• facebook
  • twitter
  • whatsapp
  • telegram

Tangents and Secants to a Circle

4 Marks Questions

1. Prove that the parallelogram circumscribing a circle is a rhombus.

Sol: ABCD is a parallelogram                                                                

AB // CD and AB = CD

AD // BC and AD = BC

We have to prove that ABCD is a rhombus

'a' is an external point of the circle                 

AP and AS are two tangents

So AP = AS, Let AP = AS = x

Similarly BP = BQ = y

CQ = CR = z, DR = DS = w

But AB = CD ⇒ x + y = z + w ....... (1)

BC = DA ⇒ y + z = x + w

⇒ x − y = z - w .......... (2)

2. A triangle ABC is drawn to circumscribe a circle of radius 3 cm. Such that the segments BD and DC into which BC is divided by the point of contact D are of length 9 cm and 3 cm respectively. Find the sides AB and AC.

Sol: From the figure,

BD = 9 cm, and DC = 3 cm

Let AE = x cm

'A' be an external point to the circle

'B' is another external point to the circle

then BE = BD = 9 cm (tangents)

'C' is another external point to the circle

then CD = CF = 3 cm (tangents)

AB = AE + EB = x + 9

BC = BD + DC = 9 + 3 = 12 cm

AC = AF + FC = x + 3

∴ AB = x + 9, BC = 12, AC = x + 3

    (c)                (a)           (b)

then AE = AF = x (AE & AF are tangents)

From (1) and (2)

27x2 + 324x  = (36 + 3x)2

                      = 1296 + 216x + 9x2

27x2 − 9x2 + 324x − 216x − 1296 = 0

18x2 + 108x − 1296 = 0

x2 + 6x − 72 = 0

x2 + 12x − 6x − 72 = 0
(x − 6) (x + 12) = 0

⇒ x = 6 or − 12

'x' never be negative, therefore x = 6

∴ AB = x + 9 = 6 + 9 = 15 cm

AC = x + 3 = 6 + 3 = 9 cm

3. In a right triangle ABC, a circle with a side AB as diameter is drawn to intersect the hypotenuse AC in P. Prove that the tangents to the circle at P bisects the side BC.

Sol: Given that,

PQ is the tangent to the circle

'P' is contact point

we have to prove that BQ = QC

∠APB = 90° (... angle in semi circle)

∴ ∠BPC = 90° (... APC line segment)

∴ ∠BPQ + ∠QPC = 90° ............. (1)    (... ∠BPC = ∠BPQ + ∠QPC)

In ΔABC,

∠BAC + ∠BCA = 90°  ..................(2)  (... ∠B = 90°)

But ∠BPQ = ∠ BAC    ..................(3)

From (1) and (2)

∠BPQ + ∠QPC = ∠BAC + ∠BCA

∠BPQ + ∠QPC = ∠BPQ + ∠BCA (by (3))

∠QPC = ∠BCA

∴ PQ = QC (sides opposite to equal angles)

Similarly PQ = QB

∴ QB = QC

Hence PQ bisects BC

4. A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding minor segment of the circleAB is diameter, ∠ABC = 90°

    (use Π = 3.14 and  =  1.732)

 

5. Find the area of the shaded region in figure, where ABCD is a square of side 10 cm and semicircles are drawn with each side of the square as diameter (use Π = 3.14)

Sol: Let us mark the four unshaded regions as I, II, III and IV.

Area of I + Area of III = Area of Square ABCD − Area of two semi circles of each of radius 5 cm

 = (100 − 25Π) cm2

Similarly, Area of II + Area of IV = (100 − 25Π) cm2

∴ Area of shaded region = Area of ABCD − (Area of I + II + III + IV)

= 100 − 2(100 − 25Π)

= 100 − 200 + 50Π

= 50Π − 100

= 50(3.14) − 100

= 157 − 100

= 57 cm2

6. Find the area of the segments shaded in figure, if PQ = 24 cm, PR = 7 cm and QR is the diameter of the circle with centre O (Take Π =  ).

Area of the segments shaded = Area of sector OQPR − Area of triangle PQR

By Pythagoras theorem

In Δ PQR, QR2 = PQ2 + PR2

                         = 242 + 72

                         = 625

        QR = 625 = 25 cm

Sol: Since QR is diameter, QPR = 90° (Angle in semi circle)

7. AB and CD are respectively arcs of two concentric circles of radii 21 cm. and 7 cm. with centre O. If ∠AOB = 30°, find the area of the shaded region.
 

Sol: ∠AOB = ∠COD = x = 30°

radii are 7 cm, 21 cm

8. Around table top has six equal designs as shown in the figure. If the radius of the table top is 14 cm. Find the cost of making the designs with paint at the rate of Rs.5 Per cm2 (use  = 1.732).
 

Sol: We know that the radius of circumscribing circle of a regular hexagon is equal to the length of its side.

Each side of regular hexagon = 14 cm

Area of six design segments = Area of circle − Area of regular hexagon Now, Area of circle = πr2

                                                                 
         = 616 cm2 ............ (1)

Hence, Area of six designs = 616 − 509.21

                          = 106.79 cm2

Cost of painting the design at the rate of Rs.5 Per cm2 = 106.79 × 5
                                  = Rs.533.95

2 Marks Questions
 

1. Write construction algorithm to draw a tangent at a point on the circle when centre of circle is not known.
 

Sol: * Draw any two chords from a point 'p' on the circle. Let they are PQ, PR.

* Now draw perpendicular bisectors to PQ, PR which meet at centre 'O'.

* Now plot the point 'T' where tangent to be drawn.

* Now joint 'T' with centre 'O'.

* Then draw perpendicular TT' to the radius OT.

* TT' is the required tangent.

2. An 8 cm long chord of an outer circle is at a distance of 3 cm from the centre of inner circle. If both circles are concentric, then find the radius of bigger circle. The chord touches inner circle.

Sol: Let 'O' is the centre of two concentric circles radius of smaller circle = 3 cm

OP is perpendicular bisector of AB

                        = 4 cm

            ∠APO = 90°

In ΔOAP, by Pythagoras theorem

OA2 = OP2 + PA2

         = 42 + 32

         = 16 + 9

         = 25

         = 5 cm

3. In the figure XY and PS are two parallel tangents to a circle with centre 'O' and another tangent AB with point of contact C intersecting XY at A and PS at B then prove that ∠AOB = 90°

Since 'A' is an external point, AD and AC are tangentsSol: Join OC, OA and OB

⇒ AD = AC .........(1)

In Δ DOA and ΔCOA

AD = AC, OD = OC (radii)

OA = OA (common side)

∴ DOA = OCOA (S.S.S. Congruency)

Let ∠DOA = ∠COA = X

Similarly, ΔBOC ≅ ΔBOE and Let ∠BOC = ∠BOE = Y

∠DOA + ∠COA + ∠BOC + ∠BOE = 180° (Straight line)

⇒ X + X + Y + Y = 180°

⇒ 2(X + Y) = 180°

⇒ X + Y = 90°

∴ AOB = 90°

Hence Proved.

4. Two concentric circles of radii 5 cm and 3 cm are drawn. Find the length of the chord of larger circle which touches the smaller circle.

Sol: Draw OC ⊥ AB

Radius of smaller circle OC = 3 cm

Radius of larger circle OA = 5 cm

In ΔOAC, by Phythagoras theorem

OA2 = OC2 + AC2

⇒ 52 = 32 + AC2

⇒ AC2 = 25 − 9

AC =  = 4 cm

We know that AC = CB (∵ AB is the chord of larger circle)

∴ AB = AC + CB

          = 4 + 4

          = 8 cm

∴ AB = 8 cm

5. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 10 minutes.

Sol: Minute hand clock made by angle in 60 minutes = 360°

6. A car has two wipers, which do not overlap. Each wiper has a blade of length 25 cm. Sweeping through an angle of 120°. Find the total area cleared at each sweep of the blade.

Sol: Here radius r = 25 cm

                  Angle x = 120°

                  = 654.76 cm2

       Area = 654.76 cm2

∴ Total area cleaned by two blades = 2 × 654.76

                             = 1309.52 cm2
 

1 Mark Questions
 

1. Calculate the length of a tangent from a point 15 cm, away from the centre of the circle of radius 9 cm.
 

Sol: Radius OA = r = 9 cm

                     OB = 15 cm

                     OA ⊥ AB

by Pythagoras theorem OB2 = OA2 + AB2

         ⇒ 152 = 92 + AB2

         ⇒ AB2 = 152 − 92

         = 225 − 81

         ⇒ AB = 

         = 12 cm
 

2. If AP and AQ are two tangents of a circle with centre 'O' so that ∠POQ = 110°, then find ∠PAQ.
 

Sol: In OPAQ

∠OPA = ∠OQA = 90° and ∠POQ = 110°

∠O + ∠P + ∠A + ∠Q = 360°

               ⇒110° + 90° + ∠A + 90° = 360°
               ⇒∠A = 360° − 290°

               = 70°

     ∴ ∠PAQ = 70°
 

3. If tangents PA and PB from a point P to a circle with centre 'O' are inclined to each other at angle of 80°. Then find ∠POA.

Sol: Given ∠APB = 80°

∴ ∠OPA = ∠OPB = 40°

∠OAP = 90°

(Angle between tangent and radius)

In ΔOAP,

∠POA = 180° − 90° − 40° = 50°
 

4. The radius of a circle is 4 cm. Then find the area of a square inscribed in the circle.

Sol: Let ABCD is a square inscribed in a circle with centre 'O' and radius 4 cm.

AC and BD are diameters and AC = BD = 4 × 2 = 8 cm                        (∵ d = 2r)

Writer: T.S.V.S. Surya Narayanamurthy

Posted Date : 27-04-2021

గమనిక : ప్రతిభ.ఈనాడు.నెట్‌లో కనిపించే వ్యాపార ప్రకటనలు వివిధ దేశాల్లోని వ్యాపారులు, సంస్థల నుంచి వస్తాయి. మరి కొన్ని ప్రకటనలు పాఠకుల అభిరుచి మేరకు కృత్రిమ మేధస్సు సాంకేతికత సాయంతో ప్రదర్శితమవుతుంటాయి. ఆ ప్రకటనల్లోని ఉత్పత్తులను లేదా సేవలను పాఠకులు స్వయంగా విచారించుకొని, జాగ్రత్తగా పరిశీలించి కొనుక్కోవాలి లేదా వినియోగించుకోవాలి. వాటి నాణ్యత లేదా లోపాలతో ఈనాడు యాజమాన్యానికి ఎలాంటి సంబంధం లేదు. ఈ విషయంలో ఉత్తర ప్రత్యుత్తరాలకు, ఈ-మెయిల్స్ కి, ఇంకా ఇతర రూపాల్లో సమాచార మార్పిడికి తావు లేదు. ఫిర్యాదులు స్వీకరించడం కుదరదు. పాఠకులు గమనించి, సహకరించాలని మనవి.

ప్రత్యేక కథనాలు