Why can a fixed point be written as a linear combination of curve if all normals pass through the fixed point? In one of my tutorials, we got the following question: "Assume that all normals of a parametrized cuve pass through a fixed point. Prove that the trace of the curve is contained in the circle" (Do Carmo, differential geometry of curves and surfaces). I was stuck on this, and got the hint that the fixed point part implies that, for every $s$, there exists a $r(s)$ such that the fixed point, denoted by $P$, can be written as a linear combination of $\alpha(s)$ and $\alpha ''(s)$, i.e. $P= \alpha(s) + r(s)\alpha''(s)$. From this point, I have been able to solve the question (by differentiating etceta). However, why is the latter statement true? Is there a geometrical or algebraic explanation why this holds?

If the curve is parametrized by arclencht, the normal vector is parallel to the normal vector. Therefore, for every $s$, the equation of the normal line to the curve passing through $\alpha(s)$ will be $$\beta(r)=\alpha(s)+r\alpha''(s), \space \space r \in \mathbb{R}.$$

All of these lines passes through $P_0$. That means that, for each line, there will be a value of $r$, let's say $r_0$ that will satisfy $$P_0=\alpha(s)+r_0\alpha''(s). \space \space \space (1)$$

Of course, this value of $r_0$ will depend of the normal line we are considering, and, therefore, will depend of the parameter $s$. We can rewrite $(1)$ to reinforce this dependence: $$P_0=\alpha(s)+r_0(s)\alpha''(s).$$

And that's just your equation.