How do I find equation of this curve? I need to find equation of the curve as shown below, for which, I need to find equation for upper part. lower part is half circle. upper part is a constant distance from circle with line passing through a point which is eccentric to the circle center as shown in the image. 
edit 1: I have updated picture with more informative drawings, in current picture, I need to figure out equation of the Green curve. 

The line $L$ has constant length, and passes through a fixed point $E$. It's lower end (point $P$) moves along the lower semi-circle. It's upper end (the point $Q$) traces out some curve. We need to find the equation of this curve.
 A: Let's work with a unit circle, centered at the origin, and suppose the "eccentric" point $E$ is at $(0,a)$, where $a<1$. Let $c$ be the length of the line $L = PQ$, where $c > 1+a$. 
A little vector reasoning shows that
$$
Q = P + \frac{c}{ \| E - P \| }(E - P)
$$
If the point $P$ is on the circle, then it can be described by coordinates $P = (\cos\theta, \sin\theta)$. If we substitute $P = (\cos\theta, \sin\theta)$ and $E = (0,a)$ into the equation above, and simplify, the $xy$ coordinates of $Q$ are given by
$$
x = \cos\theta  - \frac{c \cos\theta}{\sqrt{1 -2a\sin\theta + a^2}}  $$
$$
y = \sin\theta  + \frac{c(a - \sin\theta)}{\sqrt{1 -2a\sin\theta + a^2}}
$$
Here is a plot for the case $a=1/3$, $c=3/2$, for $\tfrac34\pi \le \theta \le \tfrac94\pi$:
 
Another one, this time showing the lines. Again with $a=1/3$, $c=3/2$, but now just the portion $\pi \le \theta \le 2\pi$:

The interesting thing is that the curve actually looks nothing like the pictures that you and I drew.
The situation is fairly simple, so this sort of mechanism (and the curve it traces out) might be well-known in the kinematics field, but I couldn't find any references.
A: A very simple equation in polar is coordinates is obtained if the center of the system of axis is not located at the center of circle, but at the "exentric" point :




