Notes on Circle | Grade 9 > Compulsory Maths > Geometry | KULLABS.COM

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### Basic Concepts on Circle

A Circle is a closed curve line, lying in a plane, whose all points are equidistant from a fixed point in the same plane. The fixed point is called the centre of the circle and the curve is called the circumference of the circle. A circle is a simple closed curve which divides the plane into two regions: an interior and an exterior. In everyday use, the term "circle" may be used interchangeably to refer to either the boundary of the figure or to the whole figure including its interior; in strict technical usage, the circle is only the boundary and the whole figure is called a disc.

#### Diameter

In geometry, the Diameter of a circle is any straight line segment that passes through the centre of the circle and whose endpoints lie on the circle. It can also be defined as the longest chord of the circle. It is denoted by'd'.

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

#### Chord

A straight line joining any two points on the circumference of a circle is called a chord but the line does not always pass through the center. Diameter is the longest chord.

#### Arc

A part of the circumference of a circle is called an arc of that circle. In general, an arc is any smooth curve joining two points. The length of an arc is known as its arc length. In particular, an arc is any portion (other than the entire curve) of the circumference of a circle. In figure, AB is an arc of the circle.

#### Semi-Circle

Exactly half of any circle is called semicircle. A straight line joining the endpoints of the semicircle is the diameter of the full circle formed the semicircle.

#### Concentric circles

The circles lying on the same plane and having a common centre are known as concentric circles. The distance between two concentric circles is the difference between the radii of those circles.

The circumference of concentric circles with different radii lie at equal distances from other circles. So they are parallel to each other. The diameter is double of the radius and the radius is half of diameter.

Radii and diameters of same circles are equal. The diameter of a circle divides the circle into two equal halves. Every diameter of a circle is the longest chord of the circle.

#### Intersecting Circles or Overlapping Circles

Two circles having common chord are known as intersecting circles.

#### Sector

A closed figure between two radii of a circle and the arc intercepted by them is called sector of that circle. In the figure, two radii OA and OB along with arc AB has bounded the shaded region which is known as sector of the circle.

#### Circle segment

A circle divided by a chord into two arcs, one of the portions of the arc of the chord is the segment. In the figure, shade pasrt is said minor segment and unshaded part is said major segment of the circle.

#### Equal circles

Circles having equal radii or diameter are called equal circles. Equal circles cover equal area.

#### Theoretical proof of theorems related to chord of a circle

Theorem 1

Perpendicular drawn from the centre of a circle to a chord bisects the chord.

Given: O is the centre of circle. AB is a Chord and OP⊥AB.

To prove: AP = PB

Construction: Join OA and OB

Proof:

 S.N. Statements Reasons 1. In ΔOAP and ΔOBP i. $$\angle$$OPA = $$\angle$$OPB (R) Both of them are right angles. ii. OA = OB (H) Radii of the same circle. iii. OP =OP (S) Common side 2. ∴ΔOAP≅ ΔOBP By R.H.S. axiom. 3. AP = PB Corresponding sides of congruent triangles are equal.

Theorem 2

A line joining the mid point of any chord and the centre of a circle is perpendicular to the chord.

Given: O is the centre of the given circle and AB is a chord with midpoint marked as P.

To prove: OP ⊥ AB

Construction: Join OA and OB

Proof:

 S.N. Statements Reasons 1. In ΔOAP and ΔOBP i. OA = OB (S) Radii of the same circle. ii. AP = PB (S) Given iii. OP = OP (S) Common side. 2. ∴ΔOAP ≅ ΔOBP By S.S.S. axiom 3. $$\angle$$OPA = $$\angle$$OPB Corresponding sides of congruent triangles are equal. 4. $$\angle$$OPA = $$\angle$$OPB = 90o Equal adjacent angles in a linear pair are 90o each 5. ∴ OP ⊥ AB From statement 3

Theorem 3

The perpendicular bisector of a circle passes through the centre of that circle.

Given: O is the centre of a circle. AB is a chord and its perpendicular bisector is PN.

To prove: Centre O lies in PN.

Construction: Let centre O does not lie in PN. Join OP.

Proof:

 S.N. Statements Reasons 1. NP ⊥ AB From given 2. OP ⊥ AB Mid point of chord AB is joined to the centre. 3. $$\angle$$NPO = $$\angle$$OPB NP and OP both are perpendicular to AB 4. $$\angle$$NPO = $$\angle$$OPB = 0o From statement 3 5. ∴ NP and OP coincide each other i.e.center O lies in PN From statement 4

Theorem 4

Equal chords of a circle are equidistant from the centre.

Given: O is the centre of a circle.Two equal chords AB and CD are drawn on either side of the centre where OP⊥ AB and OQ ⊥ CD (AB = CD).

To prove : OP = OQ

Construction: Join O, A and O, C

Proof:

 S.N. Statements Reasons 1. AB = CD Given 2. AP = CQ A perpendicular line drawn from the centre to the chord bisects the chord 3. In ΔOAP and ΔOCQ i. $$\angle$$OPA = $$\angle$$OQC (R) Both of them are right angles ii. OA = OC (H) Radii of the same circle iii. AP = CQ (S) From statement 2 4. ∴ ΔOAP≅ΔOCQ By R.H.S axiom 5. OP = OQ Corresponding sides of congruent triangles are equal.

Theorem 5

Given: O is the centre of a circle. AB and CD are two chords on their side of a centre where OP⊥ AB, OQ ⊥ CD and OP = OQ.

To prove: AB = CD

Construction: Join O, A and O, C

Proof:

 S.N. Statements Reasons 1. In ΔOAP and ΔOCQ i. $$\angle$$OPA = $$\angle$$OQC (R) Both of them are right angles ii. OA = OC (H) Radii of the same circle iii. OP = OQ (S) Given 2. ∴ ΔOAP ≅ ΔOCQ By R.H.S axiom 3. AP = CQ Corresponding sides of congruent triangles are equal. 4. AB = CD Perpendicular lines drawn from centre O bisects the chord.

• Arc: any connected part of the circle.
• Centre: the point equidistant from the points on the circle.
• Chord: a line segment whose endpoints lie on the circle.
• Circumference: the length of one circuit along the circle, or the distance around the circle.
• Diameter: a line segment whose endpoints lie on the circle and which passes through the centre; or the length of such a line segment, which is the largest distance between any two points on the circle. It is a special case of a chord, namely the longest chord, and it is twice the radius.
• Disc: the region of the plane bounded by a circle.
• Lens: the intersection of two discs.
• Passant: a coplanar straight line that does not touch the circle.
• Radius: a line segment joining the centre of the circle to any point on the circle itself; or the length of such a segment, which is half a diameter.
• Sector: a region bounded by two radii and an arc lying between the radii.
• Segment: a region, not containing the centre, bounded by a chord and an arc lying between the chord's endpoints.
• Secant: an extended chord, a coplanar straight line cutting the circle at two points.
• Semicircle: an arc that extends from one of a diameter's endpoints to the other. In non-technical common usage, it may mean the diameter, arc, and its interior, a two-dimensional region, that is technically called a half-disk. A half-disk is a special case of a segment, namely the largest one.
• Tangent: a coplanar straight line that touches the circle at a single point.
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#### Click on the questions below to reveal the answers

OX ⊥AB and OY ⊥ BC ( given)
Here , OY = 3 cm and OB radius = 5 cm ,
In rt. angled triangle OYB ,
BY = $$\sqrt{OB^{2} - OY^{2}}$$
= $$\sqrt5^{2 }- 3^{2}$$
= $$\sqrt{16}$$
= 4cm.
$$\therefore$$ Chord BC = 2 BY
= 2 $$\times$$ 4 cm = 8cm.

Given ,
OM ⊥ AB and On ⊥ CD
Costruction:Join OA
Here , OM = ON = 4cm , OA = OE = radius = 5cm
In rt. angled triangle OMA
AM = $$\sqrt{OA^{2} - OM^{2}}$$
= $$\sqrt{5^{2} - 4^{2} }$$
= $$\sqrt{9}cm^{2}$$
= 9cm.
Chord = AB = 2AM
= 2 $$\times$$ 3 = 6cm Ans.

Given ,
OB = 13cm , CD = 24cm
COnstruction : join OC
Now , OB =OC = radius = 13cm
CN = $$\frac{1}{2}$$ $$\times$$ CD.
= $$\frac{1}{2}$$ $$\times$$ 24 cm = 12cm.

In rt.angled ΔONC.

Given : OE = 17cm , OD = 8cm
Construction : join OB
Here , OB = OE = radius = 17cm
In rt. angled ΔOBDBD =$$\sqrt{OB^{2} - OD^{2}}$$
= $$\sqrt{17^{2} - 8^{2}}$$
= $$\sqrt{289 - 64}$$
= $$\sqrt{225}$$
= 15cm.
Chord = 2BD [perpendicular OD bisects BC]
=2 $$\times$$ 15 cm = 30 cm

Given : OA = 25cm , OC = 24cm
Here , In rt. angled triangle OAC
AC =$$\sqrt{OA^{2} - OC^{2}}$$
= $$\sqrt{25^{2} - 24^{2}}$$
= $$\sqrt{625 - 576}$$
= $$\sqrt{49}$$ = 7cm.

Chord AB = "AC [perpendicular OC bisects AB]
= 2 $$\times$$ 7cm = 14 cm Ans.

Given : OC = 8 cm , AB = 12cm
Construction : Join OA.
Here , AC = $$\frac{1}{2}$$AB [perpendicular OC bisetcs AB]
= $$\frac{1}{2}$$ $$\times$$ 12 cm = 6cm.

In rt. angled triangle OCA ,
OA = $$\sqrt{OC^{2} + AC^{2}}$$
= $$\sqrt{8^{2} + 6^{2}}$$
= $$\sqrt{100}$$ = 10 cm. Ans.

Given ,
O is the centre of circle ,Chords AB and CD intersect at P and OP is bisector of APD
To prove AB = CD
Construction Draw OX⊥AB and OY⊥CD

 Statements Reasons 1. In OXp and OYP(i) OXP = OYP = 90(ii) OPX = OPY(iii) OP = OP 1.(i) By construction(ii) Being OP is bisector(iii) Common side 2. OXP = OYP 2 .By A.A.S fact 3 . OX = OY 3. Corresponding sides of congruent sides. 4. AB = CD 4. Chords of circle equidistance from centre f the circle.

Given , PQ =RS
To prove OM bisects PMS
Construction Draw OX⊥ PQ and OY⊥ RS.

 Statements Reasons 1. In OMX and OMY(i) OXM = OYM(ii) OM = OM(iii) OX = OY 1.(i) Being OX⊥ PQ and OY⊥ RS(ii) Common Side(iii) Equal chords are equidistant from centre. 2. OMX≅ OMY 2. By R.H.S fact. 3. XMO = OMy 3. corresponding sides of≅ triangles. 4. OM bisects PMS 4. From statements 3.

Given : In two concentriccircle with centre O , the straight line MRSN cuts the external circle at M and N and internals circle at R and S respectively ,
To prove : MR = NS
Construction:Draw OE⊥ MN

 Statements Reasons 1. ME = NE 1. Perpendicular OE drawn from centre of the circle to the chord MN bisects MN. 2. RE = ES 2 .Perpendicular OE drawn from centre O to the chord RS bisects RS. 3. ME - RE = NE - SE$$\therefore$$ MR = NS 3. Subtracting statements 2 from statement 1.

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## ASK ANY QUESTION ON Circle

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##### Sanjok Thapa

How can we solve unseen theorems?

##### Sarina

If O is the center of circle and chord PQparallelRS PQ=16cm,RS=12cm radius of circle is 10cm find Value ofMN