Conic hull of the set $S = \{(x_1,x_2) : (x_1 - 1)^2 + x_2^2 = 1 \}$

Question

Show that the conic hull of the set $$S = \left\{(x_1,x_2) : (x_1 - 1)^2 + x_2^2 = 1 \right\}$$ is the set $$\{(x_1,x_2) : x_1 > 0\} \;\cup \; \{(0,0)\}$$

The set $$S = \{(x_1,x_2) : (x_1 - 1)^2 + x_2^2 = 1 \}$$ is a circle centered around $(1,0)$ with radius $1$ and its convex hull should be the filled circle $$\mbox{conic}(S) = \{(x_1,x_2) : (x_1 - 1)^2 + x_2^2 \leq 1 \}$$

How to prove the statement and can someone tell me how the conic hull of a closed and bounded set is an unbounded set ?

The key is to have a definition of "conic hull", evidently not to be confused with the convex hull. See this closely related Question. — hardmath, Oct 27 '21 at 17:59
Hint: given a point $p$ in space, find a point $q$ on the circle such that $p = kq$ for $k \ge 0$. — DanielV, Oct 27 '21 at 17:52

score 1 · Answer 1 · answered Oct 27 '21 at 17:59

If $x\in \mbox{cone} A$ then $tx \in \mbox{cone} A$ for all $t\geq 0.$ Hence any half line $l$ that has a origin in $(0,0)$ and intersects the set $A$ have to be contained in $A.$ In other words the $\mbox{cone} A$ is equal to the union of all half lines $l$ which has origin in $(0,0)$ and intersects the set $A.$ So you have to draw all half lines which has origin at $(0,0)$ and meets the set $A.$

Conic hull of the set $S = \{(x_1,x_2) : (x_1 - 1)^2 + x_2^2 = 1 \}$

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