Consider a line segment of length $d$ and an initial position $(0, 0)$, $(0, d)$ on a Cartesian coordinate system. Now consider an $\epsilon$ move such that the new position is $(\epsilon, 0)$, $(0, \sqrt{d^2-\epsilon^2})$ with $\epsilon \in [0, d]$.
Now, if we let $\epsilon$ vary continuously from $0$ to $d$ it seems like the segment movement is bounded by a curve which I am trying to find its equation. Any help is much appreciated!
As Paul mentions in a comment, it looks like you’re talking about what’s called the envelope of a family of curves, that is, a curve that is tangent to each member of the family at some point. In this case, the family of curves consists of the lines $\frac x\epsilon + \frac y{\sqrt{d^2-\epsilon^2}} = 1$, parameterized by $\epsilon$. The envelope satisfies the system $$F(\epsilon,x,y) = \frac x\epsilon + \frac y{\sqrt{d^2-\epsilon^2}} - 1 = 0 \\ \frac\partial{\partial\epsilon}F(\epsilon,x,y) = \frac x{\epsilon^2} - {\epsilon y\over\left(d^2-\epsilon^2\right)^{3/2}} = 0.$$
Eliminating $\epsilon$ in Mathematica produces the equation $$y^4 \left(3 x^2-3 d^2\right)+y^2 \left(3 d^4+21 d^2 x^2+3 x^4\right)+y^6=d^6-3 d^4 x^2+3 d^2 x^4-x^6.\tag{*}$$ The following plot shows the graph of this curve in the first quadrant for a few values of $d$:
The equation of this family of lines can be turned into a polynomial equation in $\epsilon$ by clearing denominators, rearranging and squaring: $$(\epsilon - x)^2(d^2-\epsilon^2)-y^2\epsilon^2 = 0,$$ which expands into $$\epsilon^2 - 2x\epsilon^3 + (x^2+y^2-d^2)\epsilon^2+2d^2x\epsilon-d^2x^2=0.$$ The definition of the envelope is equivalent to $t$ being a double root of this equation, so you can use the formula for the discriminant of a quartic to eliminate $\epsilon$. After removing an extraneous factor of $16x^2y^2$, you will be left with (*).