1.    (i) Interior

(ii) Exterior

(iii) Diameter

(iv) Semi-circle

(v) Chord

(vi) Three

2.    (i) True

(ii) False

(iii) False

(iv) True

(v) False

(vi) True

3.    I Part: Two circles are said to be congruent if and only if one of them can be superposed on the other so as to cover it exactly.

Let C (O, r) and C (O’, s) be two circles. Let us imagine that the circle C (O’, s) is superposed on C (O, r) so that O’ coincide with O. Then it can easily be seen that C (O’, s) will cover C (O, r) completely if and only if r = s.

Hence we can say that two circles are congruent, if and only if they have equal radii.

II Part: Given: In a circle (O, r), AB and CD are two equal chords, subtend ∠ AOB and ∠ COB at the centre.

To Prove:  AOB = ∠ COD

Proof: In Δ AOB and Δ COD,

AB = CD                                               [Given]

AO = CO                                               [Radii of the same circle]

BO = DO                                               [Radii of the same circle]

∴ Δ AOB ≅ Δ COD                           [By SSS axiom]

⇒ ∠ AOB = ∠ COD                             [By CPCT]                            Hence Proved.

4.   Given: In a circle (O, r), AB and CD subtend two angles at the centre such that Z AOB = Z COD

To Prove: AB = CD

Proof: : In Δ AOB and Δ COD,

AO = CO                                          [Radii of the same circle]

BO = DO                                          [Radii of the same circle]

∠ AOB = ∠ COD                             [Given]

∴ Δ AOB ≅ Δ COD                         [By SAS axiom]

⇒ AB = CD                                      [By CPCT]           Hence proved.

5.    From the figure, we observe that when different pairs of circles are drawn, each pair have two points (say A and B) in common.

Maximum number of common points are two in number.

Suppose two circles C (0, r) and C (0′, s) intersect each other in three points, say A, and C.

Then A, B and C are non-collinear points. We know that:

There is one and only one circle passing through three Therefore a unique circle passes through A, B and C.

⇒ O’ coincides with 0 and s = r.

A contradiction to the fact that C (0′, s) ≠ C (0, r)

∴ Our supposition is wrong.

Hence two different circles cannot intersect each other at more than two points.

6.    

Steps of Construction:

(a) Take any three points A, B and C on the circle.

(b) Join AB and BC.

(c) Draw perpendicular bisector say LM of AB.

(d) Draw perpendicular bisector PQ of BC.

(e) Let LM and PQ intersect at the point O. Then 0 is the centre of the circle.

Verification:

O lies on the perpendicular bisector of AB.

∴ OA = OB             … (i)

O lies on the perpendicular bisector of BC.

∴ OB = OC                  … (ii)

From eq. (i) and (ii), we observe that

OA = OB = OC = r (say)

Three non-collinear points A, B and C are at equal distance (r) from the point O inside the circle.

Hence O is the centre of the circle.

7.    Given: Let C (O, r) and C (O’, r’) be two circles intersecting at A and B. AB is the common chord.

To prove: 00′ is the perpendicular bisector of the chord AB.

Construction: Join OA, OB, O’A, O’B.

Proof: In triangles OAO’ and OBO’,

OA = OB                                              [Each radius]

O’A = O’B                                            [Each radius]

OO’ = OO’                                            [Common]

∴ Δ OAO’ ≅ Δ OBO’                            [By SSS congruency]

⇒ ∠ AOO’ = ∠ BOO’                            [By CPCT]

⇒ ∠ AOM = ∠ BOM

Now in Δ AOB,             OA = OB

And                                ∠ AOB = ∠ OBA               [Proved earlier]

Also                                ∠ AOM = ∠ BOM

∴ Remaining                 ∠ AMO = ∠ BMO

⇒ ∠ AMO = ∠ BMO = 90°                [Linear pair]

⇒ OM ⊥ AB

⇒ OO’ ⊥ AB

 Since OM ⊥ AB

∴ M is the mid-point of AB.

Hence OO’ is the perpendicular bisector of AB.

8.    Let two circles with centres O and O’ intersect each other at points A and B. On joining A and B, AB is a common chord.

Radius OA = 5 cm, Radius O’A = 3 cm,

Distance between their centers OO’ = 4 cm

In triangle AOO’,

52 = 42 + 32

⇒ 25 = 16 + 9

⇒ 25 = 25

Hence AOO’ is a right triangle, right angled at O’.

Since, perpendicular drawn from the center of the circle bisects the chord. Hence 0′ is the mid-point of the chord AB. Also 0′ is the centre of the circle II. Therefore length of chord AB = Diameter of circle II

∴ Length of chord AB = 2 x 3 = 6 cm.

9.    Given: Let AB and CD are two equal chords of a circle of centers O intersecting each other at point E within the circle.

To prove: (a) AE = CE (b) BE = DE

Construction: Draw OM ⊥ AB, ON ⊥ CD. Also join OE.

Proof: In right triangles OME and ONE,

∠ OME = ∠ ONE = 90°

OM = ON

[Equal chords are equidistance from the centre]

OE = OE                             [Common]

∴ Δ OME ≅ Δ ONE           [RHS rule of congruency]

∴ ME = NE                         [By CPCT]                          … (i)

Now, O is the centre of circle and       OM ⊥ AB

∴ AM =  AB        [Perpendicular from the centre bisects the chord]          … (ii)

Similarly,              NC = CD           … (iii)

But          AB = CD                            [Given]

From eq. (ii) and (iii),                  AM = NC       … (iv)

Also                 MB = DN            … (v)

Adding (i) and (iv), we get,

AM + ME = NC + NE

⇒ AE = CE          [Proved part (a)]

Now AB = CD                           [Given]

AE = CE                                [Proved]

⇒ AB – AE = CD – CE

⇒ BE = DE          [Proved part (b)]

10.  Given: AB and CD be two equal chords of a circle with centre 0 intersecting each other with in the circle at point E. OE is joined.

To prove: ∠ OEM = ∠ OEN

Construction: Draw OM ⊥ AB and ON ⊥ CD.

Proof: In right angled triangles OME and ONE,

∠ OME = ∠ ONE         [Each 90°]

OM = ON                      [Equal chords are equidistant from the centre]

OE = OE                       [Common]

∴ Δ OME ≅ Δ ONE         [RHS rule of congruency]

∴ ∠ OEM = ∠ OEN         [By CPCT]

11.   Given: Line l intersects two concentric circles with centre 0 at points A, B, C and D.

To prove: AB = CD

Construction: Draw OL ⊥ l

Proof: AD is a chord of outer circle and OL ⊥ AD.

∴ AL = LD           … (i) [Perpendicular drawn from the centre bisects the chord]

Now,          BC is a chord of inner circle and OL ⊥ BC

∴ BL = LC           … (ii) [Perpendicular drawn from the centre bisects the chord]

Subtracting (ii) from (i), we get,

AL – BL = LD – LC

⇒ AB = CD

12.  Let Reshma, Salma and Mandip takes the position C, A and B on the circles.

Since AB = AC

The centre lies on the bisector of ∠ BAC.

Let M be the point of intersection of BC and OA.

Again, since AB = AC and AM bisects ∠ CAB.

∴ AM ⊥ CB and M is the mid-point of CB.

Let OM = x, then MA = 5 – x

From right angled triangle OMB,      OB2 = OM2 + MB2

⇒ 52 = x2 + MB2                   … (i)

Again, in right angled triangle AMB, AB2 = AM2 + MB2

⇒ 62 = (5 – x)2 + MB2                … (ii)

Equating the value of MB2 from eq. (i) and (ii),

52x2 = 62 – (5 – x)2                            (5 – x)2 = 62 – 52

⇒ 25 – 10x + x2x2 = 36 – 25          =          10x = 25 – 11

∴ BC = 2MB = 2 × 4.8 = 9.6 cm

13.  Let position of three boys Ankur, Syed and David are denoted by the points A, B and C respectively.

∴ A = B = C = x cm                   [say]

Since equal sides of equilateral triangle are as equal chords and perpendicular distances of equal chords of a circle are equidistant from the centre.

∴ OD = OE = OF = x cm             [say]

Join OA, OB and OC.

Area of Δ AOB = Area of Δ BOC = Area of Δ AOC

And        Area of Δ ABC

= Area of Δ AOB + Area of Δ BOC + Area of Δ AOC

And   Area of Δ ABC = 3 × Area of Δ BOC

Now, CE ⊥ BC

Now in right angled triangle BEO,

OE2 + BE2 = OB2                                                 [Using Pythagoras theorem]

⇒ x2 + (√3x)2 = (20)2                    ⇒                          x2 + 3x2 = 400

⇒ 4x2 = 400                                     ⇒                          x2 = 1 00

⇒ x = 10 m

And          a = 2√3x = 2√3 × 10 = 20√3 m

Thus distance between any two boys is 20√3 m.

14.  ∠ AOC = ∠ AOB + ∠ BOC          ⇒        ∠ AOC = 60° +30° = 90°

Now ∠ AOC = 2 ∠ ADC

[∵ Angled subtended by an arc, at the centre of the circle is double the angle subtended by the same arc at any point in the remaining part of the circle]

15.  Let AB be the minor arc of circle.

∴ Chord AB = Radius OA = Radius OB

∴ Δ AOS is an equilateral triangle.

⇒ ∠ AOB = 60°

⇒ ∠ AOB + ∠ BOA = 360°

⇒ 60° + ∠ BOA = 360°

⇒ ∠ BOA = 360° – 60° = 300°

D is a point in the minor arc.

 = 2 BDA

⇒ ∠ BOA = 2 ∠ BDA

Thus angle subtended by major arc,  at any point D in the minor arc is 150°.

Let E be a point in the major arc .

= 2 ∠ AEB

⇒ ∠ AOB = 2 ∠ AEB

16.  In the figure, Q is a point in the minor arc .

 = 2 ∠ PQR                                ⇒              ∠ ROP = 2 ∠ PQR

⇒ ∠ ROP = 2 × 100° = 200°

Now  = 360°                       ⇒              ∠ POR + ∠ ROP = 360°

⇒ ∠ POR + 200° = 360°                     ⇒             ∠ POR = 360° – 200° = 160°

Now          Δ OPR is an isosceles triangle.

∴ OP = OR                       [radii of the circle]

⇒ ∠ OPR = ∠ ORP          [angles opposite to equal sides are equal]

Now in isosceles triangle OPR,

∠ OPR + ∠ ORP + ∠ POR = 180°

⇒ ∠ OPR + ∠ ORP + 160° = 180°      ⇒     2 ∠ OPR =180° – 160°      [Using (i) & (ii)]

⇒ 2 ∠ OPR = 20°                ⇒                         ∠ OPR= 10°

17.  In triangle ABC,

∠ BAC + ∠ ABC + ∠ ACB = 180°

⇒ ∠ BAC + 69° + 31° = 180°                ⇒            ∠ BAC = 180° – 69° – 31°

⇒ ∠ BAC = 80°              … (i)

Since, A and D are the points in the same segment of the circle.

∴ ∠ BDC = ∠ BAC

[Angles subtended by the same arc at any points in the alternate segment of a circle are equal]

⇒ ∠ BDC = 80°   [Using (i)]

18.  Given: ∠ BEC = 130° and ∠ ECD = 20°

∠ DEC = 180° – ∠ BEC = 180° —130° = 50°         [Linear pair]

Now in A DEC,

∠ DEC + ∠ DCE + ∠ EDC = 180°                     [Angle sum property]

⇒ 50°+20°+ ∠ EDC = 180°          ⇒           ∠ EDC = 110°

⇒ ∠ BAC = ∠ EDC = 110°                                        [Angles in same segment]

19.   Here, ∠ DBC = 70° and ∠ BAC = 30°

And        ∠ DAC = ∠ DBC = 70°             [Angles in same circle]

Now         ABCD is a cyclic quadrilateral.

∴ ∠ DAB + ∠ BCD = 180°

[Sum of opposite angles of a cyclic quadrilateral is supplementary]

⇒ 100°+ ∠ BCD = 180°          ⇒         ∠ BCD = 80°

20.  Since AC is a diameter.

∴ ∠ B = ∠ D = 90°                  … (i)

[Angle in semicircle is right angle]

Similarly    ∠ A = ∠ C = 90°          … (ii)

Now        AC = BD                       [Diameters of same circle]

⇒  AD = BC         [Chords opposite to equal arcs]           … (iii)

Similarly                     AB = DC                     … (iv)

From eq. (i), (ii), (iii) and (iv), we observe that each angles of the quadrilateral is 90° and opposite sides are equal.

Hence ABCD is a rectangle.

21.  Given: A trapezium ABCD in which AB || CD and AD = BC.

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

Construction: Draw DE || CB.

Proof: Since DE || CB and EB || DC.

∴ EBCD is a parallelogram.

∴ DE = CB and ∠ DEB = ∠ DCB Now AD = BC and DA = DE

⇒ ∠ DAE = ∠ DEB

But          ∠ DEA + ∠ DEB = 180°

⇒ ∠ DAE + ∠ DCB = 180°                         [∵ ∠ DEA = ∠ DAE and ∠ DEB = ∠ DCB]

⇒ ∠ DAB + ∠ DCB = 180°

⇒ ∠ A + ∠ C = 180°

Hence, ABCD is a cyclic trapezium.

22.  In triangles ACD and QCP,

∠ A = ∠ P and ∠ Q = ∠ D                  [Angles in same segment]

∴ ∠ ACD = ∠ QCP                                  [Third angles]                     … (i)

Subtracting ∠ PCD from both the sides of eq. (1), we get,

∠ ACD – ∠ PCD = ∠ QCP – ∠ PCD

⇒ ∠ ACPO = ∠ QCD.               Hence proved.

23.  Given: Two circles intersect each other at points A and B. AP and AQ be their respective

To prove:           Point B lies on the third side PQ.

Construction: Join A and B.

Proof: AP is a diameter.

∴  ∠ 1 = 90°

[Angle in semicircle]

Also AQ is a diameter.

∴ ∠ 2 = 90°

[Angle in semicircle]

∠ 1 + ∠ 2 = 90° + 90°

⇒ PBQ = 180°

⇒ PBQ is a line.

Thus point B. i.e. point of intersection of these circles lies on the third side i.e., on PQ.

24.  We have ABC and ADC two right triangles, right angled at B and D respectively.

⇒     ∠ ABC = ADC              [Each 90°]

If we draw a circle with AC (the common hypotenuse) as diameter, this circle will definitely passes through of an arc AC, Because B and D are the points in the alternate segment of an arc AC.

Now we have subtending ∠ CBD and ∠ CAD in the same segment.

∴  ∠ CAD = ∠ CBD                  Hence proved.

25.  Let two circles with respective centers A and B intersect each other at points C and D.

We have to prove ∠ ACB = ∠ ADB

Proof: In triangles ABC and ABD,

AC = AD = r

BC = BD = r

AB = AB                [Common]

∴ Δ ABC ≅  Δ ABD

[SSS rule of congruency]

⇒ ∠ ACB = ∠ ADB [By CPCT]

26.  Let O be the centre of the circle.

Join OA and OC.

Since perpendicular from the centre of the circle to the chord bisects the chord.

Let radius of the circle be r.

In right angled triangle AEO,

A02 = AE2 + 0E2

[Using Pythagoras theorem]

Again In right angled triangle CFO,

OC2 = CF2 + OF2

[Using Pythagoras theorem]

27.  Let AB = 6 cm and CD = 8 cm are the chords of circle with centre O.

Join OA and OC.

Since perpendicular from the centre of the circle to the chord bisects the chord.

Perpendicular distance of chord AB from the centre O is OE.

∴ OE = 4 cm.

Now in right angled triangle AOE,

OA2 = AE2 + OE2                                   [Using Pythagoras theorem]

⇒ r2 = 32 + 42                  ⇒                        r2 = 9 + 16 = 25

⇒ r = 5 cm

Perpendicular distance of chord CD from the center O is OF.

Now in right angled triangle OFC,

OC2 = CF2 + OF2                                                [Using Pythagoras theorem]

⇒ r2 = 42 + OF2                           ⇒                    52 = 16 + OF2

⇒ OF2 = 16                                   ⇒                    OF = 3cm

Hence distance of other chord from the centre is 3 cm.

28.  Vertex B of ∠ ABC is located outside the circle with centre O.

Side AB intersects chord CE at point E and side BC intersects chord AD at point D with the circle.

We have to prove that

∠ ABC =  [ A AOC – A DOE]

Join OA, OC, OE and OD.

Now ∠ AOC = 2 ∠ AEC

[Angle subtended by an arc at the centre of the circle is twice the angle subtended by the same arc at any point in the alternate segment of the circle]

Now        ∠ AEC = ∠ ADC          [Angles in same segment in circle]     … (iv)

Also         ∠ DCE = ∠ DAE          [Angles in same segment in circle]    … (v)

Using eq. (iv) and (v) in eq. (iii),

Hence proved.

29.  Let ABCD be a rhombus in which diagonals AC and BD intersect each other at point O.

As we know that diagonals of a rhombus bisect and perpendicular to each other.

∴ ∠ AOB = 90°

And if we draw a circle with side AB as diameter, it will definitely pass through point 0 (the point intersection of diagonals) because then ∠ AOB = 90° will be the angle in a semi-circle.

30.  In figure (a),

ABCD is a parallelogram.

⇒ ∠ 1 = ∠ 3                             … (i)

ABCE is a cyclic quadrilateral.

∴ ∠ 1 + ∠ 6 = 180°                 … (ii)

And ∠ 5 + ∠ 6 = 180°            … (iii)     [Linear pair]

From eq. (ii) and (iii),                             ∠ 1 = ∠ 5        … (iv)

Now, from eq. (i) and (iv),

∠ 3 = ∠ 5        ⇒          AE = AD         [Sides opposite to equal angles are equal]

In figure (b),

ABCD is a parallelogram.

∴ ∠ 1 = ∠ 3 and ∠ 2  = ∠ 4

Also AB || CD and BC meets them.

∴ ∠ 1 + ∠ 2 = 180°              … (i)

And AD || BC and EC meets them.

∴ ∠ 5 = ∠ 2                          … (ii)         [Corresponding angles]

Since ABCE is a cyclic quadrilateral.

∴ ∠ 1 + ∠ 6 = 180° …………….. (iii)

From eq. (i) and (iii),

∠ 1 + ∠ 2 = ∠ 1 + ∠ 6                 = ∠ 2 = ∠ 6

But from eq. (ii),                                                 ∠ 2 = ∠ 5

 ∠ 5 = ∠ 6

Now in triangle AED,

∠ 5 = ∠ 6                      ⇒               AE = AD       [Sides opposite to equal angles]

Hence in both the cases,           AE = AD

31.  Given: AC and BD of a circle bisect each other at O.

Then OA = OC and OB = OD

To prove:   (i) AC and BD are the diameters. In other words, 0 is the centre of the circle.

(ii) ABCD is a rectangle.

Proof: (i) In triangles AOD and BOC,

AO = OC                       [given]

∠ AOD = ∠ BOC         [Vertically opp.]

OD = OB                      [given]

∴ Δ AOD ≅ Δ COB [SAS congruency]

⇒ AD = CB                      [By CPCT]

Similarly                Δ AOB ≅ Δ COD

∴ AC and BD are the diameters as only diameters can bisects each other as the chords of the circle.

(ii) Ac is the diameter. [Proved in (i)]

∠ B = ∠ D = 90°        … (i)            [Angle in semi-circle]

Similarly BD is the diameter.

∠ A = ∠ C = 90°         … (ii)           [Angle in semi-circle]

Now diameters AC = BD

⇒ AD = BC           [Chords corresponding to the equal arcs]      … (iii)

Similarly          AB = DC           … (iv)

From eq. (i), (ii), (iii) and (iv), we observe that each angle of the quadrilateral is 90° and opposite sides are equal.

Hence ABCD is a rectangle.

32.  According to question, AD is bisector of ∠ A.

Since the angles in the same segment of a circle are equal.

∴ ∠ 9 = ∠ 3           [angles subtended by  ]           ..(i)

And    ∠ 8 = ∠ 5           [angles subtended by ]       ..(ii)

Adding both equations,

33.  Given: Two equal circles intersect in A and B.

A straight line through A meets the circles in P and Q.

To prove: BP = BQ

Construction: Join A and B.

Proof: AB is a common chord and the circles are equal.

∴ Arc about the common chord are equal,

i.e., 

Since equal arcs of two equal circles subtend equal angles at any point on the remaining part of the circle, then we have,

∠ 1 = ∠ 2

In triangle PBQ,

∠ 1 = ∠ 2         [proved]

∴ Sides opposite to equal angles of a triangle are equal.

Then we have, BP = BQ.