So I am trying to get a parametric expression of form $x = r_c \cdot \cos(\gamma), y = r_c \cdot \sin(\gamma)$ that represents the curve formed by the single intersection points of the two circles as seen in the figure.
So essentially when the two links with length $l_2$ are collinear. The main thing here is that $\vec{r_c }$ will vary in magnitude depending on the angle $\gamma$. I'm kind of stuck figuring out an expression for the magnitude of $\vec{r_c}$ in function of known parameters. I should add that the distance between the joints $M_1$ and $M_2$ is given as $W$ and these are fixed. Lengths of the lower links (directly connected to the fixed joints $M_1,M_2$) are equal to $l1$ and the upper links have a length $l_2$. Both of these are given and fixed. The joints that are located at the center of the circles are not fixed. Gamma is the independent variable for the parametric expression of the curve and the expression for $||\vec{r_c}||$.
The website isn't allowing me to comment but here's what I've done. I have shifted the graph to make $M_1$ the origin. If the angle between link 1 and the x-axis is $\alpha$. $O_1$ has coordinates $(l_1 cos(\alpha), l_1 sin(\alpha))$. Drawing a circle of radius $2l_2$ from $O_1$ and another of radius $l_1$ from $M_2$ gives two intersection points on a line $$2l_1 ysin(\alpha)=2xw-2l_1 xcos(\alpha)+2l_1^2-w^2-l_2^2$$ Finding the exact points give a really complex equation, But let $$R=\sqrt {l_1^2-2l_1 wcos(\alpha)+w^2}$$ $$O_2 = \frac{1}{2}(l_1cos(\alpha)+w,l_1sin(\alpha))+\frac{l_2^2-l_1^2}{2R^2}(w-l_1cos(\alpha),-l_1sin(\alpha)) \pm \frac{1}{2} \sqrt{2\frac{l_2^2+l_1^2}{R^2}-\frac{(l_2^2-l_1^2)^2}{R^4}-1}\cdot(-l_1sin(\alpha),l_1cos(\alpha)-w)$$ Then the midpoints can be found from there. Not really worth it in my opinion. Note that there are two midpoints so the curve you find depends on the one you choose.