This is meant to be a slightly cleaned up version of Peter's answer above. If I'm not mistaken, it is not necessary to assume the characteristic of $k$ is zero. However, I don't know the theory of reductive groups in positive characteristic—even in characteristic 0, I've only dealt with semi-simple groups—so I may be making wrong assumptions about how the representation theory works.
Let $B\subseteq G$ be a borel with torus $T$ and unipotent radical $U$. Let $V=\bigoplus V_\lambda$ be the decomposition of $V$ into weight spaces with respect to $T$.
Lemma: Suppose $v_\lambda\in V_\lambda$, $v=\sum v_\lambda$ is a point of $C$, and $\mu\in T^\vee$ is extremal among weights in the support of $v$ (i.e. weights for which $v_\mu\neq 0$). Then $C$ contains $v_\mu \def\l{\langle}\def\r{\rangle}$.
Proof: Since $\mu$ is extremal among weights in the support of $v$, there is a 1-parameter subgroup $\alpha\in \hom(\mathbb G_m,T)\cong \hom(T^\vee,\mathbb Z)$ such that $\l\alpha,\mu\r > \l\alpha,\lambda\r$ for any $\lambda\neq \mu$ in the support of $v$. Consider the map $f\colon\mathbb A^1\to V$ given by $f(t)= \sum t^{\l\alpha,\mu\r-\l\alpha,\lambda\r}v_\lambda$. Away from $t=0$, we have that $f(t)=t^{\l\alpha,\mu\r}\alpha(t^{-1})\cdot v$ is in $C$. Since $C$ is closed, we get that $f(0)=v_\mu$ is in $C$. $\square$
Now suppose $\mu$ is an extremal highest (with respect to $B$) weight appearing in the decomposition of $V$. Since $C$ spans $V$, there is a point of $C$ with a component in $V_\mu$, so by the lemma, there is a non-zero vector $v\in V_\mu\cap C$. Since $\mu$ is a highest weight, we have that $U$ stabilizes $v$. Since the stabilizer of $v$ is reductive, it must contain the derived subgroup of $G$, so the irreducible representation with highest weight $\mu$ is a character, so $\mu$ pairs to zero with every coroot. It follows that every weight in the decomposition of $V$ pairs to zero with every coroot, so $V$ is a direct sum of characters.