Now it gets hard

I formed a recurrence relation here and solved. In the diagram, the angles have been larger than they really are .

Clearly, the angle of incidence is same as angle of reflection, as the magnitude of components along common tangent and normal of ball and circle remain same.

Refer to the diagram. Here, the trajectory of particle between $n^{\text{ th }}$ and ${n+1}^{\text{ th }}$ bounce is shown.

Now, say atr $N$ bounce from side of table , the distance from $O$ is $x_{n}$ , i.e. $OA = x_{n}$ . Then, $OB = x_{n+1}$ . Clearly, $x_{0} = \epsilon$ .

Also, say $\theta_{n}$ is the angle that the motion of ball makes with vertical(the line that seems vertical in the diagram). Clearly $\theta_{0} = 0$

Using $OP = 1$ , $CP= 1$ , we have

$\displaystyle \phi = x_{n} + \theta_{n}$

$\displaystyle \theta_{n+1} = 2 \phi +\theta = 2x_{n} + 3 \theta_{n}$ ...... (i)

$\displaystyle x_{n+1} = x_{n} + \theta + 2\phi + \theta$

$\displaystyle\Rightarrow x_{n+1} = 3x_{n} +4 \theta_{n}$ .... (ii)

From (ii),

$\displaystyle \theta_{n} = \dfrac{1}{4}(x_{n+1} - 3x_{n})$

Putting this in in (i),

$\displaystyle \theta_{n+1} = 2x_{n} + \dfrac{3}{4} (x_{n+1} - 3x_{n})$

Thus, $\displaystyle \theta_{n} = 2x_{n-1} + \dfrac{3}{4} (x_{n} - 3x_{n-1})$

Putting this value in (ii),we have

$\displaystyle \boxed{ x_{n+1} = 6x_{n} - x_{n-1}}$ ,

where $x_{0} = \epsilon$ , $x_{1} = 3 \epsilon$

Hence, we can have $\displaystyle x_{n} = \dfrac{\epsilon}{2} \big((3+2 \sqrt{2})^n + (3 - 2 \sqrt{2})^n)$ , and $\displaystyle \dfrac{x_{25}}{x_{5}} \approx 2.05E+15$