Dot product with a constant 
What is $\vec{a}\cdot (\vec{b}+c)$, where $c$ is a constant? 

Am I able to just take out the constant, or do I multiply the constant by the vector as a scalar?
EDIT (for context): I was finding the work done on a particle moving on the curve $C$: $x=cos(t)$, $y=sin(t)$, from $2\pi \leq t \leq 0$, through the vector field $\vec{F}(x,y)= -y\hat{i} +x\hat{j}$. I found that the work done is $-2\pi$. I wanted to figure out how that work changes when that same circle is moved horizontally by $n$ units, so that $x=cos(t)+n$, where $n$ is a constant. Here’s my work which led me to this issue: 
$$\oint_{C} \vec{F} \cdot d\vec{r}$$
$$\oint_{2\pi}^{0} (-sin(t)\hat{i} +(cos(t) +n)\hat{j}) \cdot (-sin(t)\hat{i} +cos(t)\hat{j}) dt$$
But I didn’t know how to cross $(cos(t)+n)$ with $(cos(t)$
 A: For an expression to be defined, all of its constituting parts need to be defined as well.
The sum of a vector and a constant $\vec b + c$ is not defined, and hence the expression $\vec a \cdot (\vec b + c)$ is not defined.
This applies to any scalar, not just constants, because $c$ can depend on another variable.

Although it probably makes sense to you that
$$\vec a \cdot (\vec b + c) = \vec a \cdot \vec b + c \cdot \vec a$$
neither side of the equation is defined, which means that you cannot use distribution the way it was used above.
A: This product has no meaning since the sum in the parentheses is undefined; you can not add a vector and a scalar. 
$c$ must be a vector, then the sum is defined and the given product is a dot product. 
A: Others have already addressed your question about adding a scalar to a vector (and then taking the dot product of that with another vector), but not the underlying issue that led you to attempt that ill-defined (unless you work with fancy geometric algebras) operation.
Simply put, you say that you want to translate the object's path by $n$ units along the horizontal axis.  Thus, the constant you need to add to the position of the object is not the scalar n, but the vector $n\hat j$ (where $\hat j$ is the horizontal unit vector, as it apparently is in your question).  Once you fix that, everything should work just fine.
In particular, the dot product distributes over the sum of vectors, so that $\vec a \cdot (\vec b + \vec c) = \vec a \cdot \vec b + \vec a \cdot \vec c$.
A: This is perfectly fine.  Grassman/Clifford/Geometric algebra interpretations have a perfectly sane explanation for these objects.  For any number of dimensions, expand the vector into its components, where $e_1$ and $e_2$ are mutually perpendicular unit vectors.  Dot product with parallel is 1 (ie: cos).  Dot product with perpendicular is 0 (ie: cos)
$$
(\vec a) \cdot  (\vec b + c) = (a_1 e_1 + a_2 e_2) \cdot (b_1 e_1 + b_2 e_2 + c) \\
= a_1 e_1 \cdot (b_1 e_1 + b_2 e_2 + c) + a_2 e_2 \cdot (b_1 e_1 + b_2 e_2 + c) \\
= a_1 b_1 + c a_1 e_1 + a_2 b_2 + c a_2 e_2 \\
= (a_1 b_1 + a_2 b_2) + c (a_1 e_1 + a_2 e_2) \\
= \vec a \cdot \vec b + c \vec a
$$
This is the same as previous answer, but worked out with an explanation of why.  Notice that this is a scalar plus a vector answer.  The question also had a scalar plus a vector.  When you add and multiply these, you can end up with something that ends up being real, or imaginary, or something else with a straightforward geometric meaning.  You can use this to find things like vector plus rotation in plane $e_1 e_2$ (ie: an imaginary plane we can call $I$):
$$
s = (a_1 b_1 + a_2 b_2) \\
(s + c (a_1 e_1 + a_2 e_2)) e_1 e_2 \\
s e_1 e_2 + ( c a_1 e_2 - c a_2 e_1 ) \\
s I + c( -a_2 e_1 + a_1 e_2 )
$$
