A tangent is perpendicular to the radius

Statement

Let $C$ be a circle with the centre $O$ and the radius $r > 0$ in the Euclidean plane $\mathbb{R}^2$. Let $T$ be a point of $C$, and let $\ell$ be a line through $T$. Then $\ell$ is tangent to $C$ at $T$ if and only if $\ell$ is perpendicular to the radius $OT$. That is, $(X - T) \cdot (T - O) = 0$ for each point $X$ of $\ell$.

Remarks

A proof: for $X = T + t v$ on $\ell$, we have $\|X - O\|^2 = r^2 + 2t\, v \cdot (T - O) + t^2 \|v\|^2$. So $t = 0$ is the only point of $\ell$ on $C$ if and only if $v \cdot (T - O) = 0$. Thus each point of $C$ has exactly one tangent line. The same holds for a sphere and its tangent plane in $\mathbb{R}^3$.

Explanations

L1 human

To draw the tangent to a circle at a point, first draw the radius to that point. The tangent is the line through the point at a right angle to the radius. For the unit circle at $(1, 0)$, the radius is horizontal and the tangent is the vertical line $x = 1$.