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.
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.
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.
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.
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.
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.
Two circles having common chord are known as intersecting circles.
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.
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.
Circles having equal radii or diameter are called equal circles. Equal circles cover equal area.
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 = 90^{o} | Equal adjacent angles in a linear pair are 90^{o} 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 = 0^{o} | 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. |
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. |
Find the area of a circle having circumference 440cm.
14400 cm^{2}
16400 cm^{2}
17400 cm^{2}
15400 cm^{2}
Find the circumference of a circle whose area is 616 cm^{2}.
88 cm
53 cm
21 cm
30 cm
If the diameter of the circle is 150 cm less than the circumference, what is the length of its circumference?
320 cm
250 cm
300 cm
220 cm
A flexible wire is in the form of a circle of radius 42 cm.If the wire is bent into a square, what will be the length of a side of the square?
22 cm
36 cm
26 cm
43 cm
A semicircle has area 77 cm^{2}^{}.Find the perimeter.
22 cm
40 cm
26 cm
36 cm
A semicircle has area 1232 cm^{2}.Find the perimeter.
130 cm
144 cm
155 cm
120 cm
An ant moves 2.2 cm per second.How long will it take to go round a circle of radius 7 cm?
8 seconds
14 seconds
20 seconds
5 seconds
A wire is looped in the form of a circle of radius 28 cm.It is rebent into a square form.
25 cm
44 cm
12 cm
35 cm
A motor car is driven at the rate of 66km/hour.Each of its wheels makes 500 revolutions per minutes.What is the radius of the wheel?
20 cm
36 cm
49 cm
35 cm
A field is in the form of a circle.The cost of ploughing the field at Rs.1.50 per m^{2} is Rs.5775.Find the cost of fencing the field at Rs. 8.75 per metre.
Rs 1925
Rs 1212
Rs 1612
Rs 1313
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Sanjok Thapa
How can we solve unseen theorems?
Mar 15, 2017
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Sarina
If O is the center of circle and chord PQparallelRS PQ=16cm,RS=12cm radius of circle is 10cm find Value ofMN
Feb 04, 2017
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find the area of a circle having circumference 440 cm
kgjm
Jan 29, 2017
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Wow so many long questions
Jan 24, 2017
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