[proofplan] The proof has two halves: existence and uniqueness. Existence is a linear algebra count. The space of degree-$2$ homogeneous polynomials in three variables has $k$-dimension $\binom{2+2}{2} = 6$. Imposing vanishing at a single point is a single linear condition on the coefficient vector, so imposing vanishing at five points cuts down a subspace of dimension $\geq 6 - 5 = 1$, giving a non-zero $F$ and hence a conic $V(F)$ through all five points. Uniqueness uses [Curves of Degree at Most 2 Meet in at Most 4 Points](/theorems/2164): if two distinct conics $C \neq D$ both passed through $p_1, \ldots, p_5$, then $C, D$ share at least $5$ points; for the bound $|C \cap D| \leq 4$ to be violated, $C$ and $D$ must share an irreducible component, and we use the no-three-collinear hypothesis to derive a contradiction. Smoothness follows from irreducibility plus [Irreducible Conics Are Smooth](/theorems/2163), and irreducibility itself follows from no-three-collinear. [/proofplan] [step:Set up the linear system: vanishing of a degree-$2$ form is a linear condition on its coefficients] Let $V_2 := k[X_0, X_1, X_2]_2$ denote the $k$-vector space of homogeneous polynomials of degree $2$ in three variables. A basis is given by the $\binom{2+2}{2} = 6$ monomials \begin{align*} \{X_0^2,\ X_1^2,\ X_2^2,\ X_0 X_1,\ X_0 X_2,\ X_1 X_2\}. \end{align*} Hence $\dim_k V_2 = 6$. Every $F \in V_2$ has a unique expansion $F = \sum_{i+j+l = 2} a_{i,j,l} X_0^i X_1^j X_2^l$, identifying $V_2 \cong k^6$ via the coefficients $(a_{i,j,l})$. For each point $q \in \mathbb{P}^2_k$, choose a representative $\tilde q \in k^3 \setminus \{0\}$ with $q = [\tilde q]$. The evaluation map \begin{align*} \operatorname{ev}_q: V_2 &\to k \\ F &\mapsto F(\tilde q) \end{align*} is a $k$-linear functional on $V_2$. (It depends on the choice of representative $\tilde q$ only up to a non-zero scalar, since rescaling $\tilde q \mapsto \lambda \tilde q$ multiplies $F(\tilde q)$ by $\lambda^2$. Different scalings give the same kernel, so the condition $F(\tilde q) = 0$ is well-defined as a condition on $q \in \mathbb{P}^2_k$.) The kernel of $\operatorname{ev}_q$ is a hyperplane in $V_2$ — i.e.\ a subspace of codimension $0$ or $1$. We claim the codimension is exactly $1$, equivalently $\operatorname{ev}_q$ is non-zero. Indeed, the polynomial $F(X_0, X_1, X_2) = (\xi_0 X_0 + \xi_1 X_1 + \xi_2 X_2)^2 \in V_2$, where $(\xi_0, \xi_1, \xi_2)$ is any vector with $\xi_0 \tilde q_0 + \xi_1 \tilde q_1 + \xi_2 \tilde q_2 \neq 0$ (such vectors exist because $\tilde q \neq 0$), satisfies $F(\tilde q) = (\xi_0 \tilde q_0 + \xi_1 \tilde q_1 + \xi_2 \tilde q_2)^2 \neq 0$. So $\operatorname{ev}_q \neq 0$ as a linear functional, and $\ker(\operatorname{ev}_q) \subset V_2$ has codimension exactly $1$. [/step] [step:Existence of the conic via dimension counting] The simultaneous vanishing of $F$ at $p_1, \ldots, p_5$ is the intersection of five hyperplanes in $V_2$: \begin{align*} W := \bigcap_{j=1}^{5} \ker(\operatorname{ev}_{p_j}) \subset V_2. \end{align*} Each $\ker(\operatorname{ev}_{p_j})$ has codimension $1$. By the standard codimension inequality for intersections of subspaces, \begin{align*} \operatorname{codim}_{V_2} W \leq \sum_{j=1}^{5} \operatorname{codim}_{V_2} \ker(\operatorname{ev}_{p_j}) = 5, \end{align*} so \begin{align*} \dim_k W \geq \dim_k V_2 - 5 = 6 - 5 = 1. \end{align*} In particular $W$ contains a non-zero element $F \in V_2 \setminus \{0\}$. Then $C := V(F) \subset \mathbb{P}^2_k$ is a non-empty closed subset (non-empty because by [Hilbert's Nullstellensatz](/theorems/2124) the zero locus of a non-zero proper ideal is non-empty over an algebraically closed field, and $V(F)$ is non-empty since $F$ is a non-constant homogeneous form of degree $\geq 1$ in $\geq 2$ variables; alternatively, the construction below directly exhibits points), and $p_j \in C$ for each $j = 1, \ldots, 5$ by construction. We must check $\deg F = 2$, i.e.\ $F$ is genuinely a conic (a degree-$2$ form) and not a lower-degree form by accident. Since $F \in V_2$ and $V_2$ consists of pure degree-$2$ homogeneous polynomials, every non-zero $F \in V_2$ has degree exactly $2$ (the $0$ polynomial is excluded). Hence $C = V(F)$ is a conic through $p_1, \ldots, p_5$. [/step] [step:Show that any conic through the five points is irreducible, using the no-three-collinear hypothesis] Let $F \in V_2 \setminus \{0\}$ be any conic through $p_1, \ldots, p_5$, and set $C = V(F)$. We claim $F$ is irreducible. Suppose for contradiction $F$ is reducible. Since $\deg F = 2$, $F$ factors as $F = L_1 L_2$ with $L_1, L_2 \in V_1 = k[X_0, X_1, X_2]_1$ linear forms (possibly equal — the "double line" case $L_1 = L_2 = L$, where $V(F) = V(L)$ is a line). Set $\ell_i := V(L_i) \subset \mathbb{P}^2_k$, lines in the projective plane. Then $C = \ell_1 \cup \ell_2$ as point sets. Each $p_j \in C$ lies in $\ell_1 \cup \ell_2$, so each $p_j$ lies in $\ell_1$ or in $\ell_2$. By the pigeonhole principle, at least one of $\ell_1, \ell_2$ contains at least $\lceil 5/2 \rceil = 3$ of the points $p_1, \ldots, p_5$. Say $\ell_1$ contains three of them, denoted $p_{j_1}, p_{j_2}, p_{j_3}$ (a subset of size $3$ from $\{p_1, \ldots, p_5\}$). But then $p_{j_1}, p_{j_2}, p_{j_3}$ are three collinear points (all lying on the line $\ell_1$), contradicting the hypothesis that no three of $p_1, \ldots, p_5$ are collinear. Hence $F$ is irreducible. [guided] The pigeonhole step handles the double-line case automatically: if $L_1 = L_2 = L$, then $C = \ell$ is a single line, and all five points $p_1, \ldots, p_5$ lie on $\ell$ — in particular, the three points $p_1, p_2, p_3$ are collinear, again contradicting the hypothesis. The hypothesis "no three collinear" is essential. If three of the points were collinear on some line $\ell$, then $\ell \cup \ell'$ for any line $\ell'$ through the remaining two points would be a reducible conic through all five — and there might be many such reducible conics, destroying uniqueness as well as smoothness. The hypothesis "five distinct points" is also used implicitly: if some $p_j = p_{j'}$, then we have only four distinct conditions, and dimension counting gives only $\dim W \geq 6 - 4 = 2$, allowing a $2$-parameter family of conics — uniqueness fails. [/guided] [/step] [step:Conclude smoothness from irreducibility] By Step 3, any conic $C = V(F)$ through $p_1, \ldots, p_5$ is irreducible. By [Irreducible Conics Are Smooth](/theorems/2163), an irreducible conic over an algebraically closed field of characteristic $\neq 2$ is smooth. Both hypotheses of the cited theorem are satisfied: $\operatorname{char} k \neq 2$ by our standing assumption, and $F$ is irreducible by Step 3. Hence $C$ is smooth. [/step] [step:Uniqueness via the bound on intersection of two conics] Suppose for contradiction that $C = V(F)$ and $D = V(G)$ are two distinct conics, each passing through all of $p_1, \ldots, p_5$. By Step 3, both $F$ and $G$ are irreducible. We claim $C$ and $D$ share no common irreducible component. Suppose they did. The only irreducible component of $C$ (as a variety, or equivalently as a curve) is $C$ itself, since $C$ is irreducible. Similarly the only component of $D$ is $D$. So a shared component would force $C = D$ as varieties, hence $V(F) = V(G)$. By [Hilbert's Nullstellensatz](/theorems/2124) applied to the irreducible varieties $V(F) = V(G)$, the radical ideals $\sqrt{(F)} = I(V(F)) = I(V(G)) = \sqrt{(G)}$. Since $F$ is irreducible, $(F)$ is a prime ideal in the polynomial ring (over a UFD, irreducible elements generate prime ideals), hence $(F)$ is its own radical. Similarly $(G) = \sqrt{(G)}$. So $(F) = (G)$, meaning $G = \lambda F$ for some $\lambda \in k^*$, hence $C = V(F) = V(\lambda F) = V(G) = D$ — but we assumed $C \neq D$. Contradiction. So $C$ and $D$ share no common component. The hypotheses of [Curves of Degree at Most 2 Meet in at Most 4 Points](/theorems/2164) are now met: $C, D$ are plane curves of degree $\leq 2$ sharing no common component, and $\operatorname{char} k \neq 2$. The cited theorem gives \begin{align*} |C \cap D| \leq 4. \end{align*} But $\{p_1, \ldots, p_5\} \subseteq C \cap D$ and the $p_j$ are five distinct points, so $|C \cap D| \geq 5 > 4$, a contradiction. Hence at most one conic passes through $p_1, \ldots, p_5$, establishing uniqueness. [guided] The uniqueness argument is the converse direction of dimension counting. Existence said: at least one conic passes through five points. Uniqueness says: at most one. Combined, the conic is unique. The mechanism is intersection-theoretic: two distinct conics meet in at most four points (provided they share no component), so they cannot agree at five. The "no three collinear" hypothesis enters indirectly through Step 3 — without it, conics through the five points could be reducible, and then $C \cap D$ might contain a common line, sharing infinitely many points along it. The cited theorem [Curves of Degree at Most 2 Meet in at Most 4 Points](/theorems/2164) requires "no shared component" precisely to rule this out. A more refined statement: the space of conics through $p_1, \ldots, p_5$ in $\mathbb{P}^2_k$ corresponds to lines in $\mathbb{P}(V_2) \cong \mathbb{P}^5_k$ — a $0$-dimensional projective subspace, i.e.\ a single point. Existence shows the dimension is at least $0$ (the subspace is non-empty); uniqueness shows the dimension is at most $0$ (no $1$-parameter family). Together, exactly one conic. [/guided] [/step] [step:Combine existence, irreducibility, smoothness, and uniqueness] By Step 2, a non-zero $F \in V_2$ vanishing on $\{p_1, \ldots, p_5\}$ exists, giving a conic $C = V(F)$ through all five points. By Step 3, $F$ is irreducible (using "no three collinear"). By Step 4, $C$ is smooth (via [Irreducible Conics Are Smooth](/theorems/2163)). By Step 5, $C$ is the unique such conic. This completes the proof. [/step]