geometryterminology
Is there a mathematical name for a 2-dimensional shape in the general form of a dumbbell? That is two circular nodes connected by a center beam such as shown in this image from this answer. It could also be thought of as a circle that has been pinched together. I believe topology does not help us here since there are no holes, and thus such as shape is the same as a circle, but perhaps geometry gives us a name for it?
I don't believe such a shape has a single-word name like "sphere" or "cube" associated to it. However, in mathematics we can characterize such a shape as "the wedge of two spheres" and write it symbolically as $$S^2\vee S^2$$ $S^2$ denotes the "$2$-sphere" (Wikipedia link). Note that in mathematics this refers specifically to the "hollow" sphere; if you meant in your question to refer to a "filled-in" sphere, then the correct mathematical word is "$3$-ball" (Wikipedia link) and you would write $B^3\vee B^3$ instead.
The $\vee$ in the middle is the "wedge sum" operation (Wikipedia link). It takes two "shapes" (i.e., topological spaces) and glues them together at a single point. But of course, taking two spheres and attaching them at a single point produces the same shape as starting with one sphere and pinching it in the middle.
It's hard to tell from the picture, but from the description "The shape itself consists of lines connecting the circle to line segment", I might represent the shape thusly:
Suppose that the circle has radius $r$, and that the shape has height $h$. If the surface lines joining the segment to the circle lie in a plane perpendicular to that segment, then the figure looks like this:
Then, we can parameterize $P_\theta$ on the circle, and companion point $Q_\theta$ on the segment ...
$$P_\theta = (r \cos\theta, r \sin\theta, h) \qquad Q_\theta = ( r\cos\theta, 0, 0)$$
... so that the line between them has equation
$$P_\theta + t \; (Q_\theta - P_\theta) : \begin{cases} x = r \cos\theta \\ y = r (1-t) \sin\theta \\ z = h (1- t)\end{cases}$$
Eliminating the parameters $\theta$ and $t$ yields this formula for the surface:
$$x^2 z^2 + y^2 h^2 = r^2 z^2 \tag{1}$$
Note that the equation (as well as the figure) indicates that the level curves of the surface are ellipses, with constant major radius ($r$) and linearly-varying minor radius ($z r/h$).
Now, the volume and surface area of the shape can be determined from this function:
$$y = f(x,z) = \frac{z}{h} \sqrt{r^2-x^2} \quad \tag{2}$$ over the rectangular region determined by $-r \leq x \leq r$ and $0\leq z \leq h$.
Edit. OP has indicated, in comments with @YvesDaoust, that the desired shape is the conoid, which indeed fits the surface described by $(1)$. (The parameterization given in the Wikipedia article matches $(1)$, for $r = 1$ and $h = z_0$, and under the coordinate transformation $z \to z_0-z$.)
The Wikipedia entry states that the volume of the conoid is $\frac{\pi}{2}r^2 h$. This is easily confirmed from $(2)$ ...
$$V = \frac{2}{h} \int_{0}^{h} z \int_{-r}^{r} \sqrt{r^2-x^2}\,dx dz = \frac{2}{h} \int_{0}^{h} \frac{\pi}{2} r^2 z dz = \frac{2}{h}\cdot \frac{\pi}{4} r^2 h^2 = \frac{\pi}{2}r^2 h \tag{3}$$
(where we have recognized the $x$ integral as giving the area of the half-circle of radius $r$).
For surface area, we note that $$f_x = \frac{-xz}{h \sqrt{r^2-x^2}} \qquad\qquad f_z = \frac{1}{h}\sqrt{r^2-x^2}$$ so that $$\begin{align} S &= 2\;\int_{0}^{h} \int_{-r}^{r} \sqrt{(f_x)^2 + (f_z)^2 + 1\;}\; dx dz \\ &= \frac{2}{h}\;\int_{0}^{h} \int_{-r}^{r} \;\sqrt{\frac{ h^2 (r^2-x^2) + (r^2-x^2)^2 + x^2 z^2}{r^2 - x^2}\;} \; dx dz \end{align}$$
This is a bit trickier to evaluate symbolically. I'll have to return to it.
Best Answer
The shape reminds of a Cassini oval (http://mathworld.wolfram.com/CassiniOvals.html) but without a more precise equation, one cannot be more specific.