Let $M$ be the intersection of the segments $OT$ and $AB$. Then $|OT| = |OM| + |MT|$. Write $|AB| = |AT| = |BT| = a$. We know that $AB \perp OT$. By Pythagoras' theorem we have $|OM| = \sqrt{r^2 - |BM|^2} = \sqrt{r^2 - \frac{a^2}{4}}$ and $|MT| = \sqrt{a^2 - \frac{a^2}{4}} = \frac{\sqrt{3}a}{2}$. We can now find an upper bound for $|OT|$ by using the inequality between the arithmetic and the quadratic mean (the A-G inequality):
$$\begin{aligned} |OT| &= |OM| + |MT| \\ &= \sqrt{r^2 - \frac{a^2}{4}} + \frac{\sqrt{3}a}{2} \\ &= \sqrt{r^2 - \frac{a^2}{4}} + \frac{\sqrt{3}a}{6} + \frac{\sqrt{3}a}{6} + \frac{\sqrt{3}a}{6} \\ &\le 4\sqrt{\frac{(r^2 - \frac{a^2}{4}) + \frac{a^2}{12} + \frac{a^2}{12} + \frac{a^2}{12}}{4}} \\ &= \sqrt{4r^2} = 2r. \end{aligned}$$
We have shown that $|OT| \le 2r$, so now we must only check if the equality can hold. The equality in the A-G inequality holds when all four terms are the same, i.e. when $\sqrt{r^2 - \frac{a^2}{4}} = \frac{\sqrt{3}a}{6} \iff a = r\sqrt{3}$. Since $a = |AB|$ can reach any value from the interval $(0, 2r]$, the equality holds exactly when $TA$ and $TB$ are tangents to the circle $\mathcal{K}$.