The tangent at any point of a circle is perpendicular to the radius through the point of contact.
[ JKBOSE – 2017, 15, 13 ]
Given: A circle with centre O and a tangent AB at a point P on the circle.
To Prove: $$\quad O P \perp A B$$
Construction: Take any point Q , other than P on the tangent AB and join OQ .
Proof: Here Q is a point on tangent AB other than point of contact P. Then, Q lies outside the circle. Let OQ intersect the circle at R.
$$
\begin{array}{ll}
& O P=O R \\
\text { Now, } & O Q=O R+R Q \\
\Rightarrow & O Q>O R \\
\Rightarrow & O Q>O P \\
\text { or } & O P<O Q
\end{array}
$$
[Radii of circle]
Thus, OP is the shortest distance from the center to any point on tangent $$A B$$.
Hence$$\quad O P \perp A B$$
