Performance of Modified Dijkstra's algorithm with Binariy heap as Priority Queue
In the example, we can assume the start cell $s$ is the origin in a plane consisting of square cells with Descarte coordinates.  That is, $s$ is $(0,0)$. The coordinate of destination cell, $t$ is $(m,n)$. The distance of any two cells is given by the taxicab geometry. That is, the distance between cells $(p_1,p_2)$ and $(q_1,q_2)$ is
$$d((p_1,q_2),(q_1,q_2))=|p_1-p_2|+|q_1-q_1|$$
In particular, $d(s,t)=|m|+|n|$.

At the last step of running Dijkstra's algorithm with source cell $s$ when we visit cell $t$, all cells whose distance to $s$ is smaller than $d(s,t)$ must have been visited. Some of the cells whose distance to $s$ is $d(s,t)$ may have also been visited. If you color all cells that are no more than $d(s,t)$ away from $s$, you will get a diamond shape whose boundary cells $(p,q)$ are given by the following equations.

$$ \begin{align}
p+q = d(s,t)  &amp;\text{ where }p\ge0, q\ge0. \text{ This is the top right segment.}\\
p-q = d(s,t)  &amp;\text{ where }p\ge0, q\le0. \text{ This is the bottom right segment.}\\
-p+q = d(s,t) &amp;\text{ where }p\le0, q\ge0. \text{ This is the top left  segment.}\\
-p-q = d(s,t) &amp;\text{ where }p\le0, q\le0. \text{ This is the bottom left segment.}\\
\end{align}$$

That diameter shape is, in fact, a circular disk with center $(0,0)$ and radius $|m|+|n|$ in the taxicab geometry. Yes, as you pointed out, "radios" should be "radius". 

How many cells are there in the diamond shape? The diamond shape is actually a square the length of whose diagonals is $2(|m|+|n|)$. So its area is $(2(|m|+|n|))^2/2=2(|m|+|n|)^2$, which is about the number of cells in the it asymptotically. 



Exercise. The exact number of cells in a disk with radius $r$ in the taxicab geometry is $2r^2+2r+1$. In particular, there are 13 cells in a disk with radius 2.