# Circle Theorems

What are Circle Theorems?

Circle theorem helps understand the concepts of different elements of the circle, like sectors, tangents, angles, chord, and radius of the ring with proofs. A circle is the joining line of all the points that lie at an equal distance from a fixed focus point. This fixed point is in the middle point inside the circle. However, all the points on the circle are at equal distance, and hence this fixed point is known as the centre of the circle. The length between the circle centre point and any point that lies on the circle is known as the radius. The space occupied by the circle is its area, and the outer line of the circle is its circumference. The line that is perpendicular to the circle at any point on the circle is known as a tangent.

All Theorems Related To Circle

Now, let’s look into the circle theorems and circle theorems proof to define the relationships between different entities of the circle. Before getting into the theorems, let us discuss the chord as it will give a better understanding.

Know Chord Of A Circle

A chord is the line segment that connects two different points of the circle’s circumference. Also, the diameter is the most significant chord that transverses the centre of the circle.

Now, let us study different all theorem of circle class 9 related to the circle.

Circles Class 9 Theorems

Students from Class 9 come across the circle basics, and they will learn various theorems related to circle that helps to study the chord of the circle. Below are the topics that include in different circle class 9 theorems:

• Angle made by the chord of the circle at a point

• The line segment that is perpendicular to the chord to the centre.

• The distance of different chords from the circle’s centre and equal chords

• The angle created by the arch of the circle

All Theorems Of Circle Class 9

Theorem 1:

Chords, having equidistance

from the circle’s centre make equal angles at the circle’s centre.

Proof:

Consider ∆AOB and ∆POQ,

AB = PQ (Chords that are equal)   ……..(Equation 1)

OA = OB = OQ = OP (Radius of the circle)  …..(Equation 2)

From equation 1 and  equation 2, we can conclude;

∆AOB ≅ ∆POQ (Axiom of congruency SSS)

Hence, by Corresponding parts of the congruent triangles (CPCT), we will get;

∠AOB = ∠POQ

Hence, Proved

Theorem 1 Converse Rule:

When two angles sectioned at the circle’s centre that is made by two different chords are equal, those two chords are the same in length.

Proof:

Consider  ∆AOB and ∆POQ,

∠AOB = ∠POQ (Angles are equal given in the theorem statement)  …………( Equation 1)

OA = OB = OP = OQ (Radius of the same circle)

…………(Equation 2)

From equation 1 and equation 2, we can conclude;

∆AOB ≅ ∆POQ (Axiom of congruency SAS)

Hence,

AB = PQ  (By CPCT rule)

Theorem 2: Circle Geometry

When you draw the perpendicular line segment from the circle’s centre, it will bisect the chord, i.e., perpendicular will divide the chord into two equal parts. It is called a theorem 2 circle geometry.

In the above figure, OD ⊥ AB, as per the theorem, hence, AD = DB.

Proof:

Consider two triangles, ∆BOD and ∆AOD.

∠ADO = 90°, ∠BDO = 90° (AB ⊥OD) ………(Equation 1)

OB = OA (Radius of circle) ……….(Equation 2)

OD = OD (Common side) ………….(Equation 3)

From equation (1), equation (2) and equation (3);

∆AOB ≅ ∆POQ (Axiom of congruency R.H.S)

Hence, AD = DB (through CPCT rule)

Theorem 2 Circle Geometry: Converse Rule

A line segment that passes through the circle’s centre bisects the chord will be perpendicular to the chord.

Proof: Circle Geometry

Consider ∆BOD and ∆AOD,

DB = AD (OD is a bisector of AB) ……….(Equation 1)

OA = OB (Radius)  ……….( Equation 2)

OD = OD (Common side)  ………..( Equation 3)

From equation 1, equation 2 and equation 3;

∆AOB ≅ ∆POQ (Axiom of congruency By SSS)

Hence,

∠ADO = ∠BDO = 90° (By CPCT rule)

Theorem 3:

Equal chords of the given circle are equidistant i.e. at equal distance from the circle’s centre

Construction: Join OB and OD

Proof:

Consider ∆OQD and ∆OPB.

BP = 1/2 AB (Perpendicular bisects the chord)

…..(equation 1)

AB = CD  (given in the theorem statement)

DQ = 1/2 CD (Perpendicular bisects the chord)

…..(equation 2)

BP = DQ (from equation 1 and equation 2)

∠OQD= ∠OPB = 90° (OQ ⊥ CD and OP ⊥ AB)

∆OPB ≅ ∆OQD (Axiom of Congruency, R.H.S)

Hence,

OP = OQ ( By CPCT rule)

Theorem 3: Converse Rule

Two chords of the circle that are at equal distance from the circle’s center have the same length.

Proof:

Consider ∆OQD and ∆OPB.

OQ = OP ………….(equation 1)

∠OPB = ∠OQD = 90° ………..(equation 2)

OB = OD (Radius)  ……..(equation 3)

Hence, from equation 1, equation two and equation 3;

∆OPB ≅ ∆OQD (Axiom of Congruency By R.H.S)

BP = DQ (By CPCT rule)

1/2 AB = 1/2 CD (Perpendicular bisects the chord)

Hence,

AB = CD

Solved Example

Example 1

In the given figure of below circle, c is the centre of the

circle.

The circle’s diameter is BD.

A is the given point on the circle.

Find the angle CBA?

Diameter is given, BD

Angle BAD is confined within the circle (given).

To find the angle CBA,

CBA =180−(23−90) = 67°

CBA = 67°