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 T at a point M of the circle.

To prove: \[OM\bot T\]

Construction: Take a point P, other than M on T. Join OP.

Proof: P is a point on the tangent T, other than the point of contact M.

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

Let OP intersect the circle at N.

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

[\[\because \] a part is less than whole]

But                   \[OM=ON\]                                           ?(ii)

[Radii of the same circle]

\[\therefore OM<OF\]                                          [Using (ii)]

Thus, OM is shorter than any other line segment joining O to any point T, other than M.

But a shortest distance between a point and a. line is the perpendicular distance.

\[\therefore OM\bot T\]

Hence, OM is perpendicular on T.                         Hence Proved.