= Solution
A singular two-dimensional <affine toric variety> in the normal fan construction must come from a full-dimensional pointed cone; cones of smaller dimension give smooth charts with a torus factor. Let its primitive boundary generators be $v_0,v_{t+1}$. Choose an oriented <lattice basis> with $v_0=(1,0)$, and write the other generator as $(a,r)$, $r>0$. Primitivity gives $\gcd(a,r)=1$. Changing the second basis vector by a multiple of the first lets us write the cone as
$$
\sigma=\mathbb R_{\geq0}(1,0)+\mathbb R_{\geq0}(-q,r),\qquad 0<q<r,\quad\gcd(q,r)=1.
$$
Here $r>1$ because determinant one would give a smooth chart. The sublattice $N'=\mathbb Z(1,0)+\mathbb Z(-q,r)$ has index $r$, and its affine chart is $\mathbb C^2$. The quotient $N/N'$ is cyclic, generated by $(0,1)$, whose coordinates in the two boundary generators are $(q/r,1/r)$. Thus the kernel of the corresponding finite map of tori acts on the two affine coordinates as $(x,y)\mapsto(\zeta^qx,\zeta y)$. More explicitly, invariant monomials are exactly those $x^iy^j$ with $qi+j\equiv0\pmod r$, which are the elements of the original dual lattice in the first quadrant. Their ring is $\mathbb C[\sigma^\vee\cap M]$. Hence \b[the singularity is the cyclic quotient]
$$
\boxed{U_\sigma\cong\mathbb C^2/\mu_r= X_{r,q}.}
$$
The choice of the two boundary axes is unordered; exchanging them replaces $q$ by its inverse modulo $r$.
The <Hirzebruch–Jung resolution> is computed by the negative Euclidean algorithm. Put $r_0=r$, $r_1=q$ and recursively set
$$
b_i=\left\lceil\frac{r_{i-1}}{r_i}\right\rceil,\qquad r_{i+1}=b_ir_i-r_{i-1}.
$$
The nonzero remainders strictly decrease, $0\le r_{i+1}<r_i$, and their greatest common divisor stays one; the process terminates with $r_t=1$, $r_{t+1}=0$. All $b_i\ge2$, and unwinding the recursion gives the <negative continued fraction>
$$
\frac rq=[b_1,\ldots,b_t]^-:=b_1-\frac1{b_2-\displaystyle\frac1{\cdots-1/b_t}}.
$$
Start with $v_0=(1,0)$ and $v_1=(0,1)$ and define $v_{i+1}=b_iv_i-v_{i-1}$. The determinant of every consecutive pair is one, by induction. Moreover
$$
\det(v_i,(-q,r))=r_i,
$$
again by the same recursion and the initial values $r,q$. Each intermediate ray is therefore inside $\sigma$. To check the ordering explicitly, the first coordinates after $v_1$ are negative, the second coordinates positive, and successive determinants are positive. At the last step the determinant with the endpoint is zero. The vector $v_{t+1}$ is primitive, points in the same direction as $(-q,r)$, and hence equals it. Subdivide by all the intermediate rays $v_1,\ldots,v_t$. Every new cone is regular, and the subdivision has the same support, so it gives a proper smooth resolution.
Each internal ray has a compact invariant curve $E_i\cong\mathbb P^1$, because the quotient fan at it has the two opposite primitive rays. The relation $v_{i-1}+v_{i+1}=b_iv_i$ gives its self-intersection $E_i^2=-b_i$. This can also be checked directly: on the chart for $(v_{i-1},v_i)$ use coordinates $(x,y)$ with $E_i=\{y=0\}$; on the next chart the coordinates are $(u,v)=(x^{b_i}y,x^{-1})$, with $E_i=\{u=0\}$. The normal-coordinate transition is that of $\mathcal O_{\mathbb P^1}(-b_i)$, proving the self-intersection. Neighboring exceptional curves meet transversely once and nonneighbors do not meet.
Since $b_i\ge2$, there is no exceptional minus-one curve. Combinatorially, deleting $v_i$ gives a cone of determinant $b_i>1$. Deleting a longer consecutive block cannot give a regular cone either: determinants of successively more distant rays satisfy the same recursion with coefficients at least two and grow from $0,1$ to integers at least two. Thus no proper smooth coarsening of this subdivision exists. This proves the minimality of the <minimal resolution of a cyclic quotient surface singularity>; equivalently its exceptional chain has no curve that can be blown down to a smooth point. \b[The number of exceptional curves is the length $t$ of the negative continued fraction.]
If $t=1$, then $r/q=b_1$ is an integer, so coprimality gives $q=1$. Conversely $q=1$ gives the single coefficient $[r]^-$ and one curve. If $t=2$, then
$$
\frac rq=b_1-\frac1{b_2}=\frac{b_1b_2-1}{b_2}.
$$
The numerator and denominator on the right are coprime, so $r=b_1b_2-1$ and $q=b_2\ge2$. Therefore $q>1$ and $q\mid r+1$. Conversely, if these conditions hold, then $b_1=(r+1)/q\ge2$, $b_2=q\ge2$ give exactly this two-term expansion. Hence \b[the requested equivalences are]
$$
\boxed{t=1\iff q=1,\qquad t=2\iff q>1\text{ and }q\mid r+1.}
$$
Now consider a <complete fan> in dimension two with three rays and at most one singularity. Its three two-dimensional cones are the consecutive ray pairs. At least two cones are regular, and they share a ray. After an oriented lattice change, take their consecutive generators to be $v_1=(1,0)$, $v_2=(0,1)$ and $v_3=(-1,b)$: the first cone is a basis, and $\det(v_2,v_3)=1$ forces the first coordinate $-1$. Completeness forces $b<0$, because the third pair must span the missing sector with positive determinant $\det(v_3,v_1)=-b$. Write $b=-r$, $r\ge1$. The unique positive primitive relation is $v_1+rv_2+v_3=0$, so the fan is that of the <weighted projective plane> $\mathbb P(1,r,1)$, or $\mathbb P(1,1,r)$ after permuting weights. For $r=1$ this is $\mathbb P^2$; for $r>1$ its remaining cone has type $X_{r,1}$. This proves \b[the full three-ray classification]
$$
\boxed{X_\Sigma\cong\mathbb P(1,1,r),\quad r\ge1.}
$$
These are the <complete three-ray toric surfaces with at most one singular point>, including the smooth case.
For four rays and exactly the specified singularity, place the three regular cones consecutively. Normalize the middle one to have generators $(1,0),(0,1)$. Its neighboring regular cones then force the cyclic ray list
$$
v_1=(a,-1),\quad v_2=(1,0),\quad v_3=(0,1),\quad v_4=(-1,b).
$$
The remaining cone has determinant $\det(v_4,v_1)=1-ab=r$, so $ab=1-r$. Thus $a,b$ have opposite signs, and $s=|a|$ is a positive divisor of $r-1$. In the oriented lattice basis $v_4,(0,-1)$,
$$
v_1=-a\,v_4+r(0,-1),
$$
so the singular cone has quotient parameter $a\bmod r$. It follows that its parameter is congruent to $\pm s$. If identification with the specified singularity exchanges the two boundary rays, the parameter is inverted; this does not change the criterion, since for $st=r-1$ we have $s^{-1}\equiv-t\pmod r$, and $t$ is the complementary positive factor. The same identity with a minus sign handles $(-s)^{-1}$.
Conversely suppose $s\mid r-1$ and $s\equiv\varepsilon q\pmod r$, $\varepsilon\in\{1,-1\}$. Choose $a=\varepsilon s$ and $b=-\varepsilon(r-1)/s$ in the displayed four rays. They are primitive and cyclically ordered, with all four gaps less than a half-plane; the opposite signs of $a,b$ ensure completeness. Three consecutive determinants are one, and the fourth is $r$. Its quotient parameter is $a\equiv q\pmod r$, so the only singular point has the required type. We have proved the <four-ray completion criterion for a cyclic quotient surface singularity>:
$$
\boxed{\exists\text{ such complete four-ray surface}\iff\exists s>0:\ s\mid r-1,\ s\equiv\pm q\pmod r.}
$$
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