[Maths Class Notes] on Circle Theorems Pdf for Exam

The circle theorem helps understand the concepts of different elements of the circle, like sectors, tangents, angles, chords, 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 center of the circle. The length between the circle center 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 theorem 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

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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 center of the circle. 

Now, let us study different theorems 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 the 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 is perpendicular to the chord to the center.

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

• The angle created by the arch of the circle 

• Cyclic quadrilaterals

All Theorems of Circle Class 9

Theorem 1:

Chords, having equidistant from the circle’s center make equal angles at the circle’s center.

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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 congruence 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 center that are made by two different chords are equal, those two chords are the same in length. 

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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 congruence SAS)

Hence,

AB = PQ  (By CPCT rule)

Theorem 2: Circle Geometry

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

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

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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 congruence R.H.S)

Hence, AD = DB (through CPCT rule)

Theorem 2 Circle Geometry: Converse Rule

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

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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 congruence 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 center 

Construction: Join OB and OD

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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)

OB = OD (Radius)

∠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 an equal distance from the circle’s center have the same length. 

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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 Congruence By R.H.S)

BP = DQ (By CPCT rule)

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

Hence,

AB = CD

Theorem 4:

The four points lie on a circle if a line segment connecting two points subtends equal angles at two other points on the same side of the line containing the line segment. (i.e. they are concyclic). 

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Given: Let ABCD be a cyclic quadrilateral

We have to prove ∠A + ∠C = 180° and ∠B + ∠D = 180°

Construction: Join OB and OD.

Proof: 

∠BOD = 2∠BAD 

∠BAD = 1/2 ∠BOD

Similarly, ∠BCD = 1/2 reflex ∠BOD 

∴ ∠BAD + ∠BCD = 1/2 ∠BOD + 1/2 reflex ∠BOD

= 1/2 (∠BOD + reflex ∠BOD) = 1/2 ×360°

∴ ∠A + ∠C = 180°

Similarly, ∠B + ∠D = 180°

Solved Example 

Example 1

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

circle.

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The circle’s diameter is BD.

A is the given point on the circle.

Find the angle CBA?

Answer: Diameter is given, BD

Angle BAD is confined within the circle (given).

Angle BAD =90°.

To find the angle CBA,

CBA =180−(23−90) = 67°

CBA = 67°

Example 2

Two circles with radii 5 cm and 3 cm intersect at two points. Their centers are 4 cm in distance. Calculate the common chord’s length.

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Answer: Given:

OP = 5cm

OS = 4cm and

PS = 3cm

Also, PQ = 2PR

Now, suppose RS = x. 

Consider the ΔPOR,

OP² = OR²+PR²

⇒ 5²= (4-x)²+PR²

⇒ 25 = 16+x²-8x+PR²

∴ PR² = 9-x²+8x — (i)

Now consider ΔPRS,

PS² = PR²+RS²

⇒ 3² = PR²+x²

∴ PR² = 9-x² — (ii)

After equating both the equations, i.e. (i) and (ii). You will get,

9 –x²+8x = 9-x²

⇒ 8x = 0

⇒ x = 0

Now, in equation (i), put the value of x

PR² = 9-0²

⇒ PR = 3cm

∴ The length of the cord i.e. PQ = 2PR

So, PQ = 2×3 = 6cm

Conclusion

We hope this article on circle theorems helped the students to gain knowledge. Circle theorems are one of the important chapters. Practicing the theorems will help you grab hold of the concepts of all the theorems in a better way.