Given:
AB is a tangent to the circle with centre O. P is the point of contact. OP is the radius of the circle.
To prove:
OP ⊥ AB
Proof:
Let Q be any point (other than P) on the tangent AB.
Then Q lies outside the circle.
OQ > r
OQ > OP For any point Q on the tangent other than P.
OP is the shortest distance between the point O and the line AB.
OP ⊥ AB
(The shortest line segment drawn from a point to a given line, is perpendicular to the line)
Thus, the theorem is proved.
From the above theorem,
1. The perpendicular drawn from the centre to the tangent of a circle passes through the point of contact.
2. If OP is a radius of a circle with centre O, a perpendicular drawn on OP at P, is the tangent to the circle at P.
