Draw perpendicular segments from O to AB and BC which intersects two sides at point D and E. Apparently, OD and OE are radii of the semicircle.

OR = OE

Connect point O and B. Since \triangle ODB and \triangle OEB are right triangles and there're two pairs of corresponding sides equal,

\triangle ODB \cong \triangle OEB

Then

\angle DBO = \angle EBO

In other words, BO bisects the internal angle DBE of the triangle ABC.

Apply the Angle bisector theorem,

\dfrac{AB}{AO} = \dfrac{BC}{CO}

Then,

\dfrac{AB}{AO} = \dfrac{BC}{AC-AO}

Substitute the values of AB, BC and AC

\dfrac{12}{AO} = \dfrac{18}{25-AO}

Solve for AO

Then, AO = 10

Since the three sides of the triangle ABC are known values, we can use Heron's formula to establish the equation. Let A be the area of the triangle ABC and R is the radius of the semicircle.

A = A_{ABO}+A_{BCO}

Apply the Heron's formula

A = \sqrt{s(s-a)(s-b)(s-c)} \quad

in which s = \dfrac{a+b+c}{2} = \dfrac{12+18+25}{2} = \dfrac{55}{2}

Then

A = \sqrt{\dfrac{55}{2}\Big( \dfrac{55}{2}-12\Big) \Big( \dfrac{55}{2}-18\Big) \Big( \dfrac{55}{2}-25\Big) }

=\dfrac{5}{4}\sqrt{6479}

A_{ABO}+A_{BCO} = \dfrac{1}{2}AB\cdot R+\dfrac{1}{2}BC\cdot R

=15R

Therefore

R = \dfrac{A}{15} = \dfrac{1}{12} \sqrt{6479}\approx 6.7

The diameter is \dfrac{1}{6} \sqrt{6479}\approx 13.4