Exercise 10.1 Page: 147

 

1. How many tangents can a circle have?

Answer:

A circle can have infinitely many tangents since there are infinitely many points on the circumference of the circle and at each point of it, it has a unique tangent.

2. Fill in the blanks:

(i) A tangent to a circle intersects it in _______________ point(s).

(ii) A line intersecting a circle in two points is called a _______________.

(iii) A circle can have _______________ parallel tangents at the most.

(iv) The common point of a tangent to a circle and the circle is called _______________.

Answer:

(i) one
(ii) secant
(iii) two
(iv) point of contact

 

3. 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. Length PQ is:

(A) 12 cm

(B) 13 cm

(C) 8.5 cm

(D) √119 cm

Answer:

(D) 

We know that the line drawn from the centre of the circle to the tangent is perpendicular to the tangent.

∴OP ⊥ PQ

By applying Pythagoras theorem in ΔOPQ,

4. 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:

AB || CD || EF

AB,CD and EF are three parallel lines where EF is the tangent to the circle.

Here CD id secant (Intersecting circle at 2 points P and Q)

Exercise 10.2 Page: 151

1. 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) 

Let O be the centre of the circle.

Given that,

OQ = 25cm and PQ = 24 cm

As the radius is perpendicular to the tangent at the point of contact,

Therefore, OP ⊥ PQ

Applying Pythagoras theorem in ΔOPQ, we obtain

In right triangle OPQ,

[By Pythagoras theorem]

OP = 7 cm

Therefore, the radius of the circle is 7 cm.

Hence, alternative 7 cm is correct.

 

2. In Fig. 10.11, if TP and TQ are the two tangents to a circle with centre O so that ∠POQ = 110°, then ∠PTQ is equal to
(A) 60°
(B) 70°
(C) 80°
(D) 90°

Answer:

(B) It is given that TP and TQ are tangents.

Therefore, radius drawn to these tangents will be perpendicular to the tangents.

Thus, OP ⊥ TP and OQ ⊥ TQ

∠OPT = 90º

∠OQT = 90º

In quadrilateral POQT,

Sum of all interior angles = 360°

∠OPT + ∠POQ +∠OQT + ∠PTQ = 360°

⇒ 90°+ 110º + 90° +∠PTQ = 360°

⇒ ∠PTQ = 70°

Hence, alternative 70° is correct.

 

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 ∠ POA is equal to
(A) 50°  
(B) 60°
(C) 70°
(D) 80°

Answer:

(A)It is given that PA and PB are tangents.

Therefore, the radius drawn to these tangents will be perpendicular to the tangents.

Thus, OA ⊥ PA and OB ⊥ PB

∠OBP = 90º

∠OAP = 90º

In AOBP,

Sum of all interior angles = 360°

∠OAP + ∠APB +∠PBO + ∠BOA = 360°

90° + 80° +90º +∠BOA = 360°

∠BOA = 100°

In ΔOPB and ΔOPA,

AP = BP (Tangents from a point)

OA = OB (Radii of the circle)

OP = OP (Common side)

Therefore, ΔOPB ≅ ΔOPA (SSS congruence criterion)

A ↔ B, P ↔ P, O ↔ O

And thus, ∠POB = ∠POA

Hence, alternative 50° is correct.

 

4. Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

Answer:

Given: CD and EF are the tangents at the end points A and B of the diameter AB of a circle with centre O.

To prove: CD || EF.

Proof: CD is the tangent to the circle at the point A.

∴ ∠BAD = 90°

EF is the tangent to the circle at the point B.

∴ ∠ABE = 90°

Thus, ∠BAD = ∠ABE (each equal to 90°).

But these are alternate interior angles.

∴ CD || EF

 

5. Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.

Answer:

Let, O is the centre of the given circle.

A tangent PR has been drawn touching the circle at point P.

Draw QP ⊥ RP at point P, such that point Q lies on the circle.

∠OPR = 90° (radius ⊥ tangent)

Also, ∠QPR = 90° (Given)

∴ ∠OPR = ∠QPR

Now, the above case is possible only when centre O lies on the line QP.

Hence, perpendicular at the point of contact to the tangent to a circle passes through the centre of the circle.

 

6. The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.

Answer:

Since, the tangent at any point of a circle is perpendicular to radius through the point of contact.

Therefore, ∠OPQ = 90°

It is given that OQ = 5 cm

and    PQ = 4 cm

In right ΔOPQ, we have

OQ ² =OP ² +PQ ²
[Using Pythagoras Theorem]
OP ² = (5) ² – (4) ²

^{= 25 – 16 =9}

^{⇒ OP = 3 cm}

^{Hence, the radius of the circle is 3 cm.}

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

Answer:

Given Two circles have the same center O and AB is a chord of the larger circle touching the smaller circle at C; also. OA = 5 cm and OC = 3 cm

In Δ OAC,

⇒ AC = 4cm

∴ AB = 2AC (Since perpendicular drawn from the center of the circle bisects the chord)

∴ AB = 2 × 4 =  8cm

The length of the chord of the larger circle is 8 cm.

8. A quadrilateral ABCD is drawn to circumscribe a circle (see figure). Prove that:

AB + CD = AD + BC

Answer:

We know that the tangents from an external point to a circle are equal.

AP = AS ……….(i)

BP = BQ ……….(ii)

CR = CQ ……….(iii)

DR = DS……….(iv)

On adding eq. (i), (ii), (iii) and (iv), we get

(AP + BP) + (CR + DR)

= (AS + BQ) + (CQ + DS)

->AB + CD = (AS + DS) + (BQ + CQ)

so,AB + CD = AD + BC

 

9. In figure, XY and X′Y′ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X′Y′ at B. Prove that ∠ AOB = 90°.

Answer:

Given: In figure, XY and X’Y’ are two parallel tangents to a circle with centre O and another

tangent AB with point of contact C intersecting XY at A and X’Y’ at B.

Let us join point O to C.

In ΔOPA and ΔOCA,

OP = OC (Radii of the same circle)

AP = AC (Tangents from point A)

AO = AO (Common side)

ΔOPA ≅ ΔOCA (SSS congruence criterion)

Therefore, P ↔ C, A ↔ A, O ↔ O

∠POA = ∠COA …(i)

Similarly, ΔOQB ≅ ΔOCB

∠QOB = ∠COB …(ii)

Since POQ is a diameter of the circle, it is a straight line.

Therefore, ∠POA + ∠COA + ∠COB + ∠QOB = 180º

From equations (i) and  (ii),it can be observed that

2∠COA + 2 ∠COB = 180º

∠COA + ∠COB = 90º

∠AOB = 90°

10. Prove that the angel between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.

Answer:

 Let us Consider a circle with centre O. Let P be an external point from which two tangents PA and PB are drawn to the circle which are touching the circle at point A and B respectively and AB is the line segment, joining point of contacts A and B together such that it subtends ∠AOB at center O of the circle.

It can be observed that

OA ⊥ PA

∴ ∠OAP = 90°

Similarly, OB ⊥ PB

∴ ∠OBP = 90°

In quadrilateral OAPB,

Sum of all interior angles = 360º

∠OAP +∠APB +∠PBO +∠BOA = 360º

⇒ 90º + ∠APB + 90º + ∠BOA = 360º

⇒ ∠APB + ∠BOA = 180º

∴ The angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.

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

Answer:

Given: ABCD is a parallelogram circumscribing a circle.

To Prove: ABCD is a rhombus.

Proof: Since, the tangents from an external point to a circle are equal.

We know that the tangents drawn to a circle from an exterior point are equal in length.

∴ AP = AS, BP = BQ, CR = CQ and DR = DS.

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

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

 ∴ AB + CD = AD + BC or 2AB = 2BC            (since AB = DC and AD = BC)

∴ AB = BC = DC = AD.

Therefore, ABCD is a rhombus.

Hence, proved.

12. A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see figure). Find the sides AB and AC.

Answer:

In ΔABC,

Length of two tangents drawn from the same point to the circle are equal,

∴ CF = CD = 6cm

∴ BE = BD = 8cm

∴ AE = AF =x

We observed that,

13. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Answer:

Let ABCD be a quadrilateral circumscribing a circle with O such that it touches the circle at point P, Q, R, S. Join the vertices of the quadrilateral ABCD to the center of the circle.
In ΔOAP and ΔOAS,
AP = AS (Tangents from the same point)
OP = OS (Radii of the circle)
OA = OA (Common side)
ΔOAP ≅ ΔOAS (SSS congruence condition)
∴ ∠POA = ∠AOS
⇒∠1 = ∠8
Similarly we get,
∠2 = ∠3
∠4 = ∠5
∠6 = ∠7
Adding all these angles,
∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 +∠8 = 360º
⇒ (∠1 + ∠8) + (∠2 + ∠3) + (∠4 + ∠5) + (∠6 + ∠7) = 360º
⇒ 2 ∠1 + 2 ∠2 + 2 ∠5 + 2 ∠6 = 360º
⇒ 2(∠1 + ∠2) + 2(∠5 + ∠6) = 360º
⇒ (∠1 + ∠2) + (∠5 + ∠6) = 180º
⇒ ∠AOB + ∠COD = 180º
Similarly, we can prove that ∠ BOC + ∠ DOA = 180º
Hence, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.