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Std 9 Mathematics Part 2 · Chapter 6

6. Circle

Solutions to Practice Sets 6.1 to 6.3 and Problem Set 6, with figures, proofs and constructions. Maharashtra State Board (Balbharati), Std 9 Mathematics Part 2 (Geometry).

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Practice Set 6.1

1Distance of chord AB from the centre of a circle is 8 cm. Length of the chord AB is 12 cm. Find the diameter of the circle.
ABOP86r

OP ⊥ chord AB

Let O be the centre and seg OP ⊥ chord AB. Then OP = 8 cm.

A perpendicular drawn from the centre to a chord bisects the chord.
∴ PB = ½ × 12 = 6 cm

In right angled ΔOPB, OB² = OP² + PB² ...... Pythagoras theorem
= 8² + 6² = 64 + 36 = 100 ∴ OB = 10 cm

Radius = 10 cm ∴ diameter = 2 × 10 = 20 cm

∴ Diameter = 20 cm

2Diameter of a circle is 26 cm and length of a chord of the circle is 24 cm. Find the distance of the chord from the centre.
ABOP?1213

OP ⊥ chord AB

Radius OB = 26 ÷ 2 = 13 cm. Let seg OP ⊥ chord AB.

PB = ½ × 24 = 12 cm ...... perpendicular from the centre bisects the chord

In right angled ΔOPB, OP² = OB² − PB² = 13² − 12² = 169 − 144 = 25
∴ OP = 5 cm

∴ Distance of the chord from the centre = 5 cm

3Radius of a circle is 34 cm and the distance of the chord from the centre is 30 cm, find the length of the chord.
ABOP30?34

OP ⊥ chord AB

Let seg OP ⊥ chord AB. OP = 30 cm, OB = 34 cm.

In right angled ΔOPB, PB² = OB² − OP² = 34² − 30² = (34 + 30)(34 − 30) = 64 × 4 = 256
∴ PB = 16 cm

Perpendicular from the centre bisects the chord ∴ AB = 2 × PB = 32 cm

∴ Length of the chord = 32 cm

4Radius of a circle with centre O is 41 units. Length of a chord PQ is 80 units, find the distance of the chord from the centre of the circle.
PQOM

OM ⊥ chord PQ

Let seg OM ⊥ chord PQ. Then PM = MQ = ½ × 80 = 40 units.

In right angled ΔOMQ, OM² = OQ² − MQ² = 41² − 40² = (41 + 40)(41 − 40) = 81
∴ OM = 9 units

∴ Distance of the chord from the centre = 9 units

5In figure 6.9, centre of two circles is O. Chord AB of bigger circle intersects the smaller circle in points P and Q. Show that AP = BQ.
ABPQMO

Fig. 6.9 with construction OM ⊥ AB

Given: Two circles with common centre O; chord AB of the bigger circle meets the smaller circle at P and Q.
To prove: AP = BQ
Construction: Draw seg OM ⊥ chord AB.

Seg AB is a chord of the bigger circle and seg OM ⊥ AB.
∴ AM = BM ...... (I) perpendicular from the centre bisects the chord

Seg PQ is a chord of the smaller circle and seg OM ⊥ PQ.
∴ PM = QM ...... (II)

Subtracting (II) from (I), AM − PM = BM − QM

∴ AP = BQ ...... A − P − M, M − Q − B

6Prove that, if a diameter of a circle bisects two chords of the circle then those two chords are parallel to each other.
ABMCDNOPQ
Given: Seg PQ is a diameter of a circle with centre O; it bisects chord AB at M and chord CD at N.
To prove: chord AB ∥ chord CD

M is the midpoint of chord AB and O, M lie on diameter PQ.
The segment joining the centre and the midpoint of a chord is perpendicular to the chord.
∴ seg OM ⊥ chord AB, that is, line PQ ⊥ chord AB ...... (I)

Similarly, seg ON ⊥ chord CD, that is, line PQ ⊥ chord CD ...... (II)

From (I) and (II), chord AB and chord CD are both perpendicular to line PQ.

∴ chord AB ∥ chord CD ...... lines perpendicular to the same line are parallel

Practice Set 6.2

1Radius of circle is 10 cm. There are two chords of length 16 cm each. What will be the distance of these chords from the centre of the circle?

Take one chord AB with seg OP ⊥ AB. Then PB = ½ × 16 = 8 cm and OB = 10 cm.

In right angled ΔOPB, OP² = OB² − PB² = 10² − 8² = 100 − 64 = 36 ∴ OP = 6 cm

Congruent chords of a circle are equidistant from the centre, so the other chord is also 6 cm away.

∴ Each chord is 6 cm from the centre

2In a circle with radius 13 cm, two equal chords are at a distance of 5 cm from the centre. Find the lengths of the chords.

Take one chord AB with seg OP ⊥ AB. OP = 5 cm, OB = 13 cm.

PB² = OB² − OP² = 13² − 5² = 169 − 25 = 144 ∴ PB = 12 cm

AB = 2 × PB = 24 cm. The chords are equal, so each is 24 cm.

∴ Each chord is 24 cm long

3Seg PM and seg PN are congruent chords of a circle with centre C. Show that the ray PC is the bisector of ∠NPM.
PMNC
Given: Seg PM ≅ seg PN are chords of a circle with centre C.
To prove: Ray PC bisects ∠NPM, that is, ∠CPM ≅ ∠CPN.
Construction: Draw seg CM and seg CN.

In ΔPCM and ΔPCN,

seg PM ≅ seg PN ...... given

seg CM ≅ seg CN ...... radii of the same circle

seg PC ≅ seg PC ...... common side

∴ ΔPCM ≅ ΔPCN ...... S-S-S test

∴ ∠CPM ≅ ∠CPN ...... c.a.c.t.

∴ Ray PC is the bisector of ∠NPM.

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Practice Set 6.3

1Construct ΔABC such that ∠B = 100°, BC = 6.4 cm, ∠C = 50° and construct its incircle.

Draw BC = 6.4 cm; at B draw a ray at 100° and at C a ray at 50°; they meet at A.

Steps of construction:

  1. Construct ΔABC of the given measures.
  2. Draw the bisectors of ∠B and ∠C.
  3. Name the point of intersection of the bisectors as I.
  4. Draw seg IM ⊥ side BC.
  5. Draw a circle with centre I and radius IM. This is the required incircle.
IMABC

Fair figure (reduced in size). Dashed lines are construction lines. ∠B = 100°, ∠C = 50°, BC = 6.4 cm.

Check: the radius of the incircle is about 2.1 cm.

2Construct ΔPQR such that ∠P = 70°, ∠R = 50°, QR = 7.3 cm and construct its circumcircle.

∠Q = 180° − (70° + 50°) = 60°

So draw QR = 7.3 cm with ∠Q = 60° and ∠R = 50°; the rays meet at P.

Steps of construction:

  1. Construct ΔPQR of the given measures.
  2. Draw the perpendicular bisectors of side QR and side PQ.
  3. Name their point of intersection as C.
  4. Join seg CP.
  5. Draw a circle with centre C and radius CP. This is the required circumcircle.
CPQR

Fair figure (reduced in size). Dashed lines are construction lines. QR = 7.3 cm, ∠Q = 60°, ∠R = 50°.

Check: the radius of the circumcircle is about 3.9 cm.

3Construct ΔXYZ such that XY = 6.7 cm, YZ = 5.8 cm, XZ = 6.9 cm. Construct its incircle.

Draw XY = 6.7 cm; with centre X and radius 6.9 cm and with centre Y and radius 5.8 cm draw arcs meeting at Z.

Steps of construction:

  1. Construct ΔXYZ of the given measures.
  2. Draw the bisectors of ∠X and ∠Y.
  3. Name the point of intersection of the bisectors as I.
  4. Draw seg IM ⊥ side XY.
  5. Draw a circle with centre I and radius IM. This is the required incircle.
IMXYZ

Fair figure (reduced in size). Dashed lines are construction lines.

Check: the radius of the incircle is about 1.8 cm.

4In ΔLMN, LM = 7.2 cm, ∠M = 105°, MN = 6.4 cm, then draw ΔLMN and construct its circumcircle.

Draw MN = 6.4 cm; at M draw a ray at 105° and mark L on it with ML = 7.2 cm.

Steps of construction:

  1. Construct ΔLMN of the given measures.
  2. Draw the perpendicular bisectors of side MN and side LM.
  3. Name their point of intersection as C.
  4. Join seg CL.
  5. Draw a circle with centre C and radius CL. This is the required circumcircle.
CLMN

Fair figure (reduced in size). Dashed lines are construction lines.

Check: the radius of the circumcircle is about 5.6 cm. Since ∠M is obtuse, the circumcentre C lies outside the triangle.

5Construct ΔDEF such that DE = EF = 6 cm, ∠F = 45° and construct its circumcircle.

DE = EF ∴ ∠EDF = ∠EFD = 45° ...... isosceles triangle theorem

∴ ∠DEF = 180° − (45° + 45°) = 90°

So draw EF = 6 cm, draw ∠E = 90° at E and mark D with ED = 6 cm.

Steps of construction:

  1. Construct ΔDEF of the given measures.
  2. Draw the perpendicular bisectors of side EF and side DE.
  3. Name their point of intersection as C.
  4. Join seg CD.
  5. Draw a circle with centre C and radius CD. This is the required circumcircle.
CDEF

Fair figure (reduced in size). Dashed lines are construction lines.

Check: ΔDEF is right angled at E, so the circumcentre C is the midpoint of the hypotenuse DF. The radius is ½ DF ≈ 4.2 cm.

Problem Set 6

1Choose correct alternative answer and fill in the blanks.
  • (i) Radius of a circle is 10 cm and distance of a chord from the centre is 6 cm. Hence the length of the chord is — (A) 16 cm
    Half chord = √(10² − 6²) = 8 cm; chord = 16 cm.
  • (ii) The point of concurrence of all angle bisectors of a triangle is called the — (C) incentre
    The incircle is centred at the point where the angle bisectors meet.
  • (iii) The circle which passes through all the vertices of a triangle is called — (A) circumcircle
    The circumcircle passes through all three vertices.
  • (iv) Length of a chord of a circle is 24 cm. If distance of the chord from the centre is 5 cm, then the radius of that circle is — (B) 13 cm
    Radius = √(12² + 5²) = √169 = 13 cm.
  • (v) The length of the longest chord of the circle with radius 2.9 cm is — (D) 5.8 cm
    The longest chord is a diameter = 2 × 2.9 = 5.8 cm.
  • (vi) Radius of a circle with centre O is 4 cm. If l(OP) = 4.2 cm, say where point P will lie. — (C) outside the circle
    OP is greater than the radius.
  • (vii) The lengths of parallel chords which are on opposite sides of the centre of a circle are 6 cm and 8 cm. If radius of the circle is 5 cm, then the distance between these chords is — (D) 7 cm
    Distances from the centre: √(5² − 3²) = 4 cm and √(5² − 4²) = 3 cm. On opposite sides, 4 + 3 = 7 cm.
2Construct incircle and circumcircle of an equilateral ΔDSP with side 7.5 cm. Measure the radii of both the circles and find the ratio of radius of circumcircle to the radius of incircle.

Construct ΔDSP with DS = SP = PD = 7.5 cm (each angle 60°).

Steps of construction:

  1. Construct ΔDSP of the given measures.
  2. Draw the bisectors of ∠D and ∠S.
  3. Name the point of intersection of the bisectors as I.
  4. Draw seg IM ⊥ side DS.
  5. Draw a circle with centre I and radius IM. This is the required incircle.
IMDSP

Incircle (reduced in size)

Steps of construction:

  1. Construct ΔDSP of the given measures.
  2. Draw the perpendicular bisectors of side DS and side SP.
  3. Name their point of intersection as C.
  4. Join seg CD.
  5. Draw a circle with centre C and radius CD. This is the required circumcircle.
CDSP

Circumcircle (reduced in size)

On measuring: radius of incircle ≈ 2.2 cm, radius of circumcircle ≈ 4.3 cm.

Ratio = 4.33 : 2.17 = 2 : 1

∴ Radius of circumcircle : radius of incircle = 2 : 1

In an equilateral triangle the angle bisectors and perpendicular bisectors coincide, so the incentre and circumcentre are the same point.

3Construct ΔNTS where NT = 5.7 cm, TS = 7.5 cm and ∠NTS = 110° and draw incircle and circumcircle of it.

Draw TS = 7.5 cm; at T draw a ray at 110° and mark N on it with TN = 5.7 cm. Draw seg NS.

Steps of construction:

  1. Construct ΔNTS of the given measures.
  2. Draw the bisectors of ∠T and ∠S.
  3. Name the point of intersection of the bisectors as I.
  4. Draw seg IM ⊥ side TS.
  5. Draw a circle with centre I and radius IM. This is the required incircle.
IMNTS

Incircle (reduced in size)

Steps of construction:

  1. Construct ΔNTS of the given measures.
  2. Draw the perpendicular bisectors of side TS and side NT.
  3. Name their point of intersection as C.
  4. Join seg CN.
  5. Draw a circle with centre C and radius CN. This is the required circumcircle.
CNTS

Circumcircle (reduced in size)

Check: incircle radius ≈ 1.7 cm; circumcircle radius ≈ 5.8 cm. ∠T is obtuse, so the circumcentre lies outside the triangle.

4In the figure 6.19, C is the centre of the circle. Seg QT is a diameter, CT = 13, CP = 5, find the length of chord RS.
QTRSPC513

Fig. 6.19

CR = CT = 13 ...... radii of the same circle

Seg CP ⊥ chord RS.
In right angled ΔCPR, RP² = CR² − CP² = 13² − 5² = 169 − 25 = 144 ∴ RP = 12

Perpendicular from the centre bisects the chord ∴ RS = 2 × RP = 24

∴ RS = 24

5In the figure 6.20, P is the centre of the circle. Chord AB and chord CD intersect on the diameter at the point E. If ∠AEP ≅ ∠DEP then prove that AB = CD.
ABCDEPMN

Fig. 6.20 with construction PM ⊥ AB, PN ⊥ CD

Given: P is the centre; chords AB and CD meet the diameter at E; ∠AEP ≅ ∠DEP.
To prove: AB = CD
Construction: Draw seg PM ⊥ chord AB and seg PN ⊥ chord CD.

In ΔPME and ΔPNE, ∠PME = ∠PNE = 90° and ∠PEM ≅ ∠PEN ...... given

∴ ∠MPE = 90° − ∠PEM = 90° − ∠PEN = ∠NPE ...... sum of angles of a triangle

In ΔPME and ΔPNE,
∠PEM ≅ ∠PEN ...... given
seg PE ≅ seg PE ...... common side
∠MPE ≅ ∠NPE ...... proved above

∴ ΔPME ≅ ΔPNE ...... A-S-A test
∴ PM = PN ...... c.s.c.t.

Chords AB and CD are equidistant from the centre P.
∴ AB = CD ...... chords equidistant from the centre are congruent

6In the figure 6.21, CD is a diameter of the circle with centre O. Diameter CD is perpendicular to chord AB at point E. Show that ΔABC is an isosceles triangle.
ABCDEO

Fig. 6.21

Given: CD is a diameter of the circle with centre O; CD ⊥ chord AB at E.
To prove: ΔABC is isosceles.

Seg OE ⊥ chord AB ...... O and E lie on diameter CD
∴ AE = BE ...... perpendicular from the centre bisects the chord

In ΔCEA and ΔCEB,
seg AE ≅ seg BE ...... proved above
∠CEA ≅ ∠CEB ...... each is a right angle
seg CE ≅ seg CE ...... common side

∴ ΔCEA ≅ ΔCEB ...... S-A-S test
∴ CA = CB ...... c.s.c.t.

∴ ΔABC is an isosceles triangle.

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