Question

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

Answer 1

Solution!

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

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.

Hence Proved!

Thank You!

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Answer 2

Step-by-step explanation:

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