# CBSE Class 9 Maths Revision Notes Chapter 10

## Class 9 Mathematics Revision Notes for Circles of Chapter 10

Extramarks’ Class 9 Mathematics Chapter 10 Notes help students with a detailed understanding of the chapter. These notes are designed as per the latest CBSE curriculum by subject-matter experts. Moreover, students can score  better marks in their exams by carefully examining the CBSE Class 9 Chapter 10 Mathematics Notes

### Circle:

A circle can be defined as the locus of points at a certain distance from a fixed point.

### Chord:

• A straight line that connects any two points on a circle is known as a chord.
• It is represented by the letters AB.
• The longest chord that passes through the centre of the circle is termed the diameter.
• The diameter is twice the radius and is referred to as a CD.
• A line that divides a circle in half is termed a secant. PQR is a secant of a circle.

### Circumference:

It refers to the length of a full circle and is defined as the border curve (or perimeter) of the circle.

### Arc:

• Any section or a part of the circumference is referred to as an arc.
• A circle is divided into two equal pieces by diameter.
• A semicircle is larger than a minor arc.
• A semicircle is smaller than a major arc.
• ABC⌢ is a major arc, whereas ADC⌢ is a minor arc.

### Sector:

• The area between an arc and the two radii connecting the arc’s centre and endpoints is called a sector.
• A section of a circle a chord has cut off is known as a segment.

### Concentric Circles:

Circles with the same centre are concentric circles.

### Theorem 1:

Theorem 1:

A straight line drawn from the centre of a circle that is not a diameter to bisect a chord is always at a right angle to the chord.

• Given Data: Here, AB is a chord of a circle with the centre O. The midpoint of AB is M. OM is joined.
• To Prove: ∠AMO = ∠BMO = 90°
• Construction: Join AO and BO
• Proof: In ΔAOM and ΔBOM
 Statement Reason AO = BO radii AM = BM Data OM = OM Common ΔAOM ≅ΔBOM S.S.S ∴∠AMO = ∠BMO Statement (4) But ∠AMO + ∠BMO= 180° Linear pair ∠AMO = ∠BMO= 90° Statements 5 and 6

### Angle Properties (Angle, Cyclic Quadrilaterals and Arcs):

• ∠APB on the circumference is suspended by a straight line AB.
• On the remaining part of the circumference, ∠APB can be said to be subtended by arc AMB.
• Arc AMB subtends ∠AOB at the centre and ∠APB on the circumference.
• ∠AQB and ∠APB are in the same segment.

The quadrilateral is called a cyclic quadrilateral if the vertices of a quadrilateral lie on a circle. The vertices are called concyclic points.

### Corollary:

The exterior angle of a cyclic quadrilateral is equivalent to the opposite interior angle.

Given: ABCD is a cyclic quadrilateral. BC is extended to E.

To Prove: ∠DCE = ∠A

Proof:

 Statement Reason ∠A + ∠BCD=180ο Opposite ∠s of a cyclic quadrilateral ∠BCD + ∠DCE=180ο Linear pair ∴ ∠BCD + ∠DCE=∠A + ∠BCD Statements (1) and (2) ∴ ∠DCE = ∠A Statement (2)

### Alternate Segment Property

Theorem 10:

The angle in the alternate segment is equal to the angle between a tangent and a chord through the point of contact.

• Given Data:

A straight line SAT touches a given circle with centre O at A. AC is a chord through the point of contact A.

Angles in the alternate segments to ∠CAT and ∠CAS are denoted by ∠ADC and ∠AEC, respectively.

• To prove: ∠CAT=∠ADC and ∠CAS=∠AEC
• Construction: Draw AOB as diameter and join BC and OC.
• Proof:
 Statement Reason ∠OAC= ∠OCA= x Since OA=OC and supposition ∠CAT+ ∠x= 90° Since tangent-radius property ∠AOC + ∠x+ ∠y= 180° Sum of angles of a triangle ∠AOC= 180°- 2∠x Statement 3 ∠AOC= 2∠ADC ∠ at the centre = 2∠ on the circle ∠CAT= 90° – x Statement 2 2∠CAT= 180°- 2x Statement 6 2∠CAT= 2∠ADC Statements 4,5,7 ∠CAT= ∠ADC Statement 8 ∠CAS+∠CAT= 180° Linear pair ∠ADC= ∠AEC= 180° Opposite angles of a cyclic quadratic ∠CAS+∠CAT= ∠ADC+∠AEC Statements 10 and 11

Therefore, ∠CAS=∠AEC (Statement 9 and 12)

### Test for Concyclic Points:

(a) Conversely, one test for the concyclic points is ‘Angles in the same segment of a circle are equal’.

It states that:

The four points are concyclic if two equal angles are on the same side of a line and are subtended by it.

If ∠P=∠Q and the points P, and Q are on the same side of AB, then all the points A, B, P and Q are concyclic.

(b) Another test for concyclic points is the converse of ‘opposite angles of a cyclic quadrilateral are supplementary’.

It states that:

The vertices of a quadrilateral are concyclic if the opposing angles are supplementary.

For instance, if ∠A+∠C=180∘ then A, B, C and D are concyclic points.