Proof that a tangent is perpendicular to the radius

Proof

We intersect the line with the circle and solve for the parameter.

1. Let $v$ be a direction vector of the line $\ell$. Since $T \in \ell$, we have $\ell = \set{T + tv : t \in \mathbb{R}}$, and distinct values of $t$ give distinct points because $v \neq 0$.
2. The point $X = T + tv$ lies on the circle $C$ if and only if $\norm{T - O + tv}^2 = r^2$.
3. Expanding with the dot product and using $\norm{T - O} = r$, this is $r^2 + 2t\, v \cdot (T - O) + t^2\norm{v}^2 = r^2$.
4. So $X \in C$ if and only if $t\,(2\, v \cdot (T - O) + t\norm{v}^2) = 0$, that is, $t = 0$ or $t = t_1$ with $t_1 = -\frac{2\, v \cdot (T - O)}{\norm{v}^2}$.
5. By definition, $\ell$ is tangent to $C$ at $T$ if and only if $\ell \cap C = \set{T}$. By steps 1 and 4, this holds if and only if $t_1 = 0$, that is, $v \cdot (T - O) = 0$.
6. Each point of $\ell$ is $X = T + tv$, so $(X - T) \cdot (T - O) = t\, v \cdot (T - O)$. This is zero for all $X \in \ell$ if and only if $v \cdot (T - O) = 0$, taking $t = 1$ for one direction.
7. By steps 5 and 6, $\ell$ is tangent to $C$ at $T$ if and only if $(X - T) \cdot (T - O) = 0$ for each point $X$ of $\ell$.