Question

Prove that the tangent at any point of the circle is perpendicular to the radius through the point of contact.

Answer 1

Hint:
A tangent AB is drawn to the circle touching it at P and now we can choose any other point Q on the the tangent and we can see that $OQ = OR + RQ$as OQ intersects the circle at R. and then we know that OR = OP as they are the radii of the circle and hence we get $OQ > OP$, from which we can say that OP is the shortest distance and by using the property shortest distance of a point from a given line is the perpendicular distance from that line , we get the proof.

Complete step by step solution:
We are given a circle with centre O and radius r

Let AB be the tangent to the circle touching the circle at point P
So the radius of the circle through this point is OP

So now we need to prove $OP \bot AB$
So now let's choose a point Q other than P on the tangent AB
Join the point Q with the centre O making a line segment OQ intersecting the circle at R

From the diagram we can see that
$ \Rightarrow OP = OR$ ……….(1)
as they are the radii of the circle
Now
$ \Rightarrow OQ = OR + RQ$
Using (1) we get
$ \Rightarrow OQ = OP + RQ$
From this we get to know that
$ \Rightarrow OQ > OP$
From this we can conclude that any line drawn from the centre to a point on the tangent is greater than OP
We know that , shortest distance of a point from a given line is the perpendicular distance from that line .
Here OP is the shortest distance
So we conclude that $OP \bot AB$
Hence proved.

Note:
1) The tangent always touches the circle at a single point.
2) It never intersects the circle at two points.
3) The length of tangents from an external point to a circle are equal.