If I have two parametric curves one defined as x(t) and y(t) and another x(s)+c and y(s).
My assumption is that if I set t = s = value. I can find the two slopes at these values. i.e. dy/dx = x'(t)y'(t) but what I am after is a tangent between the two curves and this will depend on the x-axis shift, in this case c, which does not factor in the derivatives.
So my question is can the common tangent be determined? If so does it depend on setting a ratio between s and t that somehow depends on c?
Take the circle $(r_1\frac{1-t^2}{1+t^2},r_1\frac{2t}{1+t^2})$ centered at the origin and the circle $(h+r_2\frac{1-t^2}{1+t^2},k+r_2\frac{2t}{1+t^2}),$ centered at $(h,k)$ then by looking at the dual curves ($(\frac{-y'}{xy'-x'y},\frac{x'}{xy'-x'y})$) $(-\frac1{r_1}\frac{1-t^2}{1+t^2},-\frac1{r_1}\frac{2t}{1+t^2})$ and $(\frac{(t^2-1)}{((r_2-h)t^2+2kt+r_2+h)} ,-\frac{(2t)}{((r_2-h)t^2+2kt+r_2+h)})$ we get intersection points
$(-(k\sqrt{h^2+k^2-(r_2-r_1)^2}-hr_2+hr_1)/((k^2+h^2)r_1),(h\sqrt{h^2+k^2-(r_2-r_1)^2}+kr_2-kr_1)/((k^2+h^2)r_1)),((k\sqrt{h^2+k^2-(r_2-r_1)^2}+hr_2-hr_1)/((k^2+h^2)r_1),-(h\sqrt{h^2+k^2-(r_2-r_1)^2}-kr_2+kr_1)/((k^2+h^2)r_1), (-(k\sqrt{h^2+k^2-(r_2+r_1)^2}+hr_2+hr_1)/((k^2+h^2)r_1), (h\sqrt{h^2+k^2-(r_2+r_1)^2}-kr_2-kr_1)/((k^2+h^2)r_1)), ((k\sqrt{h^2+k^2-(r_2+r_1)^2}-hr_2-hr_1)/((k^2+h^2)r_1), -(h\sqrt{h^2+k^2-(r_2+r_1)^2}+kr_2+kr_1)/((k^2+h^2)r_1))$
corresponding to common tangents $-(k\sqrt{h^2+k^2-(r_2-r_1)^2}-hr_2+hr_1)/((k^2+h^2)r_1) x+(h\sqrt{h^2+k^2-(r_2-r_1)^2}+kr_2-kr_1)/((k^2+h^2)r_1) y+1=0,(k\sqrt{h^2+k^2-(r_2-r_1)^2}+hr_2-hr_1)/((k^2+h^2)r_1) x-(h\sqrt{h^2+k^2-(r_2-r_1)^2}-kr_2+kr_1)/((k^2+h^2)r_1) y +1=0, -(k\sqrt{h^2+k^2-(r_2+r_1)^2}+hr_2+hr_1)/((k^2+h^2)r_1) x+ (h\sqrt{h^2+k^2-(r_2+r_1)^2}-kr_2-kr_1)/((k^2+h^2)r_1) y+1=0, (k\sqrt{h^2+k^2-(r_2+r_1)^2}-hr_2-hr_1)/((k^2+h^2)r_1) x -(h\sqrt{h^2+k^2-(r_2+r_1)^2}+kr_2+kr_1)/((k^2+h^2)r_1) y+1=0.$