How do I find the diameters of the circles in this geometry puzzle?

The diagram you need to draw, with $r$ as the radius of the larger circle (giving $r{-}\frac 92$ as the radius of the smaller) is:

where $G$ is the centre of the smaller circle. From here you should be able to use Pythagoras to solve.

Since suitable time has now elapsed, the completion to a solution should look something like:

$$\require{cancel}\begin{align} \left( r-\frac 92 \right)^2 &= (r-5)^2+\left( \frac 92 \right)^2 \\ r^2 - 9r +\left( \frac 92 \right)^2 &= r^2 - 10r + 25 +\left( \frac 92 \right)^2 \\ \cancel{r^2} - 9r +\cancel{\left( \frac 92 \right)^2} &= \cancel{r^2} - 10r + 25 +\cancel{\left( \frac 92 \right)^2} \\ 10r-9r&=25\\ r &= 25 \end{align}$$

So the diameter of the large circle is $2\cdot 25 = \fbox{50}$ and of the smaller circle $50-9 =\fbox{41}$

Let $d$ be the length of the segment $DC$, and $r$ the radius of the larger circle. Thanks to the geometric mean theorem of elementary geometry we can write:

$$ \begin{align} d \cdot r &= (r-5)^2 \\ d+r &= 2r - 9 \enspace. \end{align}$$ This simplifies to $$ \begin{align} d \cdot r &= r^2 - 10r + 25 \\ d &= r - 9 \enspace. \end{align}$$ Therefore the radius of the larger circle is $25$ cm and the radius of the smaller circle is $(2\cdot 25 - 9) / 2 = 41/2 = 20.5$ cm.