Envelope of a family of lines in the plane 
Let $A$ and $B$ be two bugs lying on two distinct points $a_0, b_0$ on a fixed circle. They start to walk along the circle in the same direction such that at time $t$ their coordinates $a_t, b_t$ satisfy $G(a_t,b_t)=0,$ where $G$ is a given implicit function.

If $L_t$ is the line segment joining $a_t$ and $b_t,$ then how can we find the Envelope of the family $\{L_t\}_t$ ?
 A: Maybe something like this? 
I will use complex numbers for simplicity of notation. I will assume that the two bugs move along the unit circle and $t$ is the arc-length parametrization of the first bug $A$, i.e. we assume the first bug moves uniformly along the circle. Then by $\theta(t)$ we denote the angle that determines the motion of the second bug $B$ along the unit circle. Then, the motion of bug $A$ is $e^{it}$ and the motion of the second bug $B$ is $e^{t\theta(t)}$. Then, by assumption, the two motions are related by the equation 
$$G\big(e^{it}, e^{i\theta(t)}\big) = 0$$ which I am going to simply write as $$g\big(t,\theta(t)\big) = 0$$ The enveloping curve in question $z = z(t) = x(t) + i\, y(t)$ is by definition a curve such that 


*

*$z(t)$ is a point on the line determined by the points $e^{it}$ and $e^{i\theta(t)}$, and 

*its derivative $\dot{z}(t) = \cfrac{dz}{dt}(t)$ should be a vector parallel to the line determined by the points $e^{it}$ and $e^{i\theta(t)}$ 
The first condition means that there exists a function $\lambda(t)$ such that 
$$z(t) = e^{it} + \lambda(t) \, \big(e^{i\theta(t)} - e^{it}\big)$$ and the second condition implies that $\dot{z}(t)$ is parallel to $e^{i\theta(t)} - e^{it}$. The latter condition can be written in complex numbers as
$$0 = \text{Im}\Big(\, \dot{z}(t) \cdot\overline{\big(e^{i\theta(t)} - e^{it}\big)} \,\,\Big) =  \text{Im}\Big(\, \dot{z}(t) \cdot \big(e^{ - i\theta(t)} - e^{- it}\big) \,\Big) $$
Calculate the derivative 
$$\dot{z}(t) = ie^{it} + \dot{\lambda}(t)\,\big(e^{i\theta(t)} - e^{it}\big) + i \,\lambda(t) \, \big(\dot{\theta}(t) \, e^{i\theta(t)} - e^{it}\big)$$ and form the equation
\begin{align}
0 =  \text{Im}\Big(\, & \Big(ie^{it} + \dot{\lambda}(t)\,\big(e^{i\theta(t)} - e^{it}\big) + i \,\lambda(t) \, \big(\dot{\theta}(t) \, e^{i\theta(t)} - e^{it}\big)\Big) \cdot \Big(e^{ - i\theta(t)} - e^{- it}\Big) \,\Big)\\ 
= \text{Im}\Big(\, & ie^{it} \big(e^{ - i\theta(t)} - e^{- it}\big) + \dot{\lambda}(t)\,\big(e^{i\theta(t)} - e^{it}\big)\,\big(e^{ - i\theta(t)} - e^{- it}\big) + \\ &+ i \,\lambda(t) \, \big(\dot{\theta}(t) \, e^{i\theta(t)} - e^{it}\big)\Big) \big(e^{ - i\theta(t)} - e^{- it}\big) \,\Big)\\
= \text{Im}\Big(\, & e^{i(t- \theta(t))} - 1 + \dot{\lambda}(t)\,\big|e^{i\theta(t)} - e^{it}\big|^2 + i \,\lambda(t) \, \big(\dot{\theta}(t) -  \, \dot{\theta}(t) e^{i(\theta(t) - t)} - e^{i(t - \theta(t))} +1 \big) \,\Big)\\
= \text{Im}\Big(\, & e^{i(t- \theta(t))} - 1\Big) +  \text{Im}\Big(\, i \,\lambda(t) \, \big(\dot{\theta}(t) -  \, \dot{\theta}(t) e^{i(\theta(t) - t)} - e^{i(t - \theta(t))} +1 \big)  \Big)\\
= \sin\big(t&-\theta(t)\big) + \lambda(t) \Big(\,\big(1 - \cos\big(\theta(t) - t \big)\,\big)  \dot{\theta}(t) + 1 - \cos\big(t-\theta(t)\big)\Big) \\
= \sin\big(t&-\theta(t)\big) + \lambda(t) \big(1 - \cos\big(\theta(t) - t \big)\,\big) \big( \dot{\theta}(t) + 1\big)
\end{align}
Finally, if I haven't mad too many mistakes, we arrive at the equation
$$ \sin\big(t-\theta\big) + \lambda \big(1 - \cos\big(\theta - t \big)\,\big) \big( \dot{\theta} + 1\big) = 0$$ which we can solve for $\lambda$ and obtain 
$$\lambda = \frac{\sin\big(\theta - t\big)}{\big(1 - \cos\big(\theta - t \big)\,\big) \big( \dot{\theta} + 1\big)}$$ The function $\theta$
 is defined by the implicit function theorem for $g(t,\theta) = 0$ and $$\dot{\theta} = -\frac{ \partial_t\, g(t,\theta)}{\partial_{\theta}\,g(t, \theta)}$$ or more explicitly
$$\lambda = \frac{\sin\big(\theta - t\big) \, \partial_{\theta} \, g(t, \theta) }{\big(1 - \cos(\theta - t)\,\big) \big(\partial_{\theta} \,g(t, \theta) -  \partial_{t}\, g(t, \theta)\big)}$$
Finally, 
$$z = e^{it} + \frac{\sin\big(\theta - t\big) \, \partial_{\theta} \, g(t, \theta) }{\big(1 - \cos(\theta - t)\,\big) \big(\partial_{\theta} \,g(t, \theta) -  \partial_{t}\, g(t, \theta)\big)} \, \Big(e^{i\theta} - e^{it}\Big) $$
$$g(t,\theta) = 0$$
I do not guarantee that the calculations are correct.
There is another way, more in the spirit of implicit equations, but again there are calculations...
