Class 10 Exam  >  Class 10 Notes  >  Mathematics (Maths) Class 10  >  Proof of Theorems: Circles

Proof of Theorems: Circles | Mathematics (Maths) Class 10 PDF Download

Theorem 1: Tangent at any point of a circle is perpendicular to the radius through the point of contact.


Proof of Theorems: Circles | Mathematics (Maths) Class 10

According to the theorem, O is the centre and OP⊥XY.
Proof: Let Q be a point on XY.
Construction: Connect OQ.
Suppose it touches the circle at R.
Hence,
OQ > OR
OQ > OP (as OP = OR, radius).
The same will be the case for all other points on the circle.
Hence, OP is the smallest line that connects XY.

Proof of Theorems: Circles | Mathematics (Maths) Class 10

Thus, OP is the smallest line that connects XY, and the smallest line is perpendicular.
∴ OP ⊥ XY

Note:
1. By the theorem above, we can also conclude that at any point on a circle, there can be one and only one tangent.
2. The line containing the radius through the point of contact is also sometimes called the ‘normal’ to the circle at the point.

Theorem 2: The lengths of tangents drawn from an external point to a circle are equal.

Proof of Theorems: Circles | Mathematics (Maths) Class 10

Given:
Let the circle be with center O, and P be a point outside the circle.
PQ and PR are two tangents to the circle intersecting at points Q and R respectively.

Proof of Theorems: Circles | Mathematics (Maths) Class 10

To prove:
Lengths of tangents are equal, i.e., PQ = PR.
Construction:
Join OQ, OR, and OP.
Proof:
As PQ is a tangent,
OQ ⊥ PQ
(Tangent at any point of a circle is perpendicular to the radius through the point of contact).
So, ∠OQP = 90°
Hence, ΔOQP is a right triangle.
Similarly,
PR is a tangent,
& OR ⊥ PR
(Tangent at any point of a circle is perpendicular to the radius through the point of contact).
So, ∠ORP = 90°
Hence, ΔORP is a right triangle.
Using Pythagoras theorem:
(Hypotenuse)² = (Height)² + (Base)²

Proof of Theorems: Circles | Mathematics (Maths) Class 10

In right-angled triangle ΔOQP:
OP² = PQ² + OQ²
OP² − OQ² = PQ²
PQ² = OP² − OQ² …(1)
In right-angled triangle ΔORP:
OP² = PR² + OR²
OP² = PR² + OQ² (As OQ = OR, both are radii)
OP² − OQ² = PR²
PR² = OP² − OQ² …(2)

Note:

1. The theorem can also be proved by using the Pythagoras Theorem as follows:
PA2 = OP2 – OA2 = OP2 – OB2 = PB2 (As OA = OB) which gives PA = PB.
2. Note also that ∠ OPA = ∠ OPB. Therefore, OP is the angle bisector of ∠ APB,i.e., the centre lies on the bisector of the angle between the two tangents.

The document Proof of Theorems: Circles | Mathematics (Maths) Class 10 is a part of the Class 10 Course Mathematics (Maths) Class 10.
All you need of Class 10 at this link: Class 10
127 videos|683 docs|84 tests
Related Searches

Previous Year Questions with Solutions

,

past year papers

,

ppt

,

Free

,

Summary

,

Semester Notes

,

practice quizzes

,

shortcuts and tricks

,

Exam

,

Proof of Theorems: Circles | Mathematics (Maths) Class 10

,

Extra Questions

,

Important questions

,

MCQs

,

study material

,

Viva Questions

,

Proof of Theorems: Circles | Mathematics (Maths) Class 10

,

Sample Paper

,

mock tests for examination

,

video lectures

,

pdf

,

Proof of Theorems: Circles | Mathematics (Maths) Class 10

,

Objective type Questions

;