Circles

(i) Interior/exterior

(ii) Concentric

(iii) Exterior

(iv) Arc

(v) Diameter

(vi) Semi-circle

(vii) Centre

(viii) Three

(i) True: Because it is a one dimensional figure

(ii) True: Since, line segment joining the centre to any point on the circle is a radius of the circle

(iii) True: Because each arc measures equal

(iv) False: Since, a circle has only infinite number of equal chords

(v) True: Because, radius equal to times of its diameter

(vi) True: Yes, sector is the region between the chord and its corresponding arc

(vii) False: The degree measure of an arc is half of the central angle containing the arc

(viii) True: Yes, The degree measure of a semi-circle is 180°

Given that,

Radius of circle (OA) = 8 cm

Chord (AB) = 12 cm

Draw OC perpendicular to AB

We know that,

The perpendicular from centre to chord bisects the chord

Therefore,

AC = BC =

= 6 cm

Now,

In , by using Pythagoras theorem

AC² + OC² = OA²

6² + OC² = 8²

36 + OC² = 64

OC² = 64 – 36

OC² = 28

OC = 5.291 cm

Given that,

Distance (OC) = 5 cm

Radius of circle (OA) = 10 cm

In by using Pythagoras theorem

AC² + OC² = OA²

AC² + 5² = 10²

AC² = 100 – 25

AC² = 75

AC = 8.66 cm

We know that,

The perpendicular from centre to chord bisects the chord

Therefore,

AC = BC = 8.66 cm

Then,

Chord AB = 8.66 + 8.66

= 17.32 cm

Radius of circle (OA) = 6 cm

Distance (OC) = 4 cm

In , by using Pythagoras theorem

AC² + OC² = OA²

AC² + 4² = 6²

AC² = 36 – 16

AC² = 20

AC = 4.47 cm

We know that,

The perpendicular distance from centre to chord bisects the chord

AC = BC = 4.47 cm

Then,

AB = 4.47 + 4.47

= 8.94 cm

Let r be the radius of the given circle and its center be O. Draw OM ⊥ AB and ON⊥ CD.

Since, OM perpendicular AB, ON perpendicular CD.

and AB||CD

Therefore, points M, O and N are collinear.

So, MN = 6cm

Let, OM = x cm.

Then,ON = (6 - x)cm.

Join OA and OC.

Then OA = OC = r

As the perpendicular from the centre to a chord of the circle bisects the chord.

∴ AM = BM = 1/2 AB

= 1/2 x 5

= 2.5cm

CN = DN = 1/2CD

= 1/2 x 11

= 5.5cm
In right triangles OAM and OCN, we have,

OA^{2 =} OM²+ AM²and OC²= ON² + CN²







From (i) and (ii), we have






⇒ 4x² + 25 = 144 + 4x²- 48x + 121
⇒ 48x = 240
⇒ x = 240/48

⇒ x = 5
Putting the value of x in euation (i), we get
r² = 5² + (5/2)²
⇒ r² = 25 + 25/4
⇒ r²= 125/4
⇒ r = 5√5/2 cm

Steps of construction:

(i) Take three points A, B and C on the given circle

(ii) Join AB and BC

(iii) Draw the perpendicular bisectors of chord AB and BC which intersect each other at O

(iv) Point O will be required circle because we know that perpendicular bisector of chord always passes through centre.

Given that,

C is the mid-point of chord AB

To prove: D is the mid-point of arc AB

Proof: In

OA = OB (Radius of circle)

AC = OC (Common)

AC = BC (C is the mid-point of AB)

Then,

(By SSS congruence rule)

∠AOC = ∠BOC (By c.p.c.t)

m (A) = m (B)

A B

Here, D is the mid-point of arc AB.

Given that,

PQ is a diameter of circle which bisects chord AB to C

To prove: PQ bisects ∠AOB

Proof: In ,

OA = OB (Radius of circle)

OC = OC (Common)

AC = BC (Given)

Then,

(By SSS congruence rule)

∠AOC = ∠BOC (By c.p.c.t)

Hence, PQ bisects ∠AOB.

Steps of construction:

(i) Take three points A, B and C on the given arc

(ii) Join AB and BC

(iii) Draw the perpendicular bisectors of chord AB and BC which intersect each other at point O. Then, O will be the required centre of the required circle.

(iv) Join OA

(v) With centre O and radius OA, complete the circle

Suppose two circles intersect in three points A, B and C. Then A, B, C are non-collinear. So, a unique circle passes through these three points. This is contradiction to the face that two given circles are passing through A, B, C. Hence, two circles cannot intersect each other at more than two points.

(i) Draw a line segment AB of 5 cm

(ii) Draw the perpendicular bisector of AB

(iii) With centre A and radius of 4 cm draw an arc which intersects the perpendicular bisector at point O. O will be the required centre.

(iv) Join OA

(v) With centre O and radius OA draw a circle.

No, we cannot draw a circle of radius 2 cm passing through A and B because when we draw an arc of radius 2 cm with centre A, the arc will not intersect the perpendicular bisector and we will not find the centre.

Let ABC be an equilateral triangle of side 9 cm

Let, AD be one of its medians and G be the centroids of the triangle ABC

Then,

AG: GD = 2: 1

We know that,

In an equilateral triangle centroid coincides with the circumcentre

Therefore,

G is the centre of the circumference with circum radius GA

Also, G is the centre and GD is perpendicular to BC

Therefore,

In right triangle ADB, we have

AB² = AD² + DB²

9² = AD² + DB²

AD = cm

Therefore,

Radius = AG = AD

= 3 cm

Steps of construction:

(i) Take three points A, B, C on the given arc

(ii) Join AB and BC

(iii) Draw the perpendicular bisectors of chords AB and BC which intersect each other at point O. Then, O will be the required centre of the required circle.

(iv) Join OA

(v) With centre O and radius OA, complete the circle

Each pair of circles have 0, 1 or 2 points in common. The maximum number of points in common is 2.

Steps of construction:

(i) Take three points A, B and C in the given circle.

(ii) Join AB and BC

(iii) Draw the perpendicular bisector of chord AB and BC which intersect each other at O

(iv) Point O will be the required centre of the circle because we know that, the perpendicular bisector of the chord always passes through the centre

Draw OM perpendicular to AB and ON perpendicular to CD

Join OB and OC

BM = = (Perpendicular from centre bisects the chord)

ND = =

Let,

ON be r so OM will be (6 – x)

In ,

OM² + MB² = OB²

(6 – x)² + ()² = OB²

36 + x² – 12x + = OB² (i)

In

ON² + ND² = OD²

x² + ()² = OD²

x² + = OD² (ii)

We have,

OB = OD (Radii of same circle)

So from (i) and (ii), we get

36 + x² + 2x + = x² +

12x = 36 + -

= =

= 12

From (ii), we get

(1)² + () = OD²

OD² = 1 +

=

OD =

So, radius of circle is found to be cm

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

OM = 4 cm

MB = =

= 3 cm

In

OM² + MB² = OB²

(4)² + (3)² = OB²

16 + 9 = OB²

OB² = 25

OB = 5 cm

In

OD = OB = 5 cm (Radii of same circle)

ND = =

= 4 cm

ON² + ND² = OD²

ON²+ (4)² = (5)²

ON² = 25 – 16

= 9

ON = 3

SO, distance of bigger chord from circle is 3 cm.

Given that,

AB = BC = CA

So, ABC is an equilateral triangle

OA = 40 cm (Radius)

Medians of equilateral triangle pass through the circumference (O) of the equilateral triangle ABC

We also know that,

Median intersects each other at 2: 1 as AD is the median of equilateral triangle ABC, we can write:

=

=

OD = 20 m

Therefore,

AO = OA + OD

= 40 + 20

= 60 m

In

By using Pythagoras theorem

AC² = AO² + DC²

AC² = (60)² + ()²

AC² = 3600 +

AC² = 3600

² = 4800

m

So, length of string of each phone will be m

∠APB = 50^o (Given)

By degree measure theorem,

∠AOB = ∠APB

∠APB = 2 * 50

= 100^o

Since,

OA = OB (Radii)

Hence,

∠OAB = ∠OBA (Angle opposite to equal sides are equal)

Let,

∠OAB = x

In Triangle OAB,

∠OAB + ∠OBA + ∠AOB = 180^o

x + x + 100^o = 180^o

2x = 80^o

x = 40^o

∠OAB = ∠OBA = 40^o

We have,

∠AOC = 150^o

Therefore,

∠AOC + Reflex ∠AOC = 360^o

Reflex ∠AOC = 210^o

2 ∠ABC = 210^o (By degree measure theorem)

∠ABC = 105^o

We have,

∠AOB = 80^o

∠AOC = 110^o

∠AOB + ∠AOC + ∠BOC = 360^o (Complete angle)

80^o + 110^o + ∠BOC = 360^o

∠BOC = 170^o

By degree measure theorem,

∠BOC = 2 ∠BAC

170^o = 2 ∠BAC

∠BAC = 85^o

(i) ∠AOC = 135^o

Therefore,

∠AOC + ∠BOC = 180^o (Linear pair)

135^o + ∠BOC = 180^o

∠BOC = 45^o

By degree measure theorem,

∠BOC = 2 ∠COB

45^o = 2x

x = 22 ^o

(ii) We have,

∠ABC = 40^o

∠ACB = 90^o (Angle in semi-circle)

In triangle ABC, by angle sum property

∠CAB + ∠ACB + ∠ABC = 180^o

∠CAB + 90^o + 40^o = 180^o

∠CAB = 50^o

Now,

∠COB = ∠CAB (Angle on same segment)

x = 50^o

(iii) We have,

∠AOC = 120^o

BY degree measure theorem,

∠AOC = 2 ∠APC

120^o= 2 ∠APC

∠APC = 60^o

Therefore,

∠APC + ∠ABC = 180^o (Opposite angles of cyclic quadrilateral)

60^o + ∠ABC = 180^o

∠ABC = 120^o

∠ABC + ∠DBC = 180^o (Linear pair)

120^o + x = 180^o

x = 60^o

(iv) We have,

∠CBD = 65^o

Therefore,

∠ABC + ∠CBD = 180^o (Linear pair)

∠ABC + 65^o = 180⁰

∠ABC = 115^o

Therefore,

Reflex ∠AOC = 2 ∠ABC (By degree measure theorem)

x = 2 * 115^o

= 230^o

(v) We have,

∠OAB = 35^o

Then,

∠OBA = ∠OAB = 35^o (Angle opposite to equal sides are equal)

In triangle AOB, by angle sum property

∠AOB + ∠OAB + ∠OBA = 180^o

∠AOB + 35^o + 35^o =180^o

∠AOB = 110^o

Therefore,

∠AOB + Reflex ∠AOB = 360^o (Complete angle)

110^o + Reflex ∠AOB = 360^o

Reflex ∠AOB = 250^o

By degree measure theorem,

Reflex ∠AOB = 2 ∠ACB

250^o= 2x

x = 125^o

(vi) We have,

∠AOB = 60^o

By degree measure theorem,

∠AOB = 2 ∠ACB

60^o = 2 ∠ACB

∠ACB = 30^o

x = 30^o

(vii) We have,

∠BAC = 50^o

∠DBC = 70^o

Therefore,

∠BDC = ∠BAC = 50^o (Angles on same segment)

In triangle BDC, by angle sum property

∠BDC + ∠BCD + ∠DBC = 180^o

50^o + x + 70^o = 180^o

120^o + x = 180^o

x = 60^o

(viii) We have,

∠DBO = 40^o

∠DBC = 90^o (Angle in semi-circle)

Therefore,

∠DBO + ∠OBC = 90^o

40^o + ∠OBC = 90^o

∠OBC = 50^o

By degree measure theorem,

∠AOC = 2 ∠OBC

x = 2 * 50^o

x = 100^o

(ix) In triangle DAB, by angle sum property

∠ADB + ∠DAB + ∠ABD = 180^o

32^o + ∠DAB + 50^o= 180^o

∠DAB = 98^o

Now,

∠DAB + ∠DCB = 180^o (Opposite angle of cyclic quadrilateral)

98^o + x = 180^o

x = 180^o – 98^o

= 82^o

(x) We have,

∠BAC = 35^o

∠BDC = ∠BAC = 35^o (Angle on same segment)

In triangle BCD, by angle sum property

∠BDC + ∠BCD + ∠DBC = 180^o

30^o + x + 65^o = 180^o

x = 80^o

(xi) We have,

∠ABD = 40^o

∠ACD = ∠ABD = 40^o (Angle on same segment)

In triangle PCD, by angle sum property

∠PCD + ∠CPD + ∠PDC = 180^o

40^o + 110^o + x = 180^o

x = 30^o

(xii) Given that,

∠BAC = 52^o

∠BDC = ∠BAC = 52^o (Angle on same segment)

Since, OD = OC

Then,

∠ODC = ∠OCD (Opposite angles to equal radii)

x = 52^o

Given that,

O is the circumcentre of triangle ABC and OD perpendicular BC

To prove: ∠BOD = ∠A

Proof: In triangle OBD and triangle OCD, we have

∠ODB = ∠ODC (Each 90^o)

OB = OC (Radii)

OD = OD (Common)

By R.H.S rule,

∠BOD = ∠COD (By c.p.c.t) (i)

By degree measure theorem,

∠BOC = 2 ∠BAC

2 ∠BOD = 2 ∠BAC [From (i)]

∠BOD = ∠BAC

Hence, proved

Given that,

BO is the bisector of ∠ABC

To prove: AB = BC

Proof: ∠ABO = ∠CBO (BO bisector of ∠ABC) (i)

OB = OA (Radii)

Therefore,

∠ABO = ∠DAB (Opposite angle to equal sides are equal) (ii)

OB = OC (Radii)

Therefore,

∠CBO = ∠OCB (Opposite angles to equal sides are equal) (iii)

Compare (i), (ii) and (iii)

∠OAB = ∠OCB (iv)

In triangle OAB and OCB, we have

∠OAB = ∠OCB [From (iv)]

∠OBA = ∠OBC (Given)

OB = OB (Common)

By AAS congruence rule

(By c.p.c.t)

Hence, proved

We have,

∠3 = ∠4 (Angle on same segment)

By degree measure theorem,

∠x = 2 ∠3

∠x = ∠3 + ∠3

∠x = ∠3 + ∠4 (i) (Therefore, ∠3 = ∠4)

But,

∠y = ∠3 + ∠1 (By exterior angle property)

∠3 = ∠y - ∠1 (ii)

From (i) and (ii),

∠x = ∠y - ∠1 + ∠4

∠x = ∠y + ∠4 - ∠1

∠x = ∠y + ∠z + ∠1 - ∠1 (By exterior angle property)

∠x = ∠y + ∠z

Hence, proved

By degree measure theorem,

∠AOB = 2 ∠ACB

130^o = 2 ∠ACB

∠ACB = 65^o

Therefore,

∠ACB + ∠BCD = 180^o (Linear pair)

65^o + ∠BCD = 180^o

∠BCD = 115^o

By degree measure theorem,

Reflex ∠BOD = 2 ∠BCD

Reflex ∠BOD = 2 * 115^o

= 230^o

∠BOD + Reflex ∠BOD = 360^o (Complete angle)

230^o + x = 360^o

x = 130^o

Since,

PQ is diameter

Then,

∠PRQ = 90^o (Angle in semi-circle)

Therefore,

∠PRQ + ∠TRQ = 180^o (Linear pair)

90^o + ∠TRQ = 180^o

∠TRQ = 90^o

By degree measure theorem,

∠ROS = 2 ∠RQS

∠RQS = 20^o

In triangle RQT, we have

∠RQT + ∠QRT + ∠RTS = 180^o (By angle sum property)

20^o + 90^o + ∠RTS = 180^o

∠RTS = 70^o

We have,

∠ACB = 40^o

∠DPB = 120^o

∠ADB = ∠ACB = 40^o (Angle on same segment)

In triangle PDB, by angle sum property

∠PDB + ∠PBD + ∠BPD = 180^o

40^o + ∠PBD + 120^o = 180^o

∠PBD = 20^o

Therefore,

∠CBD = 20^o

We have,

Radius OA = Chord AB

OA = OB = AB

Then, triangle OAB is an equilateral triangle

Therefore,

∠AOB = 60^o (Angle of an equilateral triangle)

By degree measure theorem,

∠AOB = 2 ∠APB

60^o = 2 ∠APB

∠APB = 30^o

Now,

∠APB + ∠AQB = 180^o (Opposite angle of cyclic quadrilateral)

30^o + ∠AQB = 180^o

∠AQB = 150^o

Therefore,

Angle by chord AB at minor arc = 150^o

And, by major arc = 30^o

Since,

Triangle ABC is an equilateral triangle

∠BAC = 60^o

∠BAC + ∠BEC = 180^o (Opposite angles of quadrilateral)

60^o + ∠BEC = 180^o

∠BEC = 120^o

We have,

∠PQR = 35^o

∠PQR + ∠PRQ = 35^o (Angle opposite to equal sides)

In triangle PQR, by angle sum property

∠P + ∠Q + ∠R = 180^o

∠P + 35^o + 35^o = 180^o

∠P = 110^o

Now,

∠QSR + ∠QTR = 180^o

110^o + ∠QTR = 180^o

∠QTR = 70^o

Given that,

O is the centre of the circle

We have,

∠BOD = 160^o

By degree measure theorem,

∠BOD = 2 ∠BCD

160^o = 2x

x = 80^o

Therefore,

∠BAD + ∠BCD = 180^o (Opposite angles of Cyclic quadrilateral)

y + x = 180^o

y + 80^o = 180^o

y = 100^o

We have,

∠BCD = 100^o

∠ABD = 70^o

Therefore,

∠DAB + ∠BCD = 180^o (Opposite angles of cyclic quadrilateral)

∠DAB + 100^o = 180^o

∠DAB = 180^o – 100^o

= 80^o

In triangle DAB, by angle sum property

∠ADB + ∠DAB + ∠ABD = 180^o

∠ABD + 80^o + 70^o = 180^o

∠ABD = 30^o

Since, ABCD is a cyclic quadrilateral with AD ‖ BC

Then,

∠A + ∠C = 180^o (i) (Opposite angles of cyclic quadrilateral)

And,

∠A + ∠B = 180^o (ii) (Co. interior angles)

Comparing (i) and (ii), we get

∠B = ∠C

Hence, proved

Given that,

∠BOC = 100^o

By degree measure theorem,

∠AOC = 2 ∠APC

100^o = 2 ∠APC

∠APC = 50^o

Therefore,

∠APC + ∠ABC = 180^o (Opposite angles of a cyclic quadrilateral)

50^o + ∠ABC = 180^o

∠ABC = 130^o

Therefore,

∠ABC + ∠CBD = 180^o (Linear pair)

130^o + ∠CBD = 180^o

∠CBD = 50^o

Given that,

∠OBD = 50^o

Since,

AB and CD are the diameters of the circles and O is the centre of the circle

Therefore,

∠PBC = 90^o (Angle in the semi-circle)

∠OBD + ∠DBC = 90^o

50^o + ∠DBC = 90^o

∠DBC = 40^o

By degree measure theorem,

∠AOC = 2 ∠ABC

∠AOC = 2 * 40^o

= 80^o

We have,

∠CAB = 30^o

∠ACB = 90^o (Angle in semi-circle)

IN triangle ABC, by angle sum property

∠CAB + ∠ACB + ∠ABC = 180^o

30^o + 90^o + ∠ABC = 180^o

∠ABC = 60^o

Given that,

∠B = 70^o

Since, ABCD is a cyclic quadrilateral

Then,

∠B + ∠D = 180^o

70^o + ∠D = 180^o

∠D = 110^o

Since, AB ‖ DC

Then,

∠B + ∠C = 180^o (Co. interior angle)

70^o + ∠C = 180^o

∠C = 110^o

Now,

∠A + ∠C = 180^o (Opposite angles of cyclic quadrilateral)

∠A + 110^o = 180^o

∠A = 70^o

WE have,

∠A = 3 ∠C

Let, ∠C = x

Therefore,

∠A + ∠C = 180^o (Opposite angles of cyclic quadrilateral)

3x + x = 180^o

4x = 180^o

x = 45^o

∠A = 3x

= 3 * 45^o

= 135^o

Therefore,

∠A = 135^o

We have,

∠DAB = 50^o

By degree measure theorem

∠BOD = 2 ∠BAD

x = 2 * 50^o

= 100^o

Since, ABCD is a cyclic quadrilateral

Then,

∠A + ∠C = 180^o

50^o + y = 180^o

y = 130^o

By using angle sum property in triangle ABC,

∠B = 180^o – (60^o + 20^o)

= 100^o

In cyclic quadrilateral ABCD, we have

∠B + ∠D = 180^o

∠D = 180^o – 100^o

= 100^o

Since, ABC is an equilateral triangle

Then,

∠BAC = 60^o

Therefore,

∠BDC = ∠BAC = 60^o (Angles in the same segment)

Since, quadrilateral ABEC is a cyclic quadrilateral

Then,

∠BAC + ∠BEC = 180^o

60^o + ∠BEC= 180^o

∠BEC = 180^o – 60^o

= 120^o

We have,

∠AEC = 30^o

Since, quadrilateral ABCE is a cyclic quadrilateral

Then,

∠BAC + ∠AEC = 180^o

x + 30^o= 180^o

x = 150^o

By degree measure theorem,

∠AOC = 2 ∠AEC

y = 2 * 30^o

= 60^o

Therefore,

∠ADC = ∠AEC (Angles in same segment)

z = 30^o

We have,

∠BAD = 78^o

∠DCF = x^o

∠DEF = y^o

Since, ABCD is a cyclic quadrilateral

∠BAD + ∠BCD = 180^o

78^o + ∠BCD = 180^o

∠BCD = 102^o

Now,

∠BCD + ∠DCF = 180^o (Linear pair)

102^o = x – 180^o

x = 78^o

Since,

DCEF is a cyclic quadrilateral

x + y = 180^o

78^o + y = 180^o

y = 102^o

WE have,

∠A - ∠C = 60^o (i)

Since, ABCD is a cyclic quadrilateral

Then,

∠A + ∠C = 180^o (ii)

Adding (i) and (ii), we get

∠A - ∠C + ∠A + ∠C = 60^o + 180^o

2 ∠A = 240^o

∠A = 120^o

Put value of ∠A in (ii), we get

120^o + ∠C = 180^o

∠C = 60^o

∠FDC + ∠CDA = 180^o (Linear pair)

80^o + ∠CDA = 180^o

∠CDA = 100^o

Since, ABCD is a cyclic quadrilateral

∠ADC + ∠ABC = 180^o

100^o + ∠ABC = 180^o

∠ABC = 80^o

Now,

∠ABC + ∠ABF = 180^o (Linear pair)

80^o + x = 180^o

x = 180^o – 80^o

= 100^o

(i) Since, ABCD is a cyclic quadrilateral

Then,

∠ABC + ∠ADC = 180^o

∠ABC + 110^o = 180^o

∠ABC = 70^o

Since, AD ‖ BC

Then,

∠DAB + ∠ABC = 180^o (Co. interior angle)

∠DAC + 50^o + 70^o = 180^o

∠DAC = 180^o – 120^o

= 60^o

(ii) ∠BAC = ∠BDC = 40^o (Angle in the same segment)

In by angle sum property

∠DBC + ∠BCD + ∠BDC = 180^o

80^o + ∠BCD + 40^o = 180^o

∠BCD = 60^o

(iii) Given that,

Quadrilateral ABCD is a cyclic quadrilateral

Then,

∠BAD + ∠BCD = 180^o

∠BAD = 80^o

In triangle ABD, by angle sum property

∠ABD + ∠ADB + ∠BAD = 180^o

70^o + ∠ADB + 80^o = 180^o

∠ADB = 30^o

Let ABCD be a cyclic quadrilateral and let O be the centre of the corresponding circle

Then, each side of the equilateral ABCD is a chord of the circle and the perpendicular bisector of a chord always passes through the centre of the circle

So, right bisectors of the sides of the quadrilateral ABCD will pass through the centre O of the corresponding circle.

Let O be the circle circumscribing the cyclic rectangle ABCD.

Since, ∠ABC = 90^o and AC is the chord of the circle. Similarly, BD is a diameter

Hence, point of intersection of AC and BD is the centre of the circle.

Let ABCD be a rhombus such that its diagonals AC and BD intersects at O

Since, the diagonals of a rhombus intersect at right angle

Therefore,

∠ACB = ∠BOC = ∠COD = ∠DOA = 90^o

Now,

∠AOB = 90^o = circle described on AB as diameter will pass through O

Similarly, all the circles described on BC, AD and CD as diameter pass through O.

Given that,

ABCD is cyclic quadrilateral in which AB = DC

To prove: AC = BD

Proof: In and ,

AB = DC (Given)

∠BAP = ∠CDP (Angles in the same segment)

∠PBA = ∠PCD (Angles in the same segment)

Then,

(i) (By c.p.c.t)

(ii) (By c.p.c.t)

Adding (i) and (ii), we get

PA + PC = PD + PB

AC = BD

Given that, ABCD is a cyclic quadrilateral in which

(i) Since,

EA = ED

Then,

∠EAD = ∠EDA (i) (Opposite angles to equal sides)

Since, ABCD is a cyclic quadrilateral

Then,

∠ABC + ∠ADC = 180^o

But,

∠ABC + ∠EBC = 180^o (Linear pair)

Then,

∠ADC = ∠EBC (ii)

Compare (i) and (ii), we get

∠EAD = ∠EBC (iii)

Since, corresponding angles are equal

Then,

BC ‖ AD

(ii) From (iii), we have

∠EAD = ∠EBC

Similarly,

∠EDA = ∠ECB (iv)

Compare equations (i), (iii) and (iv), we get

∠EBC = ∠ECB

EB = EC (Opposite angles to equal sides)

Since,

AB is a diameter

Then,

∠ADB = 90^o (i) (Angle in semi-circle)

Since,’

AC is a diameter

Then,

∠ADC = 90^o (ii) (Angle in semi-circle)

Adding (i) and (ii), we get

∠ADB + ∠ADC = 90^o + 90^o

∠BDC = 180^o

Then, BDC is a line

Hence, the circles on any two sides intersect each other on the third side.

Given that,

∠ACB is an angle in minor segment

To prove: ∠ACB > 90^o

Proof: By degree measure theorem

Reflex ∠AOB = 2 ∠ACB

And,

Reflex ∠AOB > 180^o

Then,

2 ∠ACB > 180^o

∠ACB >

∠ACB > 90^o

Hence, proved

Given that,

∠ACB is an angle in major segment

To prove: ∠ACB > 90^o

Proof: By degree measure theorem,

∠AOB = 2 ∠ACB

And,

∠AOB < 180^o

Then,

2 ∠ACB < 180^o

∠ACB < 90^o

Hence, proved

Given that,

ABCD is a cyclic trapezium with AD ‖ BC and ∠B = 70^o

Since, ABCD is a quadrilateral

Then,

∠B + ∠D = 180^o

70^o + ∠D = 180^o

∠D = 110^o

Since, AD ‖ BC

Then,

∠A + ∠B = 180^o (Co. interior angle)

∠A + 70^o = 180^o

∠A = 110^o

Since, ABCD is a cyclic quadrilateral

Then, ∠A + ∠C = 180^o

110^o + ∠C = 180^o

∠C = 70^o

Let, triangle ABC be a right angle triangle at ∠B

Let P be the mid-point of hypotenuse AC

Draw a circle with centre P and AC as diameter

Since,

∠ABC = 90^o

Therefore, the circle passes through B

Therefore,

BP = Radius

Also,

AP = CP = Radius

Therefore,

AP = BP = CP

Hence, BP = AC

Since angles on the same segment of a circle are equal

Therefore,

∠CAD = ∠DBC = 55^o

∠DAB = ∠CAD + ∠BAC

= 55^o + 45^o

= 100^o

But,

∠DAB + ∠BCD = 180^o (Opposite angles of a cyclic quadrilateral)

Therefore,

∠BCD = 180^o – 100^o

∠BCD = 80^o

O is the centre of the smaller circle.

∠APB = 70°

By degree measure theorem,

∠AOB = 2 ∠APB

∠AOB = 2 × 70°

= 140°

Therefore,

AOBC is a cyclic quadrilateral

∠ACB + ∠AOB = 180°

∠ACB + 140° = 180°

∠ACB = 40°

Let AB be the chord of length 16cm.

Given that,

Distance from centre to the chord AB is OC = 15 cm

Now,

OC ⊥ AB

Therefore,

AC = CB (Since perpendicular drawn from centre of the circle bisects the chord)

Therefore,

AC = CB = 8 cm

In right ΔOCA,

OA² = AC² + OC²

= 8² + 15²

= 225 + 64

= 289

OA = 17 cm

Thus, the radius of the circle is 17 cm

∠AO'B = 50^o

Since, both the triangle are congruent so their corresponding angles are equal.

∠AOB = AO'B = 50°

Now,

∠APB =

∠APB =

= 25°

Let, O be the centre of the circle with chord AB = 8cm

And,

OC be the perpendicular bisector of AC

AO = 6cm

AC = 4cm

In AOC,

OA² = AC² + OC²

6² = 4² + OC²

OC² = 20

OC = 25

∠DBA = ∠DCA = 58° (Angles on the same segment)

In triangle DCA

∠DCA + ∠CDA + ∠DAC = 180°

58° + 77° + ∠DAC = 180°

∠DAC = 45°

∠DPC = 180° - 58° - 30°

= 92°

Let, O be the centre of the circle and r be the radius

Sin A =

=

=

Sin A =

Sin A = Sin 30°

A = 30°

Therefore,

∠BAO = ∠CAO = 30°

2 ∠OAB = 100°

∠OAB = 50°

Therefore,

∠OAB = ∠OBA = 50°

∠AOB = 2 ∠BCA (Angle subtended by any point on circle)

80° = 2 ∠BCA

∠BCA = 40°

Now,

In triangle ABC

∠A + ∠B + ∠C = 180°

∠A + 30^o + 40^o = 180°

∠A = 110°

∠CAB = ∠CAO + ∠OAB

110° = ∠CAO + 50°

∠CAO = 60°

ABCD is a cyclic quadrilateral

∠ADB = 30°

∠DCA = 80°

∠ADB = ∠ACB = 30° (Angle on the same segment)

Now,

∠BCD = ∠ACB + ∠DCA

= 30° + 80°

= 110°

∠OAB + ∠BCD = 180°

∠OAB + 110° = 180°

∠OAB = 70°

∠OBC + ∠CBA = ∠OBA

∠OBC + 30° = 50°

∠OBC = 20°

∠OBC + ∠BAC = ∠OBC + ∠CAB

= 20° + 110°

= 130°

Let AB and CD be two chords of the circle.

Draw OM perpendicular to AB and ON = CD

AB = 14 cm

OM = 6 cm

ON = 2 cm

Let,

CD = x

In AOM,

AO² = AM² + OM²

= 7² + 6²

AO² = 85 (i)

In CON,

CO² = ON² + CN²

CO² = 4 + (ii)

We Know,

AO = CO

AO² = CO²

85 = 4 +

x² = 324

x = 18 cm

It must be greater than 5cm.

Given that,

Arc AXB = Arc BYC (i)

Since,

Arc AXBYC is the arc equal to half circumference

And,

Angle subtended by half circumference at centre is 180°

Arc AXBYC = Arc AXB + Arc BYC

Arc AXBYC = Arc BYC + Arc BYC

Arc AXBYC = Arc AXBYC

Now,

∠BOC = ∠AOC

∠BOC = * 180°

∠BOC = 120°

Given: AC = radius = 5 cm

AB = 4 cm

DC is a chord passing B and C

In ABC

AC² = AB² + BC²

BC² = 9

BC = 3 cm

CD = 2 BC

= 6 cm

Given that,

A is the centre of the circle, then

AB = AD

ABCD is a parallelogram, then

AD ‖ BC, AB ‖ CD

CDE is a straight line, then

AB ‖ CE

Let,

∠BEC = ∠ABE = x’ (Alternate angle)

We know that,

The angle substended by an arc of a circle at the centre double the angle are angle substended by it at any point on the remaining part of circle

∠BAD = 2 ∠BEC

∠BAD = 2x’

In a rhombus opposite angles are equal to each other

∠BAD = ∠BCD = 2x’

Now, we have to find

=

=

=

Hence,

∠BCD: ∠ABE is 2: 1

In triangle ABQ,

∠ABQ + ∠AQB + ∠BAQ = 180^o

∠ABQ + 25^o + 35^o = 180^o

∠ABQ = 120^o

∠APB = 90^o (Angle in the semi-circle)

In triangle APB,

∠APB + ∠PBA + ∠PAB = 180^o

90^o + ∠PBA + 35^o = 180^o

∠PBA = 55^o

Now,

∠PBR = ∠PBA + ∠PBR

∠PBR = 55^o + (180^o – 120^o)

∠PBR = 115^o

Thus,

∠PBR = 115^o

We can't say that,

∠AOB = 60°, 90° or 120°

So, angle AOB is none of these.

We know that,

∠ACB =

∠ACB =

∠ACB = 75^o

Since,

ACD is a straight line, so

∠ACB + ∠BCD = 180^o

75^o + ∠BCD = 180^o

∠BCD = 180^o – 75^o

= 105^o

Now,

∠BCD = Reflex ∠BQD

105^o = (360^o - ∠BQD)

210^o = 360^o - ∠BQD

∠BQD = 360^o – 210^o

Therefore,

∠BQD = 150^o

The centre O lies in the interior of S

∠ADE = 95^o (Given)

Since,

OA = OB, so

∠OAB = ∠OBA

∠OAB = 30^o

∠ADE + ∠ADC = 180^o (Linear pair)

95^o + ∠ADC = 180^o

∠ADC = 85^o

We know that,

∠ADC = 2 ∠ADC

∠ADC = 2 * 85^o

∠ADC = 170^o

Since,

AO = OC (Radii of circle)

∠OAC = ∠OCA (Sides opposite to equal angle) (i)

In triangle OAC,

∠OAC + ∠OCA + ∠AOC = 180^o

∠OAC + ∠OAC + 170^o = 180^o [From (i)]

2 ∠OAC = 10^o

∠OAC = 5^o

Thus,

∠OAC = 5^o

Arc ACB = 3 arc AB (Given)

Central angle = 270°

Degree measures of the two arcs are 90°

Arc AB = 260° (Given)

Let a point C on the circle

We Know that,

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

∠ACB = ∠AOB

∠ACB = * 100

= 50°

Given that O is the centre of circle.

Triangle ABC is an equilateral triangle

∠A = ∠B = ∠C = 60°

We Know that,

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

∠BOC = 2 ∠BAC

∠BOC = 2 ∠A

∠BOC = 120°

Given: O is the centre of circle

AB and CD are diameters of the circle

AO = BO = CO = DO (Radius of the circle)

In right angle AOC,

Cos A =

Cos A =

Cos A = (i)

∠OMA = 90°

∠AOM = ∠MAO = 45°

Using value of angle A in (i)

Cos 45° =

=

AC = AB

Let O and O' be the centre of two circles​

OA and O'A = Radius of the circles​

AB be the common chord of both the circles

OM perpendicular to AB

And,

O'M perpendicular to AB

AOO' is an equilateral triangle.

AM = Altitude of AOO'

Height of AOO' = r

AB = 2 AM

= 2 r

= r

AB is a chord of circle P and Q are two points on circle

∠APB = ∠AQB (Angles on the same segment)

Let AB and CD are two diameters of circle

∠AOD = ∠BOD = ∠BOC = ∠AOC = 90°

AB and CD are diagonals of quadrilateral ABCD

They intersect each other at right angles

And

AB = BC = CD = DA

We know that,

Sides of a square are equal and diagonals intersect at 90°

Therefore, ABCD is a square

∠ABC= 135°

ABC is an arc

Circumference= 360°

Arc = 135°

=

=

Let AB be the chord of circle equal to radius r

OA = OB = r (Radii)

Therefore,

OA = OB = AB

OBC is equilateral triangle

Each angle = 60°

Hence,

Angle surrounded by AB at minor arc = (Reflex * ∠AOD)

= (360^o – 60^o)

= 150^o

Given that,

PQRS is a cyclic quadrilateral

∠QPR =67°

∠SPR = 72°

∠SPQ = ∠QPR + ∠SPR

= 67^o + 72^o

= 139°

∠SPQ + ∠QRS = 180° (Opposite angles of cyclic quadrilateral)

139^o + ∠QRS = 180°

∠QRS = 41°

∠BOC = 120°

∠AOC = ∠AOB + ∠BOC

= 90^o + 120^o

= 210°

Now,

∠ABC = * (Reflex ∠AOC)

= (360^o – 210^o)

= 75°

Given that,

AB || CD (Chords on same side of centre)

AO = CO (Radii)

OL and OM perpendicular bisector of CD and AB respectively

CL = LD = 6 cm

AM = MB = 3 cm

LM = 3 cm (Given)

In COL,

CO² = OL² + 6² (i)

In AOM,

AO² = AM² + OM²

= ౩² + (OL + LM)²

= 9 + OL² + 9 + 6O L

OL² = AO² - 18 - 6O L (ii)

Using (ii) in (i),

OL = 3 cm

Putting OL in (i),

AO² =

AO = 35

Given that,

AB || CD (Chords on opposite side of centre)

DO = BO (Radii)

OL and OM perpendicular bisector of CD and AB respectively

LM = 23 cm

AB = 16 cm

In OLB,

OB² = OL² + LB²

OL² = 225

OL = 15 cm

OM = LM - OL

= 8 cm

In OMD,

OD² = OM² + MD²

MD² = 225

MD = 15 cm

Now,

CD = 2 MD = 30 cm

The largest chord in any circle is its diameter.

The minor segment in a circle always forms an obtuse angle.

Only and only a single circle can be drawn passing through any three non collinear points.

We have,

∠AOC = 130°

∠ABC = * (Reflex of AOC)

= * (360^o – 130^o)

= * 230

= 115°

Given: AB and CD are two chords of the circle.

∠APC = 90°

∠ACP = ∠PBD = y (Angles on the same segment)

In ACP,

∠ACP + ∠APC + ∠PAC = 180°

y + 90° + y = 180°

x + y = 90°

We know that,

An angle subtended by an arc at the centre of the circle is double the angle subtended at any point on the circle

∠AOC = 2 ∠ABC

= 2 * 45°

= 90°

Given that,

Chords AD and BC intersect at right angles,

∠DAB = 35°

∠APC = 90°

∠APC + ∠CPD = 180°

90^o + ∠CPD = 180°

∠CPD = 90°

∠DAB = ∠PCD = 35° (Angles on the same segment)

In triangle PCD,

∠PCD + ∠PDC + ∠CPD = 180°

35° + ∠PDC + 90° = 180°

∠PDC = 45°

∠ADC = 45°

∠BDC = 42°

∠ABC = 90° (Angle in a semi-circle)

In ABC,

∠ABC + ∠BAC = 42° (Angles on the same segment)

90^o + 42^o + ∠ACB = 180°

∠ACB = 48°