× About Us How it works Pricing Student Archive Subjects Blog Contact Us

Enrich your knowledge with our informative blogs

Explain the different circle theorems

A circle is the locus of all the equidistant points from a particular fixed point.

This fixed point is referred to as the centre of a circle.

The circle theorem includes the concepts of angles, sectors, tangents, circle chords, and proofs.

Let us learn about the circle theorems here only. 

Circle Theorems 

The different circle theorems include: 

__________________________________________________________________

Theorem 1

“Two equal chords of the circle subtend equal angles at the circle’s centre.”

__________________________________________________________________

 

Proof of the theorem: 

In ∆AOB and ∆COD,

AB=CD {because both have equal chords} ————- (1)

OA= OB= OC= OD {Circle Radii} —————- (2)

From the equation (1) and (2),

∆AOB ≅ ∆COD

Therefore by CPCT, we get

∠AOB = ∠COD

Hence Proved. 

__________________________________________________________________

The Converse of Theorem 1

“If two angles subtended at the centre by two chords are equal, the chords are also of equal lengths.”

__________________________________________________________________

 

[REFERRING TO THE SAME IMAGE USED IN THE FIRST THEOREM] 

In ∆AOB and ∆COD,

∠AOB = ∠COD (Equal angles subtended at the circle’s centre “O”) ———– (1)

OA= OB= OC= OD ———— (2)

From equations 1 and 2,

∆AOB ≅ ∆COD

Therefore, By CPCT, we get AB = PQ. 

__________________________________________________________________

THEOREM 2

“The Perpendicular to a chord bisects the chord if drawn from the centre of a circle.”

__________________________________________________________________

 

According to the theorem, in the figure OP ⊥ AB.

Therefore, AP = PB 

Proof of the theorem 

In ∆AOP and ∆BOP,

∠APO = ∠BPO = 90° (OP ⊥ AB) ———— (1)

OA = OB —— (2)

OP = OP (COMMON SIDES) ——— (3)

From the equations 1, 2, and 3,

AP = PB By CPCT

__________________________________________________________________

The converse of the Theorem 2

“A straight line passing through a circle’s centre to bisect the chord is perpendicular to the chord.”

__________________________________________________________________

 

REFERRING THE SAME IMAGE USED IN THEOREM 2 

In ∆AOP and ∆ BOP,

AP = PB (As OP bisects AB) ———- (1)

OA = OB (Radii of circle) ————- (2)

OP = OP (Common side) ————— (3)

From equations 1, 2, and 3,

∠APO = ∠BPO = 90° (By CPCT) 

__________________________________________________________________

Theorem 3

“Equal chords of a circle are equidistant from the circle’s center.”

__________________________________________________________________

Construction: Join OB and OD 

[REFERENCE IMAGE ONLY: Using EF instead of PQ] 

Proof of Theorem 

In ∆OEB and ∆OFD,

BE = ½ AB (Perpendicular to a chord bisects it) ——— (1)

DF = ½ CD ——————— (2)

Given, AB = CD

BE = DF (from equations 1 and 2)

OB = OD (Radii of the same circle)

∠OEB = ∠OFD = 90°

∆OEB ≅ ∆OFD

Hence, OE = OF (By CPCT) 

__________________________________________________________________

The converse of the 3rd Theorem

“Chords of a circle that are equidistant from the centre are equal in length.”

__________________________________________________________________

REFERRING THE SAME IMAGE USED IN THIRD THEOREM 

In ∆OEB and ∆OFD,

OE = OF —————– (1)

∠OEB = ∠OFD = 90° ———– (2)

OB = OD ———– (3)

From equations 1, 2 and 3

∆OEB ≅ ∆OFD

BE = FD (By CPCT)

½ AB = ½ CD

Hence, AB = CD 

__________________________________________________________________

Theorem 4

“Measures of angles subtended to any point on the circle’s circumference from the same arc are equal to half of the angle subtended at the centre by the same arc.” __________________________________________________________________

[REFERENCE IMAGE ONLY: Using Q instead of D] 

In ∆AOP,

∠AOB = 2∠APB

Construction required: Joining PQ passing through the “O” 

Poof of the theorem 

OA = OP ————— (1)

∠OAP = ∠OPA (Angles opposite to equal sides of triangle) ——————— (2)

∠AOQ = ∠OAP + ∠OPA ——————- (3)

Therefore, from equations 2 and 3

∠AOQ = 2∠OPA ———- (4)

In ∆BOP,

∠BOQ = 2∠OPB ——— (5)

∠AOB = ∠AOQ + ∠BOQ

From equations 4 and 5,

∠AOB = 2∠OPA + 2∠OPB

∠AOP = 2 (∠OPA + ∠OPB)

∠AOB = 2∠APB

Hence proved

__________________________________________________________________

Theorem 5

“The opposite angles in a cyclic quadrilateral are supplementary.”

__________________________________________________________________ 

 

[REFERENCE IMAGE ONLY: Using PQRS instead of ABCD] 

Proof of the theorem 

For arc PQR,

∠POR = 2 ∠PQR = 2α (THEOREM 4) ————— (1)

Considering the arc PSR,

Reflex ∠POR = 2 ∠PSR = 2β (THEOREM 4) —————- (2)

∠POR + Reflex ∠POR = 360°

From equations 1 and 2

2 ∠PQR + 2 ∠PSR = 360°

2α + 2β = 360°

α + β = 180° 

__________________________________________________________________

Theorem 6

“Tangent to a circle perpendicular to the radius of the circle at the point of contact”

__________________________________________________________________

 [REFERENCE IMAGE ONLY: Using AB instead of PQ]

Here in this diagram,

O is the centre of the circle.

OA ⊥ XY. 

__________________________________________________________________

Theorem 7

“Number of tangents can be drawn from a given point.”

__________________________________________________________________

 

(1) If the point is the circle’s interior region, any line through that particular point will be a secant. Therefore, no tangent can be drawn to a circle that passes through the point that lies inside it.

 

[REFERENCE IMAGE ONLY]

(2) When the tangency point lies on a circle, there is precisely one tangent to a circle, which passes through it.

[REFERENCE IMAGE ONLY]

(3) When the point is at the circle’s outside, there are precisely two tangents to the circle.

[REFERENCE IMAGE ONLY] 

__________________________________________________________________

Theorem 8

“The length of the tangents drawn from any external point is equal.”

__________________________________________________________________

[REFERENCE IMAGE ONLY]

Discover better learning pace and Flexibility
Discover better learning pace and Flexibility

Online tuition by TEL Gurus helps you learn at your own pace and garner the flexibility to choose your preferred timeslots.

Book A Demo Class

Tel Guru
Tel Guru

Register For The Demo Class

captcha telguru