Let $R = \mathbb{R}[x,y]/(x^2+y^2-25)$ and $I$ the ideal of functions which vanish at the point $P = (3,4)$. I have proven that $I$ is generated by $(x-3,y-4)$ and that if $I= (f)$ for some $f \in R$, then $f$ cannot vanish at any point on the circle other than $P$ and $f$ cannot vanish to order $\geq 2$ at $P$.
Let $F \in \mathbb{R}[x,y]$ represent $f$. The curve representing the zero locus of $F(x,y)$, call it $\mathcal{C}$, is tangent or transversal to the circle $x^2+y^2=25$ at the point $P$. In the transversal case, $F$ would necessarily vanish at two points on the circle, implying that $f$ vanishes at two points on the circle, a contradiction. My professor told me that in the tangent case, $F$ would vanish at $P$ with order $\geq 2$, implying that $f$ would vanish at $P$ with order $\geq 2$, a contradiction. Thus, $I$ is not principal.
Could someone please explain why if $\mathcal{C}$ is tangent to the circle at $P$, then $F$ vanishes at $P$ with order $\geq 2$?
Note: We say that $f$ vanishes to order $\geq 2$ at a point $P$ if and only if $f \in I^2$. As pointed out in the comments below, I am assuming that $\mathcal{C}$ being tangent to the circle means $\mathcal{C}$ intersects the circle at exactly one point.