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 point P on circle.

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

Construction: Take another point Q on AB and join OQ.

Proof: Since Q is a point on AB (other than P)

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

Let OQ intersect the circle at R,

Then,                \[OR<OQ\]                                            ?(i)

But                   \[OP=OR\] (radii of circle)                        ?(iii)

\[\therefore OP<OQ\]                              (from (i) and (ii))

Thus, OP is shorter than any other line segment joining O to any point on AB.

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

\[\therefore \,\,\,OP\bot AB\]                                             Hence Proved.