## AP State Syllabus SSC 10th Class Maths Solutions 9th Lesson Tangents and Secants to a Circle Exercise 9.2

### 10th Class Maths 9th Lesson Tangents and Secants to a Circle Ex 9.2 Textbook Questions and Answers

Question 1.

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 ]

Question 2.

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.

Question 3.

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.)

Question 4.

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.

Question 5.

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.

Question 6.

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.

Question 7.

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.

Question 8.

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.

Question 9.

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.