Solution:

Let $k$ be the number of triangles. To have both a triangle line and a circle line, we need $3\le k\le6$.

If $k=3$, the three triangles must form a line. A row or column works, because the remaining six circles contain a full circle row or column. A diagonal does not work, because its complement contains no full line of circles. Thus there are $6$ configurations for $k=3$. By symmetry, there are also $6$ configurations for $k=6$.

If $k=4$, choose the triangle line and then the extra triangle. If the line is a row or column, all $6$ possible extra positions work, because one parallel row or column remains all circles. This gives $6\cdot6=36$ configurations. If the triangle line is a diagonal, no extra position leaves a full circle line. By symmetry, $k=5$ also gives $36$ configurations.

The total number of configurations is $6+36+36+6=84$.

Thus, the correct answer is D.