## Circles

###
**Basic Concepts of a Circle**

There are many objects around us that are circular in shape. A circle is defined as the collection of all the points on a plane that are at equal distance from a given fixed point on the plane. This fixed point is called the centre of the circle and the fixed distance is called the radius.

A circle divides the plane on which it lies into three parts. They are (i) inside the circle, which is also called the interior of the circle; (ii) the circle and (iii) outside the circle, which is also called the exterior of the circle. The circle and its interior together form the circular region.

**Circumference**

The circumference of a circle is the actual length around the circle. It is the length of the edge of the circle traced around the centre of the circle.

**Radius**

A line segment joining the centre of a circle with any point on its circumference is called the radius of the circle.

**Chord**

A line that joins two points on the circumference of a circle is called a chord.

**Diameter**

A chord that passes through the centre of a circle is called the diameter of the circle. A diameter divides a circle into two equal segments, each is called a semicircle. Diameter is the longest chord of a circle. The diameter of a circle is twice the radius.

**Arc**

The part of the circumference of a circle between two given points is called an arc.The shorter arc between two given points on the circumference of a circle is called as the minor arc whereas the longer arc is called as the major arc.

**Segment**

A chord of a circle divides the circle into two regions called the segments of the circle. The region bounded by the chord and the minor arc intercepted by the chord is called the minor segment. The region bounded by the chord and the major arc intercepted by the chord is called the major segment.

**Sector**

The region between two radii of a circle and any of the arcs between them is called a sector. Sector corresponding to the minor arc is called the minor sector. Sector corresponding to the major arc is the major segment.

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**Chords of a Circle**

The perpendicular from a point to a line segment is the shortest distance between them. A line that joins two points on the circumference of a circle is called a chord. A chord passing through the centre of a circle is called the diameter. The longest chord of a circle is the diameter. There is one and only one circle passing through three given non-collinear points.

Theorem: The perpendicular from the centre of a circle to a chord bisects the chord.

Given: A circle with centre O. AC is a chord and OB ⊥ AC.

To prove: AB = BC.

Construction: Join OA and OC.

**Proof: In triangles OBA and OBC,**

∠OBA = ∠OBC = 90

^{o}(Since OB ⊥ AC)
OA = OC (Radii of the same circle)

OB = OB (Common side)

ΔOBA ≅ ΔOBC (By RHS congruence rule)

⇒ AB = BC (Corresponding sides of congruent triangles)

Thus, OB bisects the chord AC.

Hence, the theorem is proved.

Theorem: The line drawn from the centre of a circle to bisect a chord is perpendicular to the chord.

Given: A circle with centre O. AC is a chord and AB = BC.

To prove: OB ⊥ AC.

Construction: Join OA and OC.

**Proof: In triangles OBA and OBC,**

AB = BC (Given)

OA = OC (Radii of the same circle)

OB = OB (Common side)

ΔOBA ≅ ΔOBC (SSS congruence rule)

⇒ ∠OBA = ∠OBC (Corresponding angles of congruent triangles)

But, ∠OBA + ∠OBC = ∠ABC = 180° [Linear pair]

∠OBC + ∠OBC = 180°

^{ }[Since ∠OBA = ∠OBC]
2 x ∠OBC = 180°

∠OBC = 180°2 = 90

^{o}
∠OBC = ∠OBA = 90°

∴ OB ⊥ AC

Hence, the theorem is proved.

Let AB and PQ be any two chords of a circle with the centre O. ∠AOB and ∠POQ are called the angles subtended by the chord at the centre of the circle. As the chord moves away from the centre, its length and the angle subtended by it at the centre decreases. On the other hand, if a chord moves closer to the centre, its length and the angle subtended by it at the centre increases.

Theorem: Equal chords of a circle subtend equal angles at the centre.

Given: A circle with centre O. AB and PQ are chords of the circle. AB = PQ

To prove: ∠AOB = ∠POQ

**Proof: In triangles AOB and POQ,**

AB = PQ (Given)

OA = OP (Radii of same circle)

OB = OQ (Radii of same circle)

ΔAOB ≅ ΔPOQ (SSS congruence rule)

⇒ ∠AOB = ∠POQ (Corresponding angles)

Hence, the theorem is proved.

Theorem: Chords that subtend equal angles at the centre of a circle are equal in length.

Given: ∠AOB = ∠POQ

To prove: AB = PQ

**Proof: In triangles AOB and POQ,**

∠AOB = ∠POQ (Given)

OA = OP (Radii of same circle)

OB = OQ (Radii of same circle)

ΔAOB ≅ ΔPOQ (SAS congruence rule)

⇒ AB = PQ (Corresponding sides)

Hence, the theorem is proved.

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**Arcs of a Circle**

A part of a circle is called an arc. Arcs of a circle that superimpose each other completely are called congruent arcs. If two arcs of a circle are congruent, then their corresponding chords are equal. Conversely, if two chords of a circle are equal, then their corresponding arcs are congruent.

Corresponding arcs of two equal chords of a circle are congruent.

Congruent arcs of a circle subtend equal angles at the centre.

Given: Two congruent arcs AB and CD.

To prove: ∠ AOB = ∠COD

Construction: Draw chords AB and CD.

**Proof: The angle subtended by an arc at the centre is equal to the angle subtended by its corresponding chord at the centre.**

In the given figure,

AB = CD (Chords corresponding to congruent arcs of a circle are equal)

∠AOB = ∠COD (Equal chords subtend equal angles at the centre)

Hence, the theorem is proved.

The angle subtended by an arc at the centre is double the angle subtended by the arc at any point on the remaining part of the circle.

Given: Arc AB. Point C on the circle is outside AB.

To prove: ∠AOB = 2 × ∠ACB

Construction: Draw a line CO extended till point D.

**Proof: In ΔOAC in each of these figures,**

∠AOD = ∠OAC + ∠OCA (Exterior angle of a triangle is equal to sum of two opposite interior angles)

OA = OC (Radii of same circle)

Thus, ∠ OAC = ∠ OCA (Angles opposite equal sides of a triangle are equal)

∠AOD = ∠OAC + ∠OCA

⇒∠AOD = 2 × ∠OCA

Similarly, in ΔOBC, ∠BOD = 2 × ∠OCB

∠AOD = 2 × ∠OCA and ∠BOD = 2 × ∠OCB

⇒ ∠AOD + ∠BOD = 2 ∠OCA + 2 ∠OCB

∠AOD + ∠BOD = 2 × (∠OCA + ∠OCB)

or ∠AOB = 2 × ∠ACB

Hence, the theorem is proved.

Angles subtended by an arc at all points within the same segment of the circle are equal.

Given: An arc AB. Points C and D are on the circle in the same segment.

To prove: ∠ACB = ∠ADB

**Proof: By the theorem that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the circle:**

∠AOB = 2 × ∠ACB

Also, ∠AOB = 2 × ∠ADB

∴ ∠ACB = ∠ADB

Hence, the theorem is proved.

All angles formed in a semi circle are right angles.

Given: A circle with centre O, and Q, P and R are three points on the circumference of the circle.

Construction: Join the points Q, P and R to the points A and B.

To prove: ∠AQB = ∠APB = ∠ARB

**Proof: ∠AQB =**12 ∠AOB [Angle subtended by an arc at the centre is double the angle subtended by the arc at any point on the remaining part of the circle.]

∠APB = 12 ∠AOB [Angle subtended by an arc at the centre is double the angle subtended by the arc at any point on the remaining part of the circle.]

Similarly, ∠ARB = 12 ∠AOB.

∴ ∠AQB = ∠APB = ∠ARB

Hence, the theorem is proved.

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**Cyclic Quadrilaterals**

A circle can be drawn passing through three non-collinear distinct points. The points that lie on a circle are called concyclic points. So, three non-collinear points are always concyclic.

If a line segment joining two points subtends equal angles at two other points on the same side of the line segment then all the four points are concyclic.

Given: Line segment AB.

Mark two points C and D such that ∠ACB = ∠ADB.

To prove: A, B, C and D are concyclic points.

Draw a circle through points A, B and C.

Assume that the circle drawn through points A, B and C does not pass through D, and intersects AD at D’.

**Proof: If A, B, C and D’ are concyclic:**

∠ACB = ∠AD’B (Angles subtended by a chord in the same segment of a circle)

∠ACB = ∠ADB (Given)

∴ ∠AD’B = ∠ADB

or D’ coincides with D.

Thus, A, B, C and D are concyclic points.

Hence, the theorem is proved.

In a cyclic quadrilateral, the sum of the opposite angles is always equal to 180°.

If the sum of the opposite angles of a quadrilateral is 180°, then the quadrilateral is cyclic.

The exterior angle formed by producing a side of a cyclic quadrilateral is equal to the interior opposite angle.

A cyclic parallelogram is a rectangle.