All circles are straight lines. What have I done wrong? I was solving this problem: Find all plane curves where all tangent lines intersect in a single point.
So, I reasoned that, the tangent vector must be parallel to $\mathbf p - \mathbf r(s)$, where $\mathbf p = (p_x, p_y)$ is the point of intersection and $\mathbf r(s) = (x(s), y(s))$ is the curve itself, arc length parameterized. That is:
$$
T(s)\quad||\quad (\mathbf p - \mathbf r(s)) \quad\implies\quad
T(s) = H(s) (\mathbf p - \mathbf r(s)), \quad\mbox{ for some function $H$}.
$$
Solving it gives:
$$
\mathbf r'(s) = H(s) (\mathbf p - \mathbf r(s)) \quad\implies\quad
\mathbf r(s) = \mathbf p_0 - \mathbf A\exp\int H(s)ds
$$
Hereby, dividing both vector components:
$$
\frac{p_y - y(s)}{p_x - x(s)} = \frac{A_y \exp(...)}{A_x \exp(...)} = \frac{A_y}{A_x} \quad\implies\quad
y = ax + b
$$
It gives a straight line. As it should. No matter what choice of $H(s)$ is made. 

But then, I tried the following way:
$$
\mathbf r'(s) = H(s) (\mathbf p - \mathbf r(s)) \quad\implies\quad
1 = \mathbf r'(s)^2 = H(s)^2 (\mathbf p - \mathbf r(s))^2
$$
That is:
$$
(\mathbf p - \mathbf r(s))^2 = \frac{1}{H(s)^2}  \quad\implies\quad
(p_x - x(s))^2 + (p_y - y(s))^2 = \frac{1}{H(s)^2}
$$
That is, a circle, with a "radius" of $1/H(s)$. Since $H$ is a proportion between tangent vector and distance from the curve to the point, I can simply find all curves such that $H(s) = 1/R$ constant. And this gives a true circle, in contradiction with the previous result.
What have I done wrong?
 A: You have shown that if $\mathbf{r}'(s)=H(s)(\mathbf{p}-\mathbf{r}(s))$ for all $s$, then $$(p_x - x(s))^2 + (p_y - y(s))^2 = \frac{1}{H(s)^2}$$ for all $s$.  In particular, if $H(s)$ is a constant function, then $\mathbf{r}(s)$ lies on a circle for all $s$.  However, you have not proved the converse: you have not proved that if $\mathbf{r}(s)$ is any curve whose image is contained in a circle, then it actually satisfies $\mathbf{r}'(s)=H(s)(\mathbf{p}-\mathbf{r}(s))$.  So you have not proved that an ordinary parametrization of a circle satisfies $\mathbf{r}'(s)=H(s)(\mathbf{p}-\mathbf{r}(s))$.
In fact, the equation $\mathbf{r}'(s)=H(s)(\mathbf{p}-\mathbf{r}(s))$ has no solution if $H(s)$ is a constant (and $\mathbf{r}(s)$ is parametrized by arc length).  Indeed, by your first argument, $\mathbf{r}(s)$ must be a straight line, but $H(s)$ will never be constant for a straight line parametrized by arc length.  So your entire second argument is vacuous: you are reasoning about properties a certain function must have, but no such function actually exists.
(Note that in your first argument, you found a solution to the equation for any $H(s)$.  However, this solution will not always be parametrized by arc length, as you are assuming in your second argument.  If you have a straight line which is parametrized by arc length, then $H(s)$ must be changing since the norm of $\mathbf{p}-\mathbf{r}(s)$ is changing but $\mathbf{r}'(s)$ is supposed to always have norm $1$.)
