A complete, purely mathematical approach would imply a high number of tedious passages and calculations (the solution to this problem is known to require $14$ reflections before reaching $t=10$). However, if I interpret correctly the question, it asks for a simple general idea of solution.
If this is the case, we could choose different methods (e.g. using vectors, analytic geometry, trigonometry). The analytic geometry approach might be rather intuitive. We have to consider a sequence of successive segments on a Cartesian plane, each characterized by a starting point, a running line (the line actually followed by the photon in its path), a collision point (identified by determining the first crossing point between the running line and a circle), and a reflection line (whose angle is determined by the slope of the tangent to the circle in the collision point). Once completed a step, the collision point and the reflection line become the starting point and the running line of the successive step. Since for each segment of the path we can calculate the distance covered by the photon (or equivalently the time needed to run this distance), we can continue our step-by-step calculations until we reach $t=10$ and look at the point where we have arrived. In this way, the problem reduces to find the way of determining, from a starting point and a running line of a generic step, the corresponding collision point and the reflection line, so that iterating our calculations we can arrive to the solution.
The general approach to each step may be as follows. If $(x_n,y_n) $ is the starting point at the $n^{th} $ step and $a_n x+b_n $ is the equation of the corresponding running line, we have to begin by finding the collision point, i.e. the first crossing point with a circle. The circle can be firstly identified by visual examination of the graph (although we have then to confirm that the line actually crosses the circle and that the identified circle provides the minimal distance between the starting and collision point). Because all circles correspond to a translation of the circle $x^2 + y^2=\frac {1}{9} \,\,\,$ over an integer lattice, we can consider the equation $(x-j_n)^2 + (y-k_n)^2=\frac {1}{9} \,\,\, $, where $(j_n,k_n) \,\,$ are the integers expressing the coordinates of the centre of the circle on which the $n^{th} $ reflection occurs. Solving the system
$$\begin {cases}
y=a_n x+b_n \\
(x-j_n)^2 + (y-k_n)^2=\frac {1}{9} \end {cases} $$
to find the crossing point between the circle and the running line gives the coordinates of the collision point:
$$ x_{n+1}=\frac{m^2 \pm \sqrt {m^2-(b_n-k_n)^2-j_n^2 +1/9 }}{ a_n^2+1 }$$
and
$$y_{n+1}= a_n x_{n+1} +b_n$$
where $$m=a_n k_n +j_n-a_n b_n $$ and where we have to choose, among the two solutions, the case that corresponds to the lower distance between the starting and the collision point. This distance $d_n$ can be obtained using the standard formula, knowing that it is calculated between the points $(x_n,y_n) \,\,\, $ and $(x_{n+1},y_{n+1}) \,\,\, $.
Now we have to find the slope of the tangent to the circle in $x_{n+1},y_{n+1} \,\,\,$ to calculate the reflection line. Using the derivative of the circle equation, we obtain that the slope $t_n $ of the tangent at the $n^{th} $ reflection is
$$t_n=\pm \frac {(j _n - x_{n+1})}{\sqrt{1/9 - (j_n - x_{n+1})^2}}$$
where again we have to choose, among the two solutions, the appropriate one, looking at our graph to check the sign of the searched tangent. To calculate the slope of the reflection line, we can remind that if a line with angle $\alpha$ with respect to the $x $-axis is reflected on a line with angle $\beta $, the angle of the resulting reflected line is given by $2\beta-\alpha$. In this case we have $$\alpha=\arctan {a_n} \\ \beta=\arctan {t_n} $$ so the slope of our reflection line is $$\tan ( 2 \arctan {t_n} - \arctan {a_n} ) \,\,\, $$ Because it passes through $x_{n+1},y_{n+1} \, $, its equation is
$$\small {y= \tan ( 2 \arctan {t_n} - \arctan {a_n} ) x \\ + y_{n+1} - \tan ( 2 \arctan {t_n} - \arctan {a_n} ) x_{n+1} }$$
In this way, based on a starting point with coordinates $(x_n,y_n ) $ and a running line $a_n x+b_n $ that characterize the $n^{th} $ step, we have determined the collision point $(x_{n+1},y_{n+1}) \,\,$, the distance covered between these two points, and the reflection line. Setting the collision point and the reflection line as the starting point and the running line of the successive step, we can repeat several times the whole procedure, until the sum of the distances covered in each step achieves $10$.
To make a practical example, we can apply these calculations to the first step of the problem. We start from the point $(x_1,y_1)=(0.5,0.1) \,\,\,\,$ and the running line is $y=0.1$. The first circle to be crossed is that with centre in $(1,0)$, which has equation $(x-1)^2 +y^2=\frac {1}{9} \,\,\, $. Applying the equations above with $a_1=0 \,$, $b_1=0.1\,$, $j_1=1 \,$, and $k_1=0 \,$, we get $m=1\,$, so that the coordinates of the collision point $(x_2,y_2) \,$ (knowing that we must choose, among the two solutions, that nearer to the starting point) reduce to
$$ x_{2}=1 - \sqrt {1-0.1^2-1 +1/9 } = 1 - \frac {\sqrt {0.91}}{3} \approx 0.682 \\ y_{2}=0 \cdot x_2+0.1=0.1
$$
Using the derivative of the circle equation, we obtain that the slope $t_1 $ of the tangent at the first reflection is
$$t_1= \frac{1 - ( 1 - \sqrt {0.91}/3) }{\sqrt{1/9 - [(1- (1 - 0.91/3)]^2 }} \\ = \frac {\sqrt{0.91}/3 }{\sqrt{1/9 - 0.91/9 }} = \sqrt {91}/3 \approx 3.1798 $$
Lastly, the equation of the reflected line is
$${y=\tan \left ( 2 \arctan { \sqrt {91}/3 } \right) x \\ + 0.1 - \tan \left ( 2 \arctan { \sqrt {91}/3 } \right) \cdot (1 - \sqrt {0.91}/3 )}$$
$$y=- \frac {3 \sqrt {91}}{41} x + 0.1 + \frac {3 \sqrt {91}}{41} - \frac {91}{410} $$
which in numbers is approximately
$$y=-0.6980 x +0.5761$$
So we have determined the collision point and the reflection line of the first step. As explained above, these become the starting point and the running line of the second step, allowing to repeat the procedure.
