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I have an irreducible polynomial $f \in R[x,y,z]$, and a point $p$ that is in the zero-set $Z$ of $f$. My question is, given the following properties of $p$, is it necessarily a singular point of $f$. There is a set of (one dimensional) circles, all passing through $p$ and fully contained in $Z$. Moreover, all of these circles, except for a single circle $c$, are fully contained in the same sphere $\sigma$. In case that it matters, I am only interested in the case where all of the circles that are on $\sigma$ also have a second point $b$ in common. Intuitively, it seems to me that $a$ has to be singular because of the additional circle $c$. How can I verify or contradict this?

Perhaps one way to answer this is by answering the follow-up question: If the zero set (of an irreducible $f$) contains several circles on the same sphere, all passing through two common points $a,b$, must the sphere be contained in the zero-set? I think that the answer is positive if the number of circles is sufficiently large, at least proportional to the degree of $f$, but I would like to know the answer when the number of circles is at least some (large) constant.

I apologize in advance in case this question is trivial. My knowledge in algebraic geometry is somewhat limited, and in fact, this problem arose while studying a combinatorial problem. Many thanks.

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If the additional circle is not tangent to the sphere than the point has to be singular. – rita Aug 31 '11 at 22:33
Thanks. Indeed, this case is clear to me, but I wish to know if the point is always singular under the above conditions. That is, including when the additional circle has the same tangent plane as the sphere. Also, it seems that I have a counterexample for the first part of the follow-up question. That is, any constant number of circles can be fully contained in $f$ while the sphere that contains them isn't in $f$. But I don't have an example for the case where the additional circle is also in $f$. – Adam Sheffer Sep 4 '11 at 16:56

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