question_answer

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

Answer:

Given, a circle with centre O and a tangent AB at a point M on circle.

To prove: \[OM\bot AB\].

Construction: Take point N (other than M) on AB. Join ON.

Proof: Since N is a point on the tangent AB other than P

\[\therefore \] N lies outside the circle.

Let ON passes through point P.

Then, \[OP<ON\]                                                                       ...(i)

But, \[OM=OP\]        (Radii)                                                        ...(ii)

\[\therefore \text{ }OM<ON\]         (From eq. (i) and (ii))

Thus, OM is the shortest distance between the point O and the line AB.

But, it is known that the shortest distance between a point and a line is the perpendicular distance

\[\therefore \text{ }OM\bot AB\].                                      Hence Proved.