A line is tangent to a circle if and only if the line is perpendicular to a radius of the circle.

In the diagram above, $m$ is tangent to the circle $Q.$ This can be proven using the Perpendicular Postulate.

According to the Perpendicular Postulate, there is only one segment from $Q$ that is perpendicular to $\overleftrightarrow{m}.$ It will be shown, using contradiction, that $\overline{QP}$ must be this segment.
Suppose that $\overline{QP}$ is not perpendicular to $\overleftrightarrow{m}.$ It follows then that there exists another segment from $Q$ perpendicular to $\overleftrightarrow{m}.$ Call this segment $\overline{QT}.$

Because $\overline{QT}\perp \overleftrightarrow{m},$ a right angle is created at their intersection. As a result, $\overline{QP},$ which measures $r$ units, is the hypotenuse — and the longest side — in the right triangle $△QTP.$ Therefore, the length of $\overline{QT}$ is shorter than $\overline{QP}.$ Notice that $\overline{QT}$ contains a radius of circle $Q.$ Because part of the segment lies outside the circle, the unknown portion can be assigned another variable, $b.$ By the Segment Addition Postulate, $QT=r+b.$ The relationship between $QT$ and $QP$ can be written as an inequality. By subtracting $r$ on both sides, the inequality states that $b<0.$ Since a length cannot be less than $0,$ the assumption that $QP$ is not perpendicular to $\overleftrightarrow{m}$ must be false. Therfore, $QP$ is perpendicular to the tangent $m.$
This can be summarized in the following two-column proof.