This focuses on finding the potential simply using some geometrical connection and conservation laws. The geometric arrangement is shown in the following figure:
Notations:
- From the center of force which is located $x$ to the right of the circle's center. $(r, \theta)$ is the polar coordinate from $x$. The hatted $\hat{r}$ and $\hat{\theta}$ are the corresponding unit vectors.
- Polar coordinate from the center of circle, the same point $(r, \theta)$ is denoted as $(a, \phi)$, $a$ is the radius. Their unit vectors are denoted as $\hat{a}$ and $\hat{\phi}$.
The first conserved quantity is angular momentum in the $\hat{z}$ direction (out of paper):
$$ \tag{1}
L = m r v_\theta = m r^2 \dot{\theta}.
$$
The information at $r=a + x$ and $r = a - x$ defines the value for angular moment:
$$ \tag{2}
L = m r^2 \dot{\theta} = m (a-x) v_1 = m (a+x) v_2.
$$
One of the three paramters $v_1$, $v_2$ and $x$ can be solved as the rest two's. The angular speed $\dot{\theta}$ can be written as a function of $r$ from Eq.(2):
$$ \tag{3}
\dot{\theta} = \frac{ (a-x) v_1}{r^2} = \frac{ 2 v_1 v_2}{v_1 + v_2} \frac{a}{r^2}
$$
Then, note that it is a circular motion around the circular center, therefore the velocity is always in a tangential direction $\hat\phi$, $v = v_\phi$. The angular velocity $v_\theta$ is the projection of $v$ in the $\hat\theta$ direction. Thus:
$$ \tag{4}
v_\theta = v \cos \ \beta; \text{ therefore } v = \frac{ v_\theta}{\cos \ \beta} = \frac{ r \dot\theta}{\cos \ \beta} = \frac{(a-x) v_1}{ r \cos \ \beta}
$$
From the triangle in the figure, we find the $\cos\beta$:
$$
\cos\beta = \frac{a^2 + r^2 - x^2}{2 a r}
$$
Substitute $\cos\beta$ into Eq.(4):
$$ \tag{5}
v = \frac{2 a (a-x) v_1}{a^2 + r^2 - x^2}.
$$
The kinetic energy for the speed at position $(r, \theta)$:
$$
E_K = \frac{1}{2}m v2 = m \frac{2 a^2 (a-x)^2 v_1^2}{(a^2 + r^2 - x^2)^2}
$$
Now, using the conservation of energy, assuming the potential energy at position of $r= a-x$ is $V_1$:
$$ \tag{6}
V_1 + \frac{1}{2} m v_1^2 = V(r) + m \frac{2 a^2 (a-x)^2 v_1^2}{(a^2 + r^2 - x^2)^2}
$$
Finally,
$$\tag{7}
V(r) - V_1 = \frac{1}{2} m v_1^2 \{ 1 - \frac{4 a^2 (a-x)^2}{(a^2 - x^2 + r^2 )^2} \}.
$$
It helps to build confidence by checking at $r=r_1=a - x$, both sides of Eq.(7) should be zero:
$$
V(a-x) - V_1 = \frac{1}{2} m v_1^2 \{ 1 - \frac{4 a^2 (a-x)^2}{(a^2 - x^2 + (a-x)^2)^2} \} \\
0 =\frac{1}{2} m v_1^2 \{ 1 - \frac{4 a^2 (a-x)^2}{(2 a (a-x))^2} \} =0.
$$
Another harder exercise is to prove $V(a+x) + \frac{1}{2}mv_2^2 = V_1 + \frac{1}{2}mv_1^2$.
