With respect to the dirac delta function, what does $\delta [x + 2A, y - A]$
A (two-dimensional) finite-impulse-response filter has an impulse response $g(x,y)$ 
that is nonzero only for $(x,y)$ in a region of finite area (continuous parameters) 
of for a finite number of values of $x$ and $y$ (discrete parameter). For example,
$g(x,y)$ might be nonzero only for points inside a circle of radius $r$.  Specifically,
consider
$$g(x,y) = \begin{cases}1, &amp;x^2 + y^2 \leq r^2,\\0, &amp;\text{otherwise}\end{cases}$$
describes an image filter whose response to a point source (impulse function)
at the origin is a disc of radius $r$ in the filtered image.  For input $f(x,y) = \delta (x + 2A, y - A) - \delta (x - A, y + A)$, the filter output would be the sum of a positive
disc centered at $(-2A,A)$ and a negative disc centered at $(A,-A)$ with cancellation
occurring where the discs overlapped. (Disc ovrlap would be determined by the
relationship between $A$ and $r$). More formally,
$$g(x,y)\circledast f(x,y) = \begin{cases}
1, &amp;(x+2A)^2 + (y-A)^2 \leq r^2, (x-A)^2+(y+A)^2 &gt; r^2,\\
-1, &amp;(x+2A)^2 + (y-A)^2 &gt; r^2, (x-A)^2+(y+A)^2 \leq r^2,\\
0, &amp;\text{otherwise.}\end{cases}$$
where we have combined the outer darkness of $0$ with the $0$ produced
by the cancellation of the discs in the region of overlap (if any).

Similar calculations can be done for the discrete case as well.