# Maximum connected components after removing 2 vertices

Let a 3 regular, 2 connected (but not 3 connected) , bipartite, planar graph. After removing 2 vertices, how many connected components can there exist at most? I have created an instance with 3 (below). Is it the maximum? Is it the only case where this happens?

If we remove, the 2 red vertices, all the blue edges disappear and the graph has 3 connected components.

If $$G - \{u,v\}$$ has components $$G_1, G_2, \dots, G_k$$, then each $$G_i$$ must have an edge to both $$u$$ and $$v$$ somewhere: if $$G_i$$ did not have an edge to $$u$$, for example, then merely deleting $$v$$ would make $$G_i$$ disconnected from the rest of $$G-\{v\}$$, so $$G$$ would not be $$2$$-connected.
This requires both $$u$$ and $$v$$ to have degree at least $$k$$. Conversely, if $$\deg u = \deg v = 3$$, then $$G - \{u,v\}$$ can have at most $$3$$ components.
Your example shows that $$3$$ is possible, so $$3$$ is the maximum. For a more specific example, consider the graph below:
Ignoring the rectangles and the colors on the edges, this is a $$3$$-regular planar graph; the vertex colors show that it is bipartite. The colors on the edges give an open ear decomposition that proves it is $$2$$-connected: take the red cycle, then the blue ear, then the orange ear, and then the dashed edges (as one-edge ears) in any order. Finally, deleting the two vertices in the black rectangle leaves three connected components: the ones in the red, blue, and orange squares.