This Question has a nice mathematical approach. For closed bounded convex shapes the Hyperplane Separation Theorem gives a practical method for checking that two figures do not intersect.
Of the specific cases asked about here, an affine transformation will reduce the ellipse case to one for a circle, and similarly the rectangle case to one for a square. For simplicity let's detail just those two cases.
For two convex polygons it suffices to check the edges of both polygons as candidates for weakly separating lines, i.e. weak in the sense that a convex polygon obviously intersects the line containing one of its edges but is otherwise entirely on one side of that line. If the other convex polygon lies strictly on the opposite side of that line from the first polygon, then they never intersect.
The case involving a circle is not quite so simple to describe since it doesn't have a finite number of edges, but the procedure to check is still easy.
Test for a Triangle Not Intersecting a Circle
The write-up Circle-Triangle Intersection Method aims to optimize programming by avoiding square roots and normals, but we will use them for discussion. [WARNING: My notation/labelling differs from that article.] We assume a non-degenerate triangle, i.e. the 3 vertices are not colinear.
Let $cx + sy = b$ be the normal form of equation for a line containing one edge of the triangle $AC$, i.e. using real coefficients satisfying $c^2 + s^2 = 1$. Then $f(x,y) = cx + sy - b$ represents a signed distance of a point $(x,y)$ to line $\overline{AC}$. Let $(u,v)$ be the center of the circle with radius $r$, and let $(s,t)$ be the third vertex $B$ of the triangle (the one not on that line).
Check for an edge $AC$ of the triangle where the circle's center $O = (u,v)$ is on the opposite side from the third vertex $B = (s,t)$:
$$f(u,v) \cdot f(s,t) \lt 0$$
If no such triangle edge exists, then the test FAILS: the circle's center is inside or on the triangle.
If such a triangle edge is found, then compare the perpendicular distance $p = |f(u,v)|$ from the circle's center to line $\overline{AC}$ to the circle's radius $r$. If $p \gt r$, then the test SUCCEEDS: the circle and triangle do not intersect.
Now check the distances from center $O = (u,v)$ to the triangle vertices $A,C$ which are endpoints of that edge:
$$d_A = \text{dist}(O,A)$$
$$d_C = \text{dist}(O,C)$$
If either of these is less than or equal to radius $r$, then the test FAILS: a vertex is inside the circle.
Finally we compare two projected distances along line $\overline{AC}$ with the length of edge $AC$, namely:
$$k_A = \sqrt{d_A^2 - p^2} \gt \text{dist}(A,C) $$
$$k_C = \sqrt{d_C^2 - p^2} \gt \text{dist}(A,C) $$
If either of these inequalities is true, the test SUCCEEDS: the circle and the triangle do not intersect. Otherwise the test FAILS: they do intersect along edge $AC$.
Test for a Triangle Not Intersecting a Square
As above let a line containing one edge of the triangle have normal equation $cx + sy = b$, so that $f(x,y) = cx + sy - b$ represents a signed distance from point $(x,y)$ to that line, and let $(s,t)$ be the third vertex of the triangle (not on the line).
Let the four vertices of the square be $(x_i,y_i)$ for $i=1,2,3,4$. Vertex $(x_i,y_i)$ lies strictly on the opposite side of the line from $(s,t)$ if and only if $f(x_i,y_i) \cdot f(s,t) \lt 0$.
If all four vertices of the square lie strictly on the opposite side of the line from $(s,t)$, then the test succeeds: the square and the triangle do not intersect. Otherwise continue with testing the remaining edges of the triangle.
If there are no further edges to test, perform a similar check with each edge of the square. If all three vertices of the triangle lie across an edge of the square from the rest of the square, then the test SUCCEEDS: the triangle and square do not intersect.
Otherwise if all edges of the square have been tested without success, the test FAILS: the triangle and square do intersect.
Tests of this type generalize to convex polygons with any number of sides, and also to convex figures in higher dimensions. Tests on nonconvex polygons are often handed by subdividing them into convex subpolygons, e.g. triangulation by ear clipping.
