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.