Exercise 10.1
1. How many tangents can a circle have?
Ans:
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 _______________.
Ans:
(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
Ans: (D)
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.
Ans: (D)
Exercise 10.2
Q. 1 to 3 Choose the correct option and give justification.
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
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]
Ans: (A)
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°2
Ans: (B)
Answer:
(B) It is given that TP and TQ 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
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°
Ans:
(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.
Ans:
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.
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.
Ans:
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.
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 2 = OA2 – OC2
AC 2 = 52 – 32
AC 2 = 25 – 9
AC 2 = 16
AC = 4
⇒ 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
Ans:
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°.
Ans:
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.
Ans: 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 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.
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.
Ans: Consider the Triangle ABC, We know that the length of any two tangents which are drawn from the same point to the circle is equal.
So,
(i) CF = CD = 6 cm
(ii) BE = BD = 8 cm
(iii) AE = AF = x
Now, it can be observed that,
(i) AB = EB+AE = 8+x
(ii) CA = CF+FA = 6+x
(iii) BC = DC+BD = 6+8 = 14
Now the semi-perimeter “s” will be calculated as follows
2s = AB+CA+BC
By putting the respective values, we get,
2s = 28+2x
s = 14+xBy solving this, we get,
= √(14+x)48x ……… (i)
Again, the area of △ABC = 2 × area of (△AOF + △COD + △DOB)
= 2×[(½×OF×AF)+(½×CD×OD)+(½×DB×OD)]
= 2×½(4x+24+32) = 56+4x …………..(ii)
Now from (i) and (ii), we get,
√(14+x)48x = 56+4x
Now, square both sides,
48x(14+x) = (56+4x)2
48x = [4(14+x)]2/(14+x)
48x = 16(14+x)
48x = 224+16x
32x = 224
x = 7 cm
So, AB = 8+x
i.e. AB = 15 cm
And, CA = x+6 =13 cm.
13. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
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.