# Tangents and Secants to a Circle

Chapters

- ap-scert-10th-class-maths-lesson-1-Real-Numbers
- ap-scert-10th-class-Maths-Lesson-2-Sets
- ap-scert-10th-class-Maths-Lesson-3-Polynomials
- ap-scert-10th-class-Maths-Lesson-4-Pair-of-Linear-Equations-in-Two-Variables
- ap-scert-10th-class-Maths-Lesson-5-Quadratic-Equations
- ap-scert-10th-class-Maths-Lesson-6-Progressions
- ap-scert-10th-class-Maths-Lesson-7-Coordinate-Geometry
- ap-scert-10th-class-Maths-Lesson-8-Similar-Triangles
- ap-scert-10th-class-Maths-Lesson-9-Tangents-and-Secants-to-a-Circle
- ap-scert-10th-class-Maths-Lesson-10-Mensuration
- ap-scert-10th-class-Maths-Lesson-11-Trigonometry
- ap-scert-10th-class-Maths-Lesson-12-Applications-of-Trigonometry
- ap-scert-10th-class-Maths-Lesson-13-Probability
- ap-scert-10th-class-Maths-Lesson-14-Statistics

- ap-scert-10th-class-maths-lesson-1-Real-Numbers
- ap-scert-10th-class-Maths-Lesson-2-Sets
- ap-scert-10th-class-Maths-Lesson-3-Polynomials
- ap-scert-10th-class-Maths-Lesson-4-Pair-of-Linear-Equations-in-Two-Variables
- ap-scert-10th-class-Maths-Lesson-5-Quadratic-Equations
- ap-scert-10th-class-Maths-Lesson-6-Progressions
- ap-scert-10th-class-Maths-Lesson-7-Coordinate-Geometry
- ap-scert-10th-class-Maths-Lesson-8-Similar-Triangles
- ap-scert-10th-class-Maths-Lesson-9-Tangents-and-Secants-to-a-Circle
- ap-scert-10th-class-Maths-Lesson-10-Mensuration
- ap-scert-10th-class-Maths-Lesson-11-Trigonometry
- ap-scert-10th-class-Maths-Lesson-12-Applications-of-Trigonometry
- ap-scert-10th-class-Maths-Lesson-13-Probability
- ap-scert-10th-class-Maths-Lesson-14-Statistics

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Solutions

Fill in the blanks.

- i) A tangent to a circle intersects it in ——— point(s). (one)
- ii) A line intersecting a circle in two points is called a ———. (secant)
- iii) The number of tangents drawn at the end of the diameter is ———. (two)
- iv) The common point of a tangent to a circle and the circle is called ———. (point of contact)
- v) We can draw ——— tangents to a given circle. (infinite)

A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Find length of PQ.

Answer:

Given: A circle with centre O and radius OP = 5 cm

We know that ∠OPQ = 90°

Hence in △OPQ

OQ^{2}= OP^{2}+ PQ^{2}

[∵ hypotenuse^{2}= Adj. side^{2}+ Opp. side^{2}]

12^{2}= 5^{2}+ PQ^{2}

∴ PQ^{2}= 144 - 25 .

PQ^{2}= 119

PQ = √119

Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle.

Answer:

Steps:

- Draw a circle with some radius.
- Draw a chord of the circle.
- Draw a line parallel to the chord intersecting the circle at two distinct points.
- This is secant of the circle (l).
- Draw another line parallel to the chord, just touching the circle at one point (M). This is a tangent of the circle.

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

Answer:

Given: A circle with radius OP = 9 cm

A tangent PQ from a point Q at a distance of 15 cm from the centre, i.e., OQ =15 cm

Now in △POQ, ∠P = 90°

OP^{2}+ PQ^{2}- OQ^{2}

9^{2}+ PQ^{2}= 15^{2}

PQ^{2}= 15^{2}- 9^{2}

PQ^{2}= 144

∴ PQ = √144 = 12 cm.

Hence the length of the tangent =12 cm.

Prove that the tangents to a circle at the end points of a diameter are parallel.

Answer:

A circle with a diameter AB.

PQ is a tangent drawn at A and RS is a tangent drawn at B.

R.T.P: PQ ‖ RS.

Proof: Let ‘O’ be the centre of the circle then OA is radius and PQ is a tangent.

∴ OA ⊥ PQ ……….(1)

[∵ a tangent drawn at the end point of the radius is perpendicular to the radius]

Similarly, OB ⊥ RS ……….(2)

[∵ a tangent drawn at the end point of the radius is perpendicular to the radius]

But, OA and OB are the parts of AB.

i.e., AB ⊥ PQ and AB ⊥ RS.

∴ PQ ‖ RS.

O is the centre, PQ is a tangent drawn at A.

∠OAQ = 90°

Similarly, ∠OBS = 90°

∠OAQ + ∠OBS = 90° + 90° = 180°

∴ PQ ‖ RS.

[∵ Sum of the consecutive interior angles is 180°, hence lines are parallel]

Choose the correct answer and give justification for each.

(i) The angle between a tangent to a circle and the radius drawn at the point of contact is

- a) 60°
- b) 30°
- c) 45°
- d) 90°

Answer: [ d ]

If radius is not perpendicular to the tangent, the tangent must be a secant i.e., 90°.

(ii) From a point Q, the length of the tangent to a circle is 24 cm. and the distance of Q from the centre is 25 cm. The radius of the circle is

- a) 7 cm
- b) 12 cm
- c) 15 cm
- d) 24.5 cm

Answer: [ a ]

O - centre of the circle

OP - a circle radius = ?

OQ = 25 cm

PQ = 24 cm

OQ^{2}= OP^{2}+ PQ^{2}

[∵ hypotenuse^{2}= Adj. side^{2} + Opp. side^{2}]

25^{2}= OP^{2}+ 24^{2}

OP^{2}= 625 - 576

OP^{2}= 49

OP = √49 = 7 cm.

iii) If AP and AQ are the two tangents a circle with centre O, so that ∠POQ = 110°. Then ∠PAQ is equal to

a) 60°

b) 70°

c) 80°

d) 90°

Answer: [ b ]

In □ OPAQ,

∠OPA = ∠OQA = 90°

∠POQ = 110°

∴ ∠O + ∠P + ∠A + ∠Q = 360°

⇒ 90° + 90° + 110° + ∠PAQ - 360°

⇒ ∠PAQ = 360° - 290° = 70°

iv) 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 ∠POA is equal to

- a) 50°
- b) 60°
- c) 70°
- d) 80°

Answer: [None]

If ∠APB = 80°

then ∠AOB = 180° - 80° = 100°

[∴ ∠A + ∠B = 90° + 90° = 180°]

v) In the figure XY and XV are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and XV at B then ∠AOB =

- a) 80°
- b) 100°
- c) 90°
- d) 60°

Answer: [ c ]

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

Answer:

Given: Two circles of radii 3 cm and 5 cm with common centre.

Let AB be a tangent to the inner/small circle and chord to the larger circle.

Let ‘P’ be the point of contact.

Construction: Join OP and OB.

In △OPB ;

∠OPB = 90°

[radius is perpendicular to the tangent]

OP = 3cm OB = 5 cm

Now, OB^{2}= OP^{2}+ PB^{2}

[hypotenuse^{2}= Adj. side^{2}+ Opp. side^{2}, Pythagoras theorem]

5^{2}= 3^{2}+ PB^{2}

PB^{2}= 25 - 9 = 16

∴ PB = √l6 = 4cm.

Now, AB = 2 × PB

[∵ The perpendicular drawn from the centre of the circle to a chord, bisects it]

AB = 2 × 4 = 8 cm.

∴ The length of the chord of the larger circle which touches the smaller circle is 8 cm.

Prove that the parallelogram circumscribing a circle is a rhombus.

Answer:

Given: A circle with centre ‘O’.

A parallelogram ABCD, circumscribing the given circle.

Let P, Q, R, S be the points of contact.

Required to prove: □ ABCD is a rhombus.

Proof: AP = AS …….. (1)

[∵ tangents drawn from an external point to a circle are equal]

BP = BQ ……. (2)

CR = CQ ……. (3)

DR = DS ……. (4)

Adding (1), (2), (3) and (4) we get

AP + BP + CR + DR = AS + BQ + CQ + DS

(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)

AB + DC = AD + BC

AB + AB = AD + AD

[∵ Opposite sides of a parallelogram are equal]

2AB = 2AD

AB = AD

Hence, AB = CD and AD = BC [∵ Opposite sides of a parallelogram]

∴ AB = BC = CD = AD

Thus □ ABCD is a rhombus (Q.E.D.)

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 (See below figure). Find the sides AB and AC.

Answer:

The given figure can also be drawn as

Given: Let △ABC be the given triangle circumscribing the given circle with centre ‘O’ and radius 3 cm.

i.e., the circle touches the sides BC, CA and AB at D, E, F respectively.

It is given that BD = 9 cm

CD = 3 cm

∵ Lengths of two tangents drawn from an external point to a circle are equal.

∴ BF = BD = 9 cm

CD = CE = 3 cm

AF = AE = x cm say

∴ The sides of die triangle are

12 cm, (9 + x) cm, (3 + x) cm

Perimeter = 2S = 12 + 9 + x + 3 + x

⇒ 2S = 24 + 2x

or S = 12 + x

S - a = 12 + x - 12 = x

S - b = 12 + x - 3 - x = 9

S - c = 12 + x - 9 - x = 3

∴ Area of the triangle

Squaring on both sides we get,

27 (x^{2}+ 12x) = (36 + 3x)^{2}

27x^{2}+ 324x = 1296 + 9x^{2}+ 216x

⇒ 18x^{2}+ 108x- 1296 = 0

⇒ x^{2}+ 6x - 72 = 0

⇒ x^{2}+ 12x - 6x - 72 = 0

⇒ x (x + 12) - 6 (x + 12) = 0

⇒ (x - 6) (x + 12) = 0

⇒ x = 6 or - 12

But ‘x’ can’t be negative hence, x = 6

∴ AB = 9 + 6 = 15 cm

AC = 3 + 6 = 9 cm.

Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths. Verify by using Pythagoras Theorem.

Answer:

Steps of construction:

- Draw a circle with centre ‘O’ and radius 6 cm.
- Take a point P outside the circle such that OP =10 cm. Join OP.
- Draw the perpendicular bisector to OP which bisects it at M.
- Taking M as centre and PM or MO as radius draw a circle. Let the circle intersects the given circle at A and B.
- Join P to A and B.
- PA and PB are the required tan¬gents of lengths 8 cm each.

Proof: In △OAP

OA^{2}+ AP^{2}= 6^{2}+ 8^{2}

= 36 + 64 = 100

OP^{2}= 10^{2}= 100

∴ OA^{2}+ AP^{2}= OP^{2}

Hence AP is a tangent.

Similarly BP is a tangent.

Construct, a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also verify the measurement by actual calculation.

Answer:

Steps of construction:

- Draw two concentric circles with centre ‘O’ and radii 4 cm and 6 cm.
- Take a point ‘P’ on larger circle and join O, P.
- Draw the perpendicular bisector of OP which intersects it at M.
- Taking M as centre and PM or MO as radius draw a circle which intersects smaller circle at Q.
- Join PQ, which is a tangent to the smaller circle.

Draw a circle with the help of a bangle, take a point outside the circle. Con-struct the pair of tangents from this point to the circle measure them. Write conclusion.

Answer:

Steps of construction:

- Draw a circle with the help of a bangle.
- Draw two chords AB and AC. Perpendicular bisectors of AB and AC meets at ‘O’ which is the centre of the circle.
- Taking an outside point P, join OP.
- Let M be the midpoint of OP. Taking M as centre OM as radius, draw a circle which intersects the given circle at R and S. Join PR, PS which are the required tangents.

Conclusion: Tangents drawn from an external point to a circle are equal.

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 tangent to the circle at P bisects the side BC.

Answer:

Let ABC be a right triangle right angled at P.

Consider a circle with diametere AB.

From the figure, the tangent to the circle at B meets BC in Q.

Now QB and QP are two tangents to the circle from the same point P.

QB = QP …….. (1)

Also, ∠QPC = ∠QCP

∴ PQ = QC (2)

From (1) and (2);

QB = QC Hence proved.

Draw a tangent to a given circle with center O from a point ‘R’ outside the circle. How many tangents can be drawn to the circle from that point? [Hint: The distance of two points to the point of contact is the same.

Answer:

Only two tangents can be drawn from a given point outside the circle.

A chord of a circle of radius 10 cm. subtends a right angle at the centre. Find the area of the corresponding: (use π = 3.14)

i) Minor segment ii) Major segment

Answer:

Angle subtended by the chord = 90° Radius of the circle = 10 cm

Area of the minor segment = Area of the sector POQ - Area of △POQ

Area of the sector = x/360 × πr^{2}

90/360 × 3.14 × 10 × 10 = 78.5

Area of the triangle =12× base × height

= 1/2 × 10 × 10 = 50

∴ Area of the minor segment = 78.5 - 50 = 28.5 cm^{2}

Area of the major segment = Area of the circle - Area of the minor segment

= 3.14 × 10 × 10 - 28.5

= 314 - 28.5 cm^{2}

= 285.5 cm^{2}

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 circle.

(use π = 3.14 and √3 = 1.732)

Answer:

Radius of the circle r = 12 cm.

Area of the sector = x/360 × πr^{2}

Here, x = 120°

120/360 × 3.14 × 12 × 12 = 150.72

Drop a perpendicular from ‘O’ to the chord PQ.

△OPM = △OQM [∵ OP = OQ ∠P = ∠Q; angles opp. to equal sides OP & OQ; ∠OMP = ∠OMQ by A.A.S]

∴ △OPQ = △OPM + △OQM = 2 . △OPM

Area of △OPM = 1/2 × PM × OM

= 18 × 1.732 = 31.176 cm

∴ △OPQ = 2 × 31.176 = 62.352 cm^{2}

∴ Area of the minor segment

= (Area of the sector) - (Area of the △OPQ)

= 150.72 - 62.352 = 88.368 cm^{2}

A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm. sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades. (use π =22/7)

Answer:

Angle made by the each blade = 115°

Total area swept by two blades

= Area of the sector with radius 25 cm and angle 115°+ 115° = 230°

= Area of the sector = x/360 × πr^{2}

=230/360 × 22/7 × 25 × 25

= 1254.96

≃ 1255 cm^{2}

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).

Answer:

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

Area of I + Area of II

= Area of ABCD - Areas of two semicircles with radius 5 cm

= 10 × 10 - 2 × 1/2 × π × 5^{2}

= 100 - 3.14 × 25

= 100 - 78.5 = 21.5 cm^{2}

Similarly, Area of II + Area of IV = 21.5 cm^{2}

So, area of the shaded region = Area of ABCD - Area of unshaded region

= 100 - 2 × 21.5 = 100 - 43 = 57 cm^{2}

Find the area of the shaded region in figure, if ABCD is a square of side 7 cm. and APD and BPC are semicircles. (use π = 22/7)

Answer:

Given,

ABCD is a square of side 7 cm.

Area of the shaded region = Area of ABCD - Area of two semicircles with radius 7/2 = 3.5 cm

APD and BPC are semicircles.

= 7 × 7 - 2 × 1/2 × 22/7 × 3.5 × 3.5

= 49 - 38.5

= 10.5 cm^{2}

∴ Area of shaded region = 10.5 cm

In figure, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm., find the area of the shaded region, (use π = 22/7).

Answer:

Given, OACB is a quadrant of a Circle.

Radius = 3.5 cm; OD = 2 cm.

Area of the shaded region = Area of the sector - Area of △BOD

= 9.625 - 3.5 = 6.125 cm^{2}

∴ Area of shaded region = 6.125 cm^{2}.

AB and CD are respectively arcs of two concentric circles of radii 21 cm. and 7 cm. with centre O (See figure). If ∠AOB = 30°, find the area of the shaded region. (use π = 22/7).

Answer:

Given, AB and CD are the arcs of two concentric circles.

Radii of circles = 21 cm and 7 cm and ∠AOB = 30°

We know that,

Area of the sector = x/360 × πr^{2}

Area of the shaded region = Area of the OAB - Area of OCD

∴ Area of shaded region = 102.66 cm^{2}

Calculate the area of the designed region in figure, common between the two quadrants of the circles of radius 10 cm each, {use π = 3.14)

Answer:

Mark two points P, Q on the either arcs.

Let BD be a diagonal of ABCD

Now the area of the segment

= 28.5 + 28.5 = 57 cm^{2}

Side of the square = 10 cm

Area of the square = side × side

= 10 × 10 = 100 cm^{2}

Area of two sectors with centres A and C and radius 10 cm.

= 2 × ?r^{2}/360 × x = 2 × x/360 × 22/7 × 10 × 10

=1100/7

= 157.14 cm^{2}

∴ Designed area is common to both the sectors,

∴ Area of design = Area of both sectors - Area of square

= 157 - 100 = 57 cm^{2}

(or)

1100/7 - 100 = 1100?700/7

=400/7

= 57.1 cm^{2}

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