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.
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.