How many independent quadrics should one intersect to get the canonical curve. Let $C$ be a non hyperelliptic complex algebraic curve of genus $g$, then the vector space $I_2(C)$ of quadrics containing the canonical image of $C$ is $\binom{g+1}{2}-h^0(2\omega_C) = (g-2)(g-3)/2$ dimensional. Moreover, if $g > 4$ then $C$ is the intersection of the nulls of all these quadrics (see ACGH VI.4.1 for a proof, I'm not sure how "classical" this is, or who originally proved it).

Question (edited following a comment from David Speyer) what is the least
  $d$ so that  if $V\subset I_2(C)$ is
  any $d$ dimensional vector space, and $X$ is the intersection of the
  nulls of the quadrics in $V$, then the
  only irreducible component of $X$
  which linearly spans $|\omega_C|^*$ is
  the canonical image of $C$ ?

I don't even know the generic bound, or indeed what is the bound for hyperelliptic curves (in which case the canonical curve is a rational normal curve). 
 A: Petri's theorem (see ACGH III.3) states that if $C\subset P^{g-1}$ is a canonical curve (i.e. the canonical image of a non hyperelliptic curve) of genus $g\ge 4$ then the ideal of $C$ is generated by quadrics  iff $C$ is not trigonal or a  plane quintic (for trigonal curves or plane quintics the intersection of all quadrics through $C$ is a surface). 
If $C$ is  hyperelliptic,  the dimension of the space of quadrics through the canonical image $\Gamma$ is $(g+1)g/2-2g+1=(g-1)(g-2)/2$. The ideal of $\Gamma$ is also generated by quadrics. 
If $g-2$ is not a power of 2,  a trivial lower bound is $d\ge g-1$. Indeed by Bezout's theorem (see Fulton, Intersection theory, 8.4) if the intersection of  $g-2$ quadrics of $P^{g-1}$ is  a curve of degree $d$, then $d$ divides $2^{g-2}$.
The following $2g-5$ quadrics cut the rational normal curve in $P^{g-1}$ set theoretically: $x_0x_2-x_1^2,x_0x_3-x_1x_2, \dots, x_0x_{g-1}-x_{g-2}x_1, x_2^2-x_1x_3, x_3^2-x_2x_4,\dots,x_{g-2}^2-x_{g-3}x_{g-1}$. 
I don't know if one can do better.
A: I guess I am not getting something, as it seems elementary the maximum number of quadrics needed is usually $g$.  I.e. given a canonical curve $C$ in $P^{g-1}$ which can be cut out by quadrics, it seems any general choice of $g-2$ quadrics containing it cuts out a union of curves including $C$.  Then any general quadric containing $C$ cuts out $C$ and a finite set of points on the other curves.  Then another general quadric through $C$ omits those points.  Is this nonsense? I see now the question in the title is no longer the same as the edited question.
edit:
David, do you really want the property in your question or do you just want to know when every $d$ dimensional subspace of $I_2(C)$ determines the canonical curve somehow?  i.e. a Torelli result.
Here is an example suggesting d may be large, a plane sextic, re embedded canonically in $P^9$ via plane cubics.  The image is a del Pezzo surface $S$ of degree $9$, on which any one quadric cuts out the canonical curve, unless the quadric contains the del Pezzo.  But the $55$ dimensional space of quadrics in $P^9$ cuts out the $28$ diml space of  plane sextics, hence a $27$ diml space of quadrics contains the del Pezzo. Since the ideal $I_2$ has dimension $28$, we actually need the whole space $I_2$ to get the curve, or to get any set with the curve as a component.
Is this right?  If so, plane curves of other degrees may be problematic as well....The situation seems to improve as the degree goes up.  A plane septic seems to lie canonically on an embedded copy of $P^2$ that is contained in only $75$ independent quadrics among the $78$ containing the curve itself, so d seems to equal at least $76$, maybe $77$ since it seems to need two more quadrics this time.  For a plane octic $d$ seems to be at least $166$, out of a space $I_2$ of dim = $171$.  well we're gaining on it, but somewhat slowly.
A: Amplifying on of Speyer's comments, if p is a point on a secant line of C, then the quadrics vanishing on C and p are of codimension one in the space of all quadrics vanishing on C.  Such a quadric vanishes at 3 points of the secant line ( p and the two points of C defining the line as a secant) and hence vanished on L.  Am I doing something silly?
