Option A should be the correct answer.
The matrix $A$ can be written as a product of two matrices, namely matrix $B$ and $B^{T}$, as follows:
$A = \begin{bmatrix} x_{1} & y_{1} \\ x_{2} & y_{2}\\ x_{3} & y_{3}\\ x_{4} & y_{4} \end{bmatrix} \times \begin{bmatrix} x_{1} & x_{2} & x_{3} & x_{4}\\ y_{1} & y_{2} & y_{3} & y_{4} \end{bmatrix}$
where $B = \begin{bmatrix} x_{1} & y_{1} \\ x_{2} & y_{2}\\ x_{3} & y_{3}\\ x_{4} & y_{4} \end{bmatrix}$.
Rank of matrix $B$ is the dimension of the vector space spanned by the columns of $B$.
$B$ has two columns.
If both of the columns vectors are independent then their linear combinations can represent every vector on the plane in which both of them are contained in and hence the span of the column vectors of $B$ would be a two dimensional plane in the four dimensional space and hence the rank of $B$ would be 2.
If both the column vectors of $B$ are linearly dependent then their linear combinations will only span the one dimensional line(on which both of the columns vectors lie) in the four dimesional space.Thus in this case rank would be 1.
It is given that not all x's & y's can be zero which implies $B$ can not be a zero matrix so rank of B can not be zero.
So rank of $B$ could be 1 or 2.
FACT: A matrix and its transpose have the same rank. (for proof see http://linear.ups.edu/html/section-PD.html)
So $B$ and $B^{T}$ will have the same rank.
Now, $A = BB^{T}$
FACT: Rank($A$) <= min{Rank($B$), Rank($B^{T}$)} ( for proof see 1.4 b) of http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-241j-dynamic-systems-and-control-spring-2011/assignments/MIT6_241JS11_assn02_sol.pdf )
So Rank($A$) <= 1 or 2
but all x's & y's can not be zero so $A$ is not a zero matrix for sure and hence rank of $A$ can not be zero.
So Rank of $A$ could be 1 or 2.