Std 10 Mathematics Part 2 · Chapter 3

3. Circle

Solutions to Practice Sets 3.1 to 3.5 and Problem Set 3, with figures and proofs. Maharashtra State Board (Balbharati), Std 10 Mathematics Part 2 (Geometry).

Practice Set 3.1

1In the adjoining figure the radius of a circle with centre C is 6 cm, line AB is a tangent at A. Answer the following questions.

Fig. 3.19

(1) What is the measure of ∠ CAB? Why?

∠ CAB = 90°

Reason: a tangent is perpendicular to the radius at the point of contact (tangent theorem).

(2) What is the distance of point C from line AB? Why?

6 cm

Reason: seg CA ⊥ line AB, so the distance of C from line AB is CA, the radius, which is 6 cm.

(3) d(A, B) = 6 cm, find d(B, C).

In Δ CAB, ∠ CAB = 90°.

BC² = CA² + AB² … Pythagoras theorem

= 6² + 6² = 72

∴ BC = √72 = 6√2

∴ d(B, C) = 6√2 cm

(4) What is the measure of ∠ ABC? Why?

∠ ABC = 45°

Reason: in Δ CAB, ∠ A = 90° and CA = AB = 6 cm. So ∠ ACB = ∠ ABC, and each is 45°.

2In the adjoining figure, O is the centre of the circle. From point R, seg RM and seg RN are tangent segments touching the circle at M and N. If OR = 10 cm and radius of the circle = 5 cm, then (1) What is the length of each tangent segment? (2) What is the measure of ∠ MRO? (3) What is the measure of ∠ MRN?

Fig. 3.20

(1)

Draw radius OM. ∠ OMR = 90° … tangent theorem

RM² = OR² − OM² … Pythagoras theorem

= 100 − 25 = 75

∴ RM = √75 = 5√3

RN = RM … tangent segment theorem

∴ Each tangent segment is 5√3 cm

(2)

In Δ OMR, ∠ OMR = 90° and OM = 12 OR (5 = 12 × 10).

∴ ∠ MRO = 30° … converse of 30° − 60° − 90° theorem

∴ ∠ MRO = 30°

(3)

Similarly, ∠ NRO = 30°.

∠ MRN = ∠ MRO + ∠ NRO = 30° + 30° = 60°

∴ ∠ MRN = 60°

3Seg RM and seg RN are tangent segments of a circle with centre O. Prove that seg OR bisects ∠ MRN as well as ∠ MON.

Fig. 3.21

Proof: In Δ OMR and Δ ONR,

seg OM ≅ seg ON … radii of the same circle

seg OR ≅ seg OR … common side

∠ OMR = ∠ ONR = 90° … tangent theorem

∴ Δ OMR ≅ Δ ONR … hypotenuse-side test

∴ ∠ MRO ≅ ∠ NRO and ∠ MOR ≅ ∠ NOR … corresponding angles of congruent triangles

∴ seg OR bisects ∠ MRN as well as ∠ MON.

4What is the distance between two parallel tangents of a circle having radius 4.5 cm? Justify your answer.

Line a ∥ line b are tangents at P and Q.

Let the parallel tangents touch the circle at P and Q.

Tangents at the ends of a diameter are parallel, and parallel tangents touch the circle at the ends of a diameter. So PQ is a diameter.

Seg PQ ⊥ both tangents … tangent theorem

∴ Distance between the tangents = PQ = 2 × 4.5 = 9 cm

∴ The distance is 9 cm

Practice Set 3.2

1Two circles having radii 3.5 cm and 4.8 cm touch each other internally. Find the distance between their centres.

For circles touching internally, distance between centres = difference of radii.

= 4.8 − 3.5 = 1.3

∴ The distance is 1.3 cm

2Two circles of radii 5.5 cm and 4.2 cm touch each other externally. Find the distance between their centres.

For circles touching externally, distance between centres = sum of radii.

= 5.5 + 4.2 = 9.7

∴ The distance is 9.7 cm

3If radii of two circles are 4 cm and 2.8 cm. Draw figure of these circles touching each other - (i) externally (ii) internally.

(i) Externally

Distance between centres = 4 + 2.8 = 6.8 cm.

Draw seg AB = 6.8 cm. Draw a circle with centre A and radius 4 cm, and a circle with centre B and radius 2.8 cm. They touch at T on seg AB.

AB = 6.8 cm, AT = 4 cm, TB = 2.8 cm

(ii) Internally

Distance between centres = 4 − 2.8 = 1.2 cm.

Draw seg AB = 1.2 cm and extend it to T so that AT = 4 cm. Draw a circle with centre A and radius 4 cm, and a circle with centre B and radius 2.8 cm. They touch at T.

AB = 1.2 cm, AT = 4 cm, BT = 2.8 cm

4In fig 3.27, the circles with centres P and Q touch each other at R. A line passing through R meets the circles at A and B respectively. Prove that - (1) seg AP ∥ seg BQ, (2) Δ APR ∼ Δ RQB, and (3) Find ∠ RQB if ∠ PAR = 35°.

Fig. 3.27

(1)

Proof: Points P, R, Q are collinear … theorem of touching circles

∠ PRA ≅ ∠ QRB …… (I) … vertically opposite angles

seg PA ≅ seg PR … radii

∴ ∠ PAR ≅ ∠ PRA …… (II) … isosceles triangle theorem

seg QR ≅ seg QB … radii

∴ ∠ QRB ≅ ∠ QBR …… (III) … isosceles triangle theorem

∴ ∠ PAR ≅ ∠ QBR … from (I), (II) and (III)

These are alternate angles made by transversal AB.

∴ seg AP ∥ seg BQ … alternate angles test

(2)

In Δ APR and Δ RQB,

∠ PAR ≅ ∠ QRB … from (I) and (II)

∠ PRA ≅ ∠ QBR … from (I) and (III)

∴ Δ APR ∼ Δ RQB … AA test of similarity

(3)

∠ PAR = 35°, so ∠ PRA = 35° … from (II)

In Δ APR, ∠ APR = 180° − 35° − 35° = 110°

∠ RQB ≅ ∠ APR … corresponding angles of similar triangles

∴ ∠ RQB = 110°

5★In fig 3.28 the circles with centres A and B touch each other at E. Line l is a common tangent which touches the circles at C and D respectively. Find the length of seg CD if the radii of the circles are 4 cm, 6 cm.

Fig. 3.28: AM ⊥ BD is drawn.

Construction: Draw seg AC, seg BD and seg AB. Draw AM ⊥ BD.

AB = 4 + 6 = 10 cm … circles touch externally

AC ⊥ CD and BD ⊥ CD … tangent theorem

∴ □ ACDM is a rectangle. So AM = CD and MD = AC = 4 cm.

BM = BD − MD = 6 − 4 = 2 cm

In Δ AMB, ∠ AMB = 90°.

AM² = AB² − BM² = 100 − 4 = 96 … Pythagoras theorem

∴ AM = √96 = 4√6

∴ CD = 4√6 cm

Practice Set 3.3

1In figure 3.37, points G, D, E, F are concyclic points of a circle with centre C. ∠ ECF = 70°, m(arc DGF) = 200°. Find m(arc DE) and m(arc DEF).

Fig. 3.37

m(arc EF) = ∠ ECF = 70° … measure of a minor arc

m(arc DEF) = 360° − m(arc DGF) = 360° − 200° = 160°

m(arc DE) + m(arc EF) = m(arc DEF) … arc addition property

∴ m(arc DE) = 160° − 70° = 90°

∴ m(arc DE) = 90° and m(arc DEF) = 160°

2★In fig 3.38, Δ QRS is an equilateral triangle. Prove that, (1) arc RS ≅ arc QS ≅ arc QR (2) m(arc QRS) = 240°.

Fig. 3.38

(1)

Proof: Δ QRS is equilateral.

∴ chord RS ≅ chord QS ≅ chord QR

∴ arc RS ≅ arc QS ≅ arc QR … arcs corresponding to congruent chords

(2)

The three congruent arcs together make the whole circle.

∴ m(arc RS) = m(arc QS) = m(arc QR) = 360°3 = 120°

m(arc QRS) = m(arc QR) + m(arc RS) … arc addition property

= 120° + 120° = 240°

3In fig 3.39, chord AB ≅ chord CD. Prove that, arc AC ≅ arc BD.

Fig. 3.39

Proof: chord AB ≅ chord CD … given

∴ arc ACB ≅ arc CBD … arcs corresponding to congruent chords

∴ m(arc ACB) = m(arc CBD)

∴ m(arc AC) + m(arc CB) = m(arc CB) + m(arc BD) … arc addition property

∴ m(arc AC) = m(arc BD)

∴ arc AC ≅ arc BD … arcs of the same circle with equal measures

Practice Set 3.4

1In figure 3.56, in a circle with centre O, length of chord AB is equal to the radius of the circle. Find measure of each of the following. (1) ∠ AOB (2) ∠ ACB (3) arc AB (4) arc ACB.

Fig. 3.56

OA = OB = AB (radius), so Δ AOB is equilateral.

(1) ∠ AOB = 60°

(2) ∠ ACB = 12 ∠ AOB = 12 × 60° = 30° … inscribed angle theorem

(3) m(arc AB) = ∠ AOB = 60°

(4) m(arc ACB) = 360° − 60° = 300°

2In figure 3.57, □ PQRS is cyclic. side PQ ≅ side RQ. ∠ PSR = 110°. Find - (1) measure of ∠ PQR (2) m(arc PQR) (3) m(arc QR) (4) measure of ∠ PRQ.

Fig. 3.57

(1)

∠ PQR + ∠ PSR = 180° … opposite angles of a cyclic quadrilateral

∴ ∠ PQR = 180° − 110° = 70°

(2)

∠ PSR = 12 m(arc PQR) … inscribed angle theorem

∴ m(arc PQR) = 2 × 110° = 220°

(3)

chord PQ ≅ chord RQ, so m(arc PQ) = m(arc QR) … arcs corresponding to congruent chords

m(arc PQ) + m(arc QR) = 220°

∴ m(arc QR) = 110°

(4)

∠ PRQ = 12 m(arc PQ) … inscribed angle theorem

= 12 × 110° = 55°

3□ MRPN is cyclic, ∠ R = (5x − 13)°, ∠ N = (4x + 4)°. Find measures of ∠ R and ∠ N.

∠ R and ∠ N are opposite angles of cyclic □ MRPN.

∴ (5x − 13) + (4x + 4) = 180 … opposite angles of a cyclic quadrilateral

∴ 9x − 9 = 180

∴ 9x = 189

∴ x = 21

∠ R = 5 × 21 − 13 = 92° and ∠ N = 4 × 21 + 4 = 88°

∴ ∠ R = 92° and ∠ N = 88°

4In figure 3.58, seg RS is a diameter of the circle with centre O. Point T lies in the exterior of the circle. Prove that ∠ RTS is an acute angle.

Seg TR meets the circle at P.

Construction: Let seg TR intersect the circle at P. Draw seg PS.

Proof: ∠ RPS = 90° … angle inscribed in a semicircle

∠ RPS is an exterior angle of Δ PTS.

∴ ∠ RPS > ∠ PTS … exterior angle is greater than a remote interior angle

∴ 90° > ∠ RTS

∴ ∠ RTS is an acute angle.

5Prove that, any rectangle is a cyclic quadrilateral.

Given: □ ABCD is a rectangle.

To prove: □ ABCD is cyclic.

Proof: ∠ A = ∠ C = 90° … angles of a rectangle

∴ ∠ A + ∠ C = 180°

A pair of opposite angles of □ ABCD is supplementary.

∴ □ ABCD is a cyclic quadrilateral. … converse of cyclic quadrilateral theorem

6In figure 3.59, altitudes YZ and XT of Δ WXY intersect at P. Prove that, (1) □ WZPT is cyclic. (2) Points X, Z, T, Y are concyclic.

Fig. 3.59

(1)

Proof: ∠ WZP = 90° and ∠ WTP = 90° … YZ and XT are altitudes

∴ ∠ WZP + ∠ WTP = 180°

∴ □ WZPT is cyclic. … converse of cyclic quadrilateral theorem

(2)

Proof: ∠ XZY = 90° and ∠ XTY = 90°

∴ ∠ XZY ≅ ∠ XTY

Points Z and T lie on the same side of line XY, and seg XY subtends equal angles at them.

∴ Points X, Z, T, Y are concyclic. … theorem of concyclic points

7In figure 3.60, m(arc NS) = 125°, m(arc EF) = 37°, find the measure of ∠ NMS.

Fig. 3.60

Lines MN and MS contain chords and intersect outside the circle at M.

∠ NMS = 12 [m(arc NS) − m(arc EF)]

= 12 [125° − 37°] = 12 × 88° = 44°

∴ ∠ NMS = 44°

8In figure 3.61, chords AC and DE intersect at B. If ∠ ABE = 108°, m(arc AE) = 95°, find m(arc DC).

Fig. 3.61

Chords AC and DE intersect inside the circle at B.

∠ ABE = 12 [m(arc AE) + m(arc DC)]

∴ 108° = 12 [95° + m(arc DC)]

∴ 216° = 95° + m(arc DC)

∴ m(arc DC) = 121°

Practice Set 3.5

1In figure 3.77, ray PQ touches the circle at point Q. PQ = 12, PR = 8, find PS and RS.

Fig. 3.77

PQ² = PR × PS … tangent secant segments theorem

∴ 12² = 8 × PS

∴ PS = 1448 = 18

RS = PS − PR = 18 − 8 = 10

∴ PS = 18 and RS = 10

2In figure 3.78, chord MN and chord RS intersect at point D. (1) If RD = 15, DS = 4, MD = 8 find DN. (2) If RS = 18, MD = 9, DN = 8 find DS.

Fig. 3.78

(1)

RD × DS = MD × DN … theorem of chords intersecting inside the circle

∴ 15 × 4 = 8 × DN

∴ DN = 608 = 7.5

∴ DN = 7.5

(2)

Let DS = x. Then RD = 18 − x.

RD × DS = MD × DN … theorem of chords intersecting inside the circle

∴ (18 − x)x = 9 × 8 = 72

∴ x² − 18x + 72 = 0

∴ (x − 6)(x − 12) = 0

∴ x = 6 or x = 12

∴ DS = 6 or DS = 12

Both values are possible. If DS = 6 then RD = 12, and if DS = 12 then RD = 6.

3In figure 3.79, O is the centre of the circle and B is a point of contact. seg OE ⊥ seg AD, AB = 12, AC = 8, find (1) AD (2) DC (3) DE.

Fig. 3.79

(1)

AB² = AC × AD … tangent secant segments theorem

∴ 144 = 8 × AD

∴ AD = 18

(2)

DC = AD − AC = 18 − 8 = 10

(3)

seg OE ⊥ chord DC, so E is the midpoint of DC … perpendicular from the centre bisects the chord

DE = 12 × DC = 12 × 10 = 5

4In figure 3.80, if PQ = 6, QR = 10, PS = 8 find TS.

Fig. 3.80

PR = PQ + QR = 6 + 10 = 16

PQ × PR = PS × PT … theorem of chords intersecting outside the circle

∴ 6 × 16 = 8 × PT

∴ PT = 968 = 12

TS = PT − PS = 12 − 8 = 4

∴ TS = 4

5In figure 3.81, seg EF is a diameter and seg DF is a tangent segment. The radius of the circle is r. Prove that, DE × GE = 4r².

Fig. 3.81

Construction: Draw seg GF.

Proof: ∠ EGF = 90° … angle inscribed in a semicircle

∠ EFD = 90° … tangent theorem

In Δ EGF and Δ EFD,

∠ EGF ≅ ∠ EFD … each 90°

∠ GEF ≅ ∠ FED … common angle

∴ Δ EGF ∼ Δ EFD … AA test of similarity

∴ EGEF = EFED … corresponding sides of similar triangles

∴ EF² = DE × GE

But EF = 2r … diameter

∴ DE × GE = (2r)² = 4r²

Problem Set 3

1Four alternative answers for each of the following questions are given. Choose the correct alternative.

(1) Two circles of radii 5.5 cm and 3.3 cm respectively touch each other. What is the distance between their centers? — (D) 8.8 or 2.2 cm
They may touch externally (5.5 + 3.3 = 8.8) or internally (5.5 − 3.3 = 2.2).
(2) Two circles intersect each other such that each circle passes through the centre of the other. If the distance between their centres is 12, what is the radius of each circle? — (B) 12 cm
The centre of each circle lies on the other circle, so the radius equals the distance between the centres.
(3) A circle touches all sides of a parallelogram. So the parallelogram must be a …… — (B) rhombus
By tangent segments, AB + CD = AD + BC. In a parallelogram this gives 2AB = 2AD, so all sides are equal.
(4) Length of a tangent segment drawn from a point which is at a distance 12.5 cm from the centre of a circle is 12 cm, find the diameter of the circle. — (C) 7 cm
Radius² = 12.5² − 12² = 12.25, so radius = 3.5 cm and diameter = 7 cm.
(5) If two circles are touching externally, how many common tangents of them can be drawn? — (C) Three
One at the point of contact and two others.
(6) ∠ ACB is inscribed in arc ACB of a circle with centre O. If ∠ ACB = 65°, find m(arc ACB). — (D) 230°
Intercepted m(arc AB) = 2 × 65° = 130°, so m(arc ACB) = 360° − 130° = 230°.
(7) Chords AB and CD of a circle intersect inside the circle at point E. If AE = 5.6, EB = 10, CE = 8, find ED. — (A) 7
5.6 × 10 = 8 × ED, so ED = 7.
(8) In a cyclic □ ABCD, twice the measure of ∠ A is thrice the measure of ∠ C. Find the measure of ∠ C. — (B) 72
∠ A + ∠ C = 180° and ∠ A = 1.5 ∠ C, so 2.5 ∠ C = 180° and ∠ C = 72°.
(9)★ Points A, B, C are on a circle, such that m(arc AB) = m(arc BC) = 120°. No point, except point B, is common to the arcs. Which is the type of Δ ABC? — (A) Equilateral triangle
m(arc AC) = 360° − 240° = 120°. Equal arcs give equal chords, so AB = BC = CA.
(10) Seg XZ is a diameter of a circle. Point Y lies in its interior. How many of the following statements are true? (i) It is not possible that ∠ XYZ is an acute angle. (ii) ∠ XYZ can’t be a right angle. (iii) ∠ XYZ is an obtuse angle. (iv) Can’t make a definite statement for measure of ∠ XYZ. — (C) Only three
For a point inside the circle, ∠ XYZ is always obtuse. So (i), (ii) and (iii) are true and (iv) is false.

2Line l touches a circle with centre O at point P. If radius of the circle is 9 cm, answer the following. (1) What is d(O, P)? Why? (2) If d(O, Q) = 8 cm, where does the point Q lie? (3) If d(O, R) = 15 cm, how many locations of point R are there on line l? At what distance will each of them be from point P?

Fig. 3.82: the two positions of R on line l

(1)

d(O, P) = 9 cm

Reason: OP is a radius of the circle.

(2)

d(O, Q) = 8 cm is less than the radius 9 cm.

∴ Point Q lies in the interior of the circle.

(3)

OP ⊥ line l … tangent theorem

In Δ OPR, PR² = OR² − OP² = 225 − 81 = 144

∴ PR = 12 cm

∴ There are two locations of R, one on each side of P, each at a distance of 12 cm from P.

The textbook prints “If d(PQ) = 15 cm” in part (3). The question is about point R, so it is solved here as d(O, R) = 15 cm.

3In figure 3.83, M is the centre of the circle and seg KL is a tangent segment. If MK = 12, KL = 6√3 then find - (1) Radius of the circle. (2) Measures of ∠ K and ∠ M.

Fig. 3.83

(1)

∠ MLK = 90° … tangent theorem

ML² = MK² − KL² = 144 − 108 = 36 … Pythagoras theorem

∴ ML = 6

∴ Radius = 6

(2)

ML = 6 = 12 × MK, so the angle opposite ML is 30° … converse of 30° − 60° − 90° theorem

∴ ∠ K = 30° and ∠ M = 180° − 90° − 30° = 60°

∴ ∠ K = 30° and ∠ M = 60°

4In figure 3.84, O is the centre of the circle. Seg AB, seg AC are tangent segments. Radius of the circle is r and l(AB) = r. Prove that, □ ABOC is a square.

Fig. 3.84

Proof: AB = AC … tangent segment theorem

AB = r … given

OB = OC = r … radii

∴ AB = AC = OB = OC. So □ ABOC is a rhombus.

∠ ABO = 90° … tangent theorem

A rhombus with one right angle is a square.

∴ □ ABOC is a square.

5In figure 3.85, □ ABCD is a parallelogram. It circumscribes the circle with centre T. Point E, F, G, H are touching points. If AE = 4.5, EB = 5.5, find AD.

Fig. 3.85

AB = AE + EB = 4.5 + 5.5 = 10

Tangent segments from a point are congruent … tangent segment theorem

AE = AH, BE = BF, CF = CG, DG = DH

Adding, AE + BE + CG + DG = AH + DH + BF + CF

∴ AB + CD = AD + BC

But AB = CD and AD = BC … opposite sides of a parallelogram

∴ 2AB = 2AD

∴ AD = AB = 10

∴ AD = 10

6In figure 3.86, circle with centre M touches the circle with centre N at point T. Radius RM touches the smaller circle at S. Radii of circles are 9 cm and 2.5 cm. Find the answers to the following questions hence find the ratio MS : SR. (1) Find the length of segment MT (2) Find the length of seg MN (3) Find the measure of ∠ NSM.

Fig. 3.86

(1)

MT is a radius of the bigger circle.

∴ MT = 9 cm

(2)

The circles touch internally, so MN = 9 − 2.5 = 6.5 cm

(3)

Seg MR touches the smaller circle at S, and NS is its radius.

∴ ∠ NSM = 90° … tangent theorem

MS : SR

In Δ NSM, MS² = MN² − NS² = 6.5² − 2.5² = 42.25 − 6.25 = 36

∴ MS = 6 cm

SR = MR − MS = 9 − 6 = 3 cm

∴ MS : SR = 6 : 3 = 2 : 1

7In the adjoining figure circles with centres X and Y touch each other at point Z. A secant passing through Z intersects the circles at points A and B respectively. Prove that, radius XA ∥ radius YB. Fill in the blanks and complete the proof.

Fig. 3.87

Construction: Draw segments XZ and YZ.

Proof: By theorem of touching circles, points X, Z, Y are collinear.

∴ ∠ XZA ≅ ∠ BZY … opposite angles

Let ∠ XZA = ∠ BZY = a …… (I)

Now, seg XA ≅ seg XZ … radii of the same circle

∴ ∠ XAZ = ∠ XZA = a … isosceles triangle theorem (II)

Similarly, seg YB ≅ seg YZ … radii of the same circle

∴ ∠ BZY = ∠ YBZ = a … isosceles triangle theorem (III)

∴ from (I), (II), (III), ∠ XAZ = ∠ YBZ

∴ radius XA ∥ radius YB … alternate angles test

8In figure 3.88, circles with centres X and Y touch internally at point Z. Seg BZ is a chord of bigger circle and it intersects smaller circle at point A. Prove that, seg AX ∥ seg BY.

Fig. 3.88

Construction: Draw seg XZ and seg YZ.

Proof: Points Z, X, Y are collinear … theorem of touching circles

So ∠ XZA and ∠ YZB are the same angle.

seg XA ≅ seg XZ … radii of the smaller circle

∴ ∠ XAZ ≅ ∠ XZA … isosceles triangle theorem

seg YB ≅ seg YZ … radii of the bigger circle

∴ ∠ YBZ ≅ ∠ YZB … isosceles triangle theorem

∴ ∠ XAZ ≅ ∠ YBZ

These are corresponding angles made by transversal BZ.

∴ seg AX ∥ seg BY … corresponding angles test

9In figure 3.89, line l touches the circle with centre O at point P. Q is the mid point of radius OP. RS is a chord through Q such that chord RS ∥ line l. If RS = 12 find the radius of the circle.

Fig. 3.89

Let the radius be r. Draw seg OR.

OP ⊥ line l … tangent theorem

Chord RS ∥ line l, so OQ ⊥ chord RS.

∴ Q is the midpoint of RS, so RQ = 6 … perpendicular from the centre bisects the chord

OQ = r2

In Δ OQR, OR² = OQ² + RQ² … Pythagoras theorem

∴ r² = r²4 + 36

∴ 3r²4 = 36

∴ r² = 48

∴ r = √48 = 4√3

∴ Radius = 4√3

10★In figure 3.90, seg AB is a diameter of a circle with centre C. Line PQ is a tangent, which touches the circle at point T. seg AP ⊥ line PQ and seg BQ ⊥ line PQ. Prove that, seg CP ≅ seg CQ.

Fig. 3.90

Construction: Draw seg CT, seg CP and seg CQ.

Proof: seg CT ⊥ line PQ … tangent theorem

seg AP, seg CT and seg BQ are all perpendicular to line PQ.

∴ seg AP ∥ seg CT ∥ seg BQ

∴ ACCB = PTTQ … property of three parallel lines and their transversals

But AC = CB … radii

∴ PT = TQ

In Δ CTP and Δ CTQ,

seg PT ≅ seg QT … proved above

∠ CTP ≅ ∠ CTQ … each 90°

seg CT ≅ seg CT … common side

∴ Δ CTP ≅ Δ CTQ … SAS test

∴ seg CP ≅ seg CQ … corresponding sides of congruent triangles

11★Draw circles with centres A, B and C each of radius 3 cm, such that each circle touches the other two circles.

Circles touching externally have their centres at a distance equal to the sum of the radii, 3 + 3 = 6 cm.

So the centres A, B, C are the vertices of an equilateral triangle of side 6 cm.

Steps: Draw Δ ABC with AB = BC = CA = 6 cm. With A, B and C as centres, draw circles of radius 3 cm. Each circle touches the other two at the midpoints of the sides.

AB = BC = CA = 6 cm; each radius = 3 cm

12★Prove that any three points on a circle cannot be collinear.

Proof (indirect method): Suppose points A, B, C on a circle with centre O are collinear, with A − B − C.

Seg AB and seg BC are chords of the circle.

The centre O lies on the perpendicular bisector of chord AB, and also on the perpendicular bisector of chord BC.

Both these bisectors are perpendicular to the same line AC, and they pass through different midpoints.

∴ the two perpendicular bisectors are distinct parallel lines.

Two parallel lines cannot have the common point O. This is a contradiction.

∴ Our assumption is wrong. Any three points on a circle cannot be collinear.

13In figure 3.91, line PR touches the circle at point Q. Answer the following questions with the help of the figure.

Fig. 3.91

(1) What is the sum of ∠ TAQ and ∠ TSQ?

□ ATSQ is cyclic, and ∠ TAQ and ∠ TSQ are its opposite angles.

∴ ∠ TAQ + ∠ TSQ = 180°

(2) Find the angles which are congruent to ∠ AQP.

∠ AQP is between tangent QP and chord QA. It is congruent to angles inscribed in the arc opposite to arc AQ.

∴ ∠ AQP ≅ ∠ ATQ ≅ ∠ ASQ

(3) Which angles are congruent to ∠ QTS?

∠ QTS intercepts arc QS.

∴ ∠ QTS ≅ ∠ QAS ≅ ∠ SQR … same arc; angle between tangent and secant

(4) ∠ TAS = 65°, find the measure of ∠ TQS and arc TS.

∠ TQS ≅ ∠ TAS … angles inscribed in the same arc

∴ ∠ TQS = 65°

m(arc TS) = 2 × ∠ TAS = 2 × 65° = 130° … inscribed angle theorem

(5) If ∠ AQP = 42° and ∠ SQR = 58°, find measure of ∠ ATS.

∠ AQS = 180° − 42° − 58° = 80° … angles on line PR

∠ ATS + ∠ AQS = 180° … opposite angles of cyclic □ ATSQ

∴ ∠ ATS = 180° − 80° = 100°

14In figure 3.92, O is the centre of a circle, chord PQ ≅ chord RS. If ∠ POR = 70° and m(arc RS) = 80°, find - (1) m(arc PR) (2) m(arc QS) (3) m(arc QSR).

Fig. 3.92

(1)

m(arc PR) = ∠ POR = 70° … measure of a minor arc

(2)

chord PQ ≅ chord RS, so m(arc PQ) = m(arc RS) = 80°

m(arc QS) = 360° − [m(arc PQ) + m(arc PR) + m(arc RS)]

= 360° − [80° + 70° + 80°] = 130°

(3)

m(arc QSR) = m(arc QS) + m(arc SR) = 130° + 80° = 210°

15In figure 3.93, m(arc WY) = 44°, m(arc ZX) = 68°, then (1) Find the measure of ∠ ZTX. (2) If WT = 4.8, TX = 8.0, YT = 6.4, find TZ. (3) If WX = 25, YT = 8, YZ = 26, find WT.

Fig. 3.93

(1)

∠ ZTX = 12 [m(arc ZX) + m(arc WY)] … chords intersecting inside the circle

= 12 [68° + 44°] = 56°

(2)

WT × TX = YT × TZ … theorem of chords intersecting inside the circle

∴ 4.8 × 8.0 = 6.4 × TZ

∴ TZ = 38.46.4 = 6

(3)

TZ = YZ − YT = 26 − 8 = 18. Let WT = x. Then TX = 25 − x.

x(25 − x) = 8 × 18 = 144

∴ x² − 25x + 144 = 0

∴ (x − 9)(x − 16) = 0

∴ WT = 9 or WT = 16

16In figure 3.94, (1) m(arc CE) = 54°, m(arc BD) = 23°, find measure of ∠ CAE. (2) If AB = 4.2, BC = 5.4, AE = 12.0, find AD. (3) If AB = 3.6, AC = 9.0, AD = 5.4, find AE.

Fig. 3.94

(1)

∠ CAE = 12 [m(arc CE) − m(arc BD)] … lines containing chords intersect outside the circle

= 12 [54° − 23°] = 12 × 31° = 15.5°

(2)

AC = AB + BC = 4.2 + 5.4 = 9.6

AB × AC = AD × AE … theorem of chords intersecting outside the circle

∴ 4.2 × 9.6 = AD × 12.0

∴ AD = 40.3212 = 3.36

(3)

AB × AC = AD × AE

∴ 3.6 × 9.0 = 5.4 × AE

∴ AE = 32.45.4 = 6

17In figure 3.95, chord EF ∥ chord GH. Prove that, chord EG ≅ chord FH. Fill in the blanks and write the proof.

Fig. 3.95

Proof: Draw seg GF.

∠ EFG = ∠ FGH … alternate angles (I)

∠ EFG = 12 m(arc EG) … inscribed angle theorem (II)

∠ FGH = 12 m(arc FH) … inscribed angle theorem (III)

∴ m(arc EG) = m(arc FH) … from (I), (II), (III)

chord EG ≅ chord FH … chords corresponding to congruent arcs

18In figure 3.96, P is the point of contact. (1) If m(arc PR) = 140°, ∠ POR = 36°, find m(arc PQ). (2) If OP = 7.2, OQ = 3.2, find OR and QR. (3) If OP = 7.2, OR = 16.2, find QR.

Fig. 3.96: O is a point outside the circle.

(1)

Draw seg PQ and seg PR.

∠ PQR = 12 m(arc PR) = 12 × 140° = 70° … inscribed angle theorem

∠ OPQ = 12 m(arc PQ) … angle between tangent and secant

∠ PQR is an exterior angle of Δ OPQ.

∴ ∠ PQR = ∠ POQ + ∠ OPQ

∴ 70° = 36° + 12 m(arc PQ)

∴ 12 m(arc PQ) = 34°

∴ m(arc PQ) = 68°

(2)

OP² = OQ × OR … tangent secant segments theorem

∴ 7.2² = 3.2 × OR

∴ OR = 51.843.2 = 16.2

QR = OR − OQ = 16.2 − 3.2 = 13

∴ OR = 16.2 and QR = 13

(3)

OP² = OQ × OR

∴ 51.84 = OQ × 16.2

∴ OQ = 3.2

QR = OR − OQ = 16.2 − 3.2 = 13

19In figure 3.97, circles with centres C and D touch internally at point E. D lies on the inner circle. Chord EB of the outer circle intersects inner circle at point A. Prove that, seg EA ≅ seg AB.

Fig. 3.97

Construction: Draw seg ED and seg DA.

Proof: Points E, C, D are collinear … theorem of touching circles

C is the centre of the inner circle and E, D lie on it.

∴ seg ED is a diameter of the inner circle.

∴ ∠ EAD = 90° … angle inscribed in a semicircle

∴ seg DA ⊥ chord EB of the outer circle, whose centre is D.

∴ A is the midpoint of chord EB … perpendicular from the centre bisects the chord

∴ seg EA ≅ seg AB

20In figure 3.98, seg AB is a diameter of a circle with centre O. The bisector of ∠ ACB intersects the circle at point D. Prove that, seg AD ≅ seg BD. Complete the following proof by filling in the blanks.

Fig. 3.98

Proof: Draw seg OD.

∠ ACB = 90° … angle inscribed in semicircle

∠ DCB = 45° … CD is the bisector of ∠ C

m(arc DB) = 90° … inscribed angle theorem

∠ DOB = 90° … definition of measure of an arc (I)

seg OA ≅ seg OB … radii of the same circle (II)

∴ line OD is perpendicular bisector of seg AB … from (I) and (II)

∴ seg AD ≅ seg BD … every point on the perpendicular bisector is equidistant from the ends

21In figure 3.99, seg MN is a chord of a circle with centre O. MN = 25, L is a point on chord MN such that ML = 9 and d(O, L) = 5. Find the radius of the circle.

Fig. 3.99: OP ⊥ MN is drawn.

Construction: Draw OP ⊥ chord MN and seg OM.

P is the midpoint of MN … perpendicular from the centre bisects the chord

∴ MP = 252 = 12.5 and LP = MP − ML = 12.5 − 9 = 3.5

In Δ OPL, OP² = OL² − LP² = 25 − 12.25 = 12.75 … Pythagoras theorem

In Δ OPM, OM² = OP² + MP² = 12.75 + 156.25 = 169

∴ OM = 13

∴ Radius = 13

22★In figure 3.100, two circles intersect each other at points S and R. Their common tangent PQ touches the circles at points P, Q. Prove that, ∠ PRQ + ∠ PSQ = 180°.

Fig. 3.100

Construction: Draw seg SR.

Proof: In the circle through P, R, S, line PQ is a tangent at P and PR is a chord.

∴ ∠ QPR = ∠ PSR …… (I) … angle between tangent and chord equals angle in the opposite arc

Similarly, in the circle through Q, R, S,

∠ PQR = ∠ QSR …… (II)

In Δ PRQ, ∠ PRQ + ∠ QPR + ∠ PQR = 180° … sum of angles of a triangle

∴ ∠ PRQ + ∠ PSR + ∠ QSR = 180° … from (I) and (II)

But ∠ PSR + ∠ QSR = ∠ PSQ … angle addition

∴ ∠ PRQ + ∠ PSQ = 180°

23★In figure 3.101, two circles intersect at points M and N. Secants drawn through M and N intersect the circles at points R, S and P, Q respectively. Prove that: seg SQ ∥ seg RP.

Fig. 3.101

Construction: Draw seg MN.

Proof: □ RMNP is cyclic.

∴ ∠ SMN ≅ ∠ RPN …… (I) … exterior angle of a cyclic quadrilateral

□ MSQN is cyclic.

∴ ∠ SMN + ∠ SQN = 180° …… (II) … opposite angles of a cyclic quadrilateral

∴ ∠ RPN + ∠ SQN = 180° … from (I) and (II)

These are interior angles on the same side of transversal PQ.

∴ seg SQ ∥ seg RP … interior angles test

24★In figure 3.102, two circles intersect each other at points A and E. Their common secant through E intersects the circles at points B and D. The tangents of the circles at points B and D intersect each other at point C. Prove that □ ABCD is cyclic.

Fig. 3.102

Construction: Draw seg AB, seg AE and seg AD.

Proof: In the first circle, CB is a tangent and BE is a chord.

∴ ∠ CBE ≅ ∠ BAE …… (I) … angle between tangent and chord equals angle in the opposite arc

In the second circle, CD is a tangent and DE is a chord.

∴ ∠ CDE ≅ ∠ DAE …… (II) … same reason

In Δ BCD, ∠ BCD + ∠ CBD + ∠ CDB = 180° … sum of angles of a triangle

∴ ∠ BCD + ∠ BAE + ∠ DAE = 180° … from (I) and (II)

∴ ∠ BCD + ∠ BAD = 180° … angle addition

∴ □ ABCD is cyclic. … converse of cyclic quadrilateral theorem

25★In figure 3.103, seg AD ⊥ side BC, seg BE ⊥ side AC, seg CF ⊥ side AB. Point O is the orthocentre. Prove that, point O is the incentre of Δ DEF.

Fig. 3.103

Proof: ∠ BDO + ∠ BFO = 90° + 90° = 180°

∴ □ BDOF is cyclic. … converse of cyclic quadrilateral theorem

∴ ∠ ODF ≅ ∠ OBF …… (I) … angles inscribed in the same arc

Similarly, □ CDOE is cyclic.

∴ ∠ ODE ≅ ∠ OCE …… (II) … angles inscribed in the same arc

In Δ ABE, ∠ AEB = 90°. So ∠ ABE = 90° − ∠ A, that is ∠ OBF = 90° − ∠ A.

In Δ ACF, ∠ AFC = 90°. So ∠ ACF = 90° − ∠ A, that is ∠ OCE = 90° − ∠ A.

∴ ∠ OBF = ∠ OCE

∴ ∠ ODF = ∠ ODE … from (I) and (II)

∴ ray DO bisects ∠ EDF.

In the same way, ray EO bisects ∠ DEF and ray FO bisects ∠ DFE.

The angle bisectors of Δ DEF meet at O.

∴ O is the incentre of Δ DEF.

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