Circles

(i) Interior

(ii) Exterior

(iii) Diameter

(iv) Semi-circle

(v) Chord

(vi) Three

(i) True. All the points on the circle are at equal distances from the centre of the circle, and this equal distance is called as radius of the circle

(ii) False. There are infinite points on a circle. Therefore, we can draw infinite number of chords of given length. Hence, a circle has infinite number of equal chords

(iii) False. Consider three arcs of same length as AB, BC, and CA. It can be observed that for minor arc BDC, CAB is a major arc. Therefore, AB, BC, and CA are minor arcs of the circle

(iv) True. Let AB be a chord which is twice as long as its radius. It can be observed that in this situation, our chord will be passing through the centre of the circle.

Therefore, it will be the diameter of the circle

(v) False. Sector is the region between an arc and two radii joining the centre to the end points of the arc. For example, in the given figure, OAB is the sector of the circle

(vi) True. A circle is a two-dimensional figure and it can also be referred to as a plane figure

A circle is a collection of points which are equidistant from a fixed point. This fixed point is called as the centre of the circle and this equal distance is called as radius of the circle.

And thus, the shape of a circle depends on its radius. Therefore, it can be observed that if we try to superimpose two circles of equal radius, then both circles will cover each other

In ΔAOB and ΔCO'D

AB = CD (Chords of same length)

OA = O'C (Radii of congruent circles)

OB = O'D (Radii of congruent circles)

ΔAOB ΔCO'D (SSS congruence rule)

AOB = CO'D (By CPCT)

Hence, equal chords of congruent circles subtend equal angles at their centers

Let us consider two congruent circles (circles of same radius) with centers as O and O

In ΔAOB and ΔCO'D,

AOB = CO'D (Given)

OA = O'C (Radii of congruent circles)

OB = O'D (Radii of congruent circles)

ΔAOB ΔCO'D (SAS congruence rule)

AB = CD (By CPCT)

Hence, if chords of congruent circles subtend equal angles at their centers, then the chords are equal.

Consider the following pair of circles

The above circles do not intersect each other at any point. Therefore, they do not have any point in common

The above circles touch each other only at one point Y. Therefore, there is 1 point in common

The above circles touch each other at 1 point X only. Therefore, the circles have 1 point in common

These circles intersect each other at two points G and H.
Therefore, the circles have two points in common.
It can be observed that there can be a maximum of 2 points in common.

The below given steps will be followed to find the centre of the given circle

Step 1. Take the given circle

Step 2. Take any two different chords AB and CD of this circle and draw perpendicular bisectors of these chords

Step 3. Let these perpendicular bisectors meet at point O.

Hence, O is the centre of the given circle

Let two circles with centres O and O' respectively intersect at two points A and B
such that AB is the common chord of two circles and
OO' is the line segment joining the centres, as shown in the figure:

Also, AB is the chord of the circle centered at O. Therefore, the perpendicular bisector of AB will pass through O.

Let OO' intersect AB at M. Also, draw line segments OA, OB, O'A and O'B.

Now, In ΔOAO' and OBO',
OA = OB (radii of same circle)
O'A = O'B (radii of same circle)
O'O ( common side)
⇒ ΔOAO' ≅ ΔOBO' (Side-Side-Side congruency)
⇒ ∠AOO' = ∠BOO'
⇒ ∠AOM = ∠BOM ......(i)

Now in ΔAOM and ΔBOM we have
OA = OB (radii of same circle)
∠AOM = ∠BOM (from (i))
OM = OM (common side)
⇒ ΔAOM ≅ ΔBOM (Side-Angle-Side congruency)
⇒ AM = BM and ∠AMO = ∠BMO
But
∠AMO + ∠BMO = 180°
⇒ 2∠AMO = 180°
⇒ ∠AMO = 90°
Thus, AM = BM and ∠AMO = ∠BMO = 90°
Hence OO' is the perpendicular bisector of AB.

Let the radius of the circle centered at O and O' be 5 cm and 3 cm respectively.

OA = OB = 5 cm

O'A = O'B = 3 cm
Also distance between their centers is 4 cm.
So OO^' = 4 cm

The diagram is shown below:

If two circles intersect each other at two points, then the line joining their centres is the perpendicular bisector of the common cord.

OO' will be the perpendicular bisector of chord AB

⇒ AC = CB

Let OC be x. Therefore, O'C will be 4 − x

In ΔOAC,

OA² = AC² + OC²

⇒ 5² = AC² + x²

⇒ 25 – x² = AC² ..........(i)

Now, In ΔO'AC,

O'A² = AC² + O'C²

⇒ 3² = AC² + (4 − x)²

⇒ 9 = AC² + 16 + x² − 8x

⇒ AC² = − x² − 7 + 8x ........(ii)

From (i) and (ii), we get

25 – x² = − x² − 7 + 8x

8x = 32

x = 4

Hence, O'C = 4 - 4 = 0 cm i.e. the common chord will pass through the centre of the smaller circle i.e., O'
and hence, it will be the diameter of the smaller circle.

Also, AC² = 25 – x²

= 25 – 4²

= 25 − 16

= 9

⇒ AC = 3 cm

⇒ Length of the common chord AB = 2 AC
= (2 × 3) cm
= 6 cm

Let PQ and RS be two equal chords of a given circle and they are intersecting each other at point T

To prove: PT = RT
QT = ST

Draw perpendiculars OV and OU on these chords.

In ΔOVT and ΔOUT,

OV = OU (Equal chords of a circle are equidistant from the centre)

∠OVT = ∠OUT (Each 90°)

OT = OT (Common)

ΔOVT ΔOUT (RHS congruence rule)

VT = UT (By CPCT) (i)

It is given that,

PQ = RS (ii)

PQ = RS

PV = RU (iii)

On adding (i) and (iii), we get

PV + VT = RU + UT

PT = RT

On subtracting equation (iv) from equation (ii), we obtain

PQ − PT = RS − RT

QT = ST

Let PQ and RS are two equal chords of a given circle and they are intersecting each other at point T

Draw perpendiculars OV and OU on these chords.

In ΔOVT and ΔOUT,

OV = OU (Equal chords of a circle are equidistant from the centre)

∠OVT = ∠OUT (Each 90°)

OT = OT (Common)

ΔOVT ≅ ΔOUT (RHS congruence rule)

∠OTV = ∠OTU (By congruent parts of congruent triangles )

Therefore, it is proved that the line joining the point of intersection to the centre makes equal angles with the chords.

Let us draw a perpendicular OM on line AD

It can be observed that BC is the chord of the smaller circle and AD is the chord of the bigger circle

We know that perpendicular drawn from the centre of the circle bisects the chord

BM = MC (i) And,

AM = MD (ii)

On subtracting (ii) from (i), we get

AM − BM = MD − MC

AB = CD

To Find: Distance between Reshma and Mandip
Given: Distance between Reshma and Salma and between Salma and Mandip is 6m
Diagram:

Explanation:

Draw perpendicular OA on RS.

As OA ⊥ RS,

RA = AS

Let R, S and M be the positions of Reshma, Salma and Mandip.

Now in ΔOAR by Pythagoras theorem,

OR² = OA² + RA²

OA² = 5² - 3²

OA² = 25 – 9

OA² = 16

OA = 4

Now we know,

Area of triangle = ½ × B × H

Area ΔORS = ½ × OS × RK

= ½ × 5 × RK ….. (1)

Also,

Area ΔORS = ½ × OA × RS

= ½ × 4 × 6
= 12 ….. (2)

From (1) and (2),

½ × 5 × RK = 12


RK = 4.8 m
As the perpendicular from the centre of a circle bisects the chord.

As RM = 2RK

= 2× 4.8

= 9.6 m

The figure is shown as:

According to the given question,

AS = AD = SD

Let AS = AD = SD = 2x

In ASD, all sides are equal,

∴ ASD is an equilateral triangle.

Now draw OP⊥ SD

So, SP = DP = ½ SD

⇒ SP = DP = x

Join OS and AO.

In ΔOPS,

OS² = OP² + PS²

(20)² = OP² + x²

400 = OP² + x²

In ΔAPS,

AS² = AP² + PS²

(2x)² = AP² + x²

4x² = AP² + x²

Now,

AP = AO + OP

√3x = 20 + √400 – x²

√3x - 20 = √400 – x²

Squaring both sides, we get,

(√3x – 20)² = (√400 – x²)²

(√3x)² + (20)² – 2×(√3x)×(20) = 400 – x²

3x² + 400 – 40√3x = 400 – x²

40√3x = 4x²

x = 10 √3 m

∠AOC = ∠AOB + ∠BOC

= 60° + 30°

= 90°

We know that angle subtended by an arc at the centre is double the angle subtended by it any point on the remaining part of the circle

∠ADC = ∠AOC

= × 90^o

= 45^o

In ΔOAB,

AB = OA = OB = Radius

ΔOAB is an equilateral triangle

Therefore, each interior angle of this triangle will be of 60°

∠AOB = 60°

∠ACB = × AOB

= × 60^o

= 30^o

In cyclic quadrilateral ACBD,

∠ACB + ∠ADB = 180° (Opposite angle in cyclic quadrilateral)

∠ADB = 180° − 30°

= 150°

Therefore, angle subtended by this chord at a point on the major arc and the minor arc are 30° and 150° respectively

Method 1:

We know angle subtended at centre by an arc is double the angle subtended
by it at any other point.

Reflex angle ∠POR= 2∠PQR

= 2 × 100°
= 200°

Now, ∠POR = 360° – 200 = 160°

Also,

PO = OR [Radii of a circle]

∠OPR = ∠ORP [Opposite angles of isosceles triangle]

In ∆OPR, ∠POR = 160°

∴ ∠OPR = ∠ORP = 10°

Method 2:
Consider PR as a chord of the circle. Take any point S on the major arc of the circle

PQRS is a cyclic quadrilateral

∠PQR + ∠PSR = 180° (Opposite angles of a cyclic quadrilateral)

∠PSR = 180° − 100° = 80°

We know that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle

∠POR = 2 ∠PSR

= 2 (80°)

= 160°

In ΔPOR,

OP = OR (Radii of the same circle)

∠OPR = ∠ORP (Angles opposite to equal sides of a triangle)

∠OPR + ∠ORP + ∠POR = 180° (Angle sum property of a triangle)

∠OPR + 160° = 180°

2 ∠OPR = 180° − 160°

2 ∠OPR = 20°

∠OPR = 10°

∠BAC = ∠BDC (Angles in the segment of the circle)

In triangle ABC,

∠BAC + ∠ABC + ∠ACB = 180^o (Sum of all angles in a triangle)

∠BAC + 69^o + 31^o = 180^o

∠BAC = 180^o – 100^o

∠BAC = 80^o

Thus,

∠BDC = 80^o

In ΔCDE,

∠CDE + ∠DCE = CEB (Exterior angle)

∠CDE + 20° = 130°

∠CDE = 110°

However, ∠BAC = ∠CDE (Angles in the same segment of a circle)

∠BAC = 110°
As angles in same segment are equal.
So,

METHOD 1:
For chord CD,

∠CBD = ∠CAD (Angles on the same segment are equal)

∠CAD = 70^o

∠BAD = ∠BAC + ∠CAD

= 30^o + 70^o

= 100^o

∠BCD + ∠BAD = 180° (Opposite angles of a cyclic quadrilateral)

∠BCD + 100^o = 180^o

∠BCD = 80^o

Now, In ΔABC,

AB = BC (Given)

∠BCA = ∠CAB (Angles opposite to equal sides of a triangle)

∠BCA = 30°

We have,

∠BCD = 80°

∠BCA + ∠ACD = 80°

30° + ∠ACD = 80°

∠ACD = 50°

∠ECD = 50°
METHOD 2:
On chord AD,
∠ABD = ∠ECD.....(1) [Since angles on the same segment are equal]
In ΔABC,
Given: AB = BC
Therefore,
∠BAC = ∠BCA = 30º
Sum of angles of a triangle = 180º
30º + 30º + ∠ABC = 180º
∠ABC = 120º.....(2)
And we can see that
∠ABC = ∠ABD + ∠DBC
120º = ∠ABD + 70º
∠ABD = 50º
And from equation 1, we can see that,
∠ECD = 50º

Let ABCD be a cyclic quadrilateral having diagonals BD and AC, intersecting each other at point O.

n ∆AOD and ∆COB

AO = CO [Radii of a circle]

OD = OB [Radii of a circle]

∠AOD = ∠COB [Vertically opposite angles]

∴ ∆AOD ≅ ∆COB [SAS congruency]

∴ ∠OAD = ∠OCB [CPCT]

But these are alternate interior angles made by the transversal AC, intersecting AD and BC.

∴ AD || BC

Similarly, AB || CD.

Hence, quadrilateral ABCD is a parallelogram.

Also, ∠ABC = ∠ADC ..(i) [Opposite angles of a ||gm are equal]

And, ∠ABC + ∠ADC = 180° ...(ii)

[Sum of opposite angles of a cyclic quadrilateral is 180°]

⇒ ∠ABC = ∠ADC = 90° [From (i) and (ii)]

∴ ABCD is a rectangle.

[A ||gm one of whose angles are 90° is a rectangle]
Hence Proved.

Consider a trapezium ABCD with

AB | |CD and

BC = AD

Draw AM perpendicular to CD and BN perpendicular to CD

In ΔAMD and ΔBNC,

AD = BC (Given)

∠AMD = ∠BNC (By construction, each is 90°)

AM = BN (Perpendicular distance between two parallel lines is same)

ΔAMD ΔBNC (RHS congruence rule)

∠ADM = ∠BCN (CPCT)
⇒∠ADC = ∠BCD (i)

BAD and ADC are on the same side of transversal AD

∠BAD + ∠ADC = 180° (ii)

∠BAD + ∠BCD = 180° [Using equation (i)]

This equation shows that the opposite angles are supplementary

Therefore, ABCD is a cyclic quadrilateral.

Join chords AP and DQ

For chord AP,

∠PBA = ∠ACP (Angles in the same segment) (i)

For chord DQ,

∠DBQ = ∠QCD (Angles in the same segment) (ii)

ABD and PBQ are line segments intersecting at B

∠PBA = ∠DBQ (Vertically opposite angles) (iii)

From (i), (ii), and (iii), we get

∠ACP = ∠QCD
hence proved

Consider a ΔABC

Two circles are drawn while taking AB and AC as the diameter

Join AD

∠ADB = 90° (Angle subtended by semi-circle)

∠ADC = 90° (Angle subtended by semi-circle)

∠BDC =∠ ADB + ∠ADC = 90° + 90° = 180°

Therefore, BDC is a straight line

Thus, Point D lies on the third side BC of ΔABC.

It is given in the question that,

Ac is the common hypotenuse

So, ∠B = ∠D = 90^o

We have to prove that,

∠CAD = ∠CBD

Proof: We know that,

∠ABC and ∠ADC are 90^o as these angles are in a circle

Thus, both the triangles are lying in the circle and both have same diameter i.e. AC

Points A, B, C and D are concyclic

Therefore, CD is the chord

So,

∠CAD = ∠CBD (As, angles made by a chord in the same segment are equal)

Given that: ABCD is a cyclic quadrilateral

We have to prove that: ABCD is a rectangle

Proof: ∠1 + ∠2 = 180^o (Sum of opposite angles of a cyclic parallelogram)

We also know that, opposite angles of a cyclic parallelogram are equal

Therefore,

∠1 = ∠2

∠1 + ∠1 = 180^o

∠1 = 90^o

As,

One of the interior angle of the parallelogram is 90^o

Therefore, ABCD is a rectangle

Let two circles with centers as O and O’ intersect each other at point A and B respectively Join OO’.

Now,

In ΔAOO’ and BOO’

O’A = O’B (radius of same circle)

OA = OB (Radius of same circle)

OO’ = OO’ (Common)

Hence, by SSS congruence

ΔOAO’ ΔOBO’

∠OAO’ = ∠OBO’ (By CPCT)

Therefore,

Line of centers of two intersecting circles subtends equal angles at the two points of intersection.

Draw OM perpendicular AB and ON perpendicular CD

Join OB and OD

BM = (The perpendicular drawn from the center bisects the chord)

Let ON be x

Therefore, OM will be 6 –x

In ΔMOD,

OM² + MB² = OB²

(6 – x)² + ² = OB²

36 + x² -12x + OB² (i)

In ΔNOD,

ON² + ND² = OD²

(x)² + ² = OD²

(x)² + = OD² (ii)

We have OB = OD (radii of the same circle)

Therefore,

From (i) and (ii),

36 + x² -12x + (x)² +

36 -12x +

x = 1

From (ii),

(1)² + = OD²

OD = cm

Let AB and CD be two parallel chords in a circle centered at O

Join OB and OD

Distance of the smaller chord AB from the centre of the circle = 4cm

OM = 4cm

MB = =

In ΔOMB,

OM² + MB² = OB²

(4)² + ² = OB²

16 + 9 = OB²

OB = 5cm

In ΔOND,

ON² + ND² = OD²

(ON)² + ² = 5²

(ON)²= 9

ON = 3 cm

In ΔAOD and ΔCOE,

OA = OC (radii)

OD = OE (Radii)

AD = CE (Given)

Therefore,

By SSS congruence,

ΔAOD ΔCOE

∠OAD = ∠ OCE (By CPCT) (i)

∠ODA = ∠OEC (By CPCT) (ii)

∠OAD = ∠ODA (iii)

Using (i), (ii) and (iii), we get

∠OAD = ∠OCE = ∠ODA = ∠OEC = x

Now,

In triangle ODE,

OD = OE

Hence, ABCD is a quadrilateral

∠CAD + ∠DEC = 180^o

a + x + x + y = 180^o

2x + a + y = 180^o (iv)

∠DOE = 180^o – 2y

And,

∠COA = 180^o - 2a

∠DOE - ∠COA = 2a – 2y

= 2a – 2 (180^o - 2x – a)

= 4a + 4x – 360^o (v)

∠BAC = 180^o – (a + x)

And

∠ACB = 180 – (a - x)

In ΔABC,

∠ABC + ∠BAC + ∠ACB = 180^o (Angle sum property of triangle)

∠ABC = 2a + 2x – 180^o

∠ABC = 1/2 (4a + ax – 360⁰) [using (v)]

Let ABCD be a rhombus in which diagonals intersect at point O and a circle is drawn is drawn taking side CD as diameter

Now,

∠COD = 90⁰

(Since, diagonals of rhombus intersect at 90⁰)

Hence,

∠AOB = ∠BOC = ∠COD = ∠DOA = 90⁰

Hence, point O has to lie on the circle

It is visible that ABCE is a cyclic quadrilateral and in a cyclic quadrilateral, the sum of opposite angle is 180⁰

∠AEC = ∠CBA = 180

∠AED = ∠CBA (i)

We know that: Opposite angles of a parallelogram are equal

Hence,

∠ADE = ∠CBA (ii)

Using (i) and (ii)

∠ADE = ∠AED

AE = AD (angles opposite to equal sides are equal)

Let two chords AB and CD are intersecting each other at point O

In ΔAOB and ΔCOD,

OA = OC (Given)

OB = OD (Given)

∠AOB = ∠COD (Vertically opposite angles)

ΔAOB ΔCOD (SAS congruence rule)

AB = CD (By CPCT)

Similarly, it can be proved that ΔAOD ΔCOB

AD = CB (By CPCT)

Since in quadrilateral ACBD, opposite sides are equal in length, ACBD is a parallelogram

∠A = ∠C

However,

∠A + ∠C = 180° (ABCD is a cyclic quadrilateral)

∠A + ∠A = 180°

2 ∠A = 180°

∠A = 90°

As ACBD is a parallelogram and one of its interior angles is 90°, therefore, it is a rectangle

A is the angle subtended by chord BD and BD should be the diameter of the circle

Similarly, AC is the diameter of the circle

It is given in the question that,

∠ABE =

However,

= ∠ADE

Like this, similarly

∠ACF = ∠ADF =

D = ∠ADE + ∠ADF

= +

= (∠B + ∠C)

= (180^o – ∠A)

= 90^o - ∠A

Similarly,

E = 90^o - ∠B

And,

F = 90^o - ∠C

AB is common chord in the given both circles

Also, circles are congruent

∠APB = ∠AQB

Now, in triangle BPQ

∠APB = ∠AQB

BQ = BP (Angles opposite to equal sides are equal)

Let perpendicular bisector of side BC and angle bisector of A meet at point D. Let the perpendicular bisector of side BC intersect it at E

Perpendicular bisector of side BC will pass through circum centre O of the circle

BOC and BAC are the angles subtended by arc BC at the centre and a point A on the remaining part of the circle respectively

∠BOC = 2 ∠BAC = 2 ∠A (i)

In ΔBOE and ΔCOE,

OE = OE (Common)

OB = OC (Radii of same circle)

∠OEB = ∠OEC (Each 90° as OD perpendicular to BC)

ΔBOE COE (RHS congruence rule)

∠BOE = ∠COE (By CPCT) (ii)

However,

∠BOE + ∠COE = ∠BOC

∠BOE + ∠BOE = 2 ∠A [From (i) and (ii)]

2 ∠BOE = 2 ∠A

∠BOE = ∠A

∠BOE = ∠COE = ∠A

The perpendicular bisector of side BC and angle bisector of A meet at point D

∠BOD = ∠BOE = ∠A (iii)

Since AD is the bisector of angle A

∠BAD =

2 ∠BAD = A (IV)

From (iii) and (iv), we get

∠BOD = 2 ∠BAD

Hence, proved