-k)² = r², where r is the radius.
Since the circle touches the x-axis at (1,0), the radius is k and the center is (1,k).
The equation of the circle becomes (x-1)² + (y-k)² = k².
Since the point (2,3) lies on the circle, we substitute x=2 and y=3 into the equation:
(2-1)² + (3-k)² = k²
1 + (3-k)² = k²
1 + 9 - 6k + k² = k²
10 - 6k = 0
6k = 10
k = 5/3
Therefore, the radius is 5/3 and the diameter is 2k = 2(5/3) = 10/3.
However, this value is not among the options.
Let's reconsider the problem. The circle touches the x-axis at (1,0), so the center is (1,r), where r is the radius. The distance from the center (1,r) to the point (2,3) is r. Using the distance formula:
√((2-1)² + (3-r)²) = r
√(1 + (3-r)²) = r
1 + (3-r)² = r²
1 + 9 - 6r + r² = r²
10 - 6r = 0
6r = 10
r = 5/3
Diameter = 2r = 2(5/3) = 10/3
This is still not an option. Let's use a different approach.
Let the center be (1, r). The distance from (1, r) to (2, 3) is r.
(2-1)² + (3-r)² = r²
1 + 9 - 6r + r² = r²
10 - 6r = 0
r = 10/6 = 5/3
Diameter = 2r = 10/3 ≈ 3.33
This is not among the given options. There seems to be an issue with the question or the provided options.