= Solution
Cut the second circle factor at $p,q$. This divides the <torus> into two annuli whose boundary circles are exactly the two collapsed circles. Collapsing each <annulus>'s two boundaries separately produces a two-sphere; the two spheres then share their two poles. In particular, the two collapsed circles give two distinct vertices, not a single common collapsed point.
For an explicit <cellular homology> calculation, each suspended circle contributes one edge from the first vertex to the second and one two-cell attached with zero cellular boundary. Thus $C_0\cong\mathbb Z^2$, $C_1\cong\mathbb Z^2$, $C_2\cong\mathbb Z^2$, with no higher cells. Both edges have boundary $v_q-v_p$, so $\partial_1$ has rank one and kernel generated by the difference of the edges. Each two-cell's attaching loop traverses its edge in opposite directions, so $\partial_2=0$. Consequently the <homology of a torus with two parallel circles collapsed> is
$$
\boxed{H_0\cong\mathbb Z,\quad H_1\cong\mathbb Z,\quad
H_2\cong\mathbb Z^2,\quad H_k=0\ (k\geq3).}
$$
Collapsing either connecting edge as a maximal tree also exhibits the <homotopy> type $S^1\vee S^2\vee S^2$.
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