Class 10 Maths Chapter 10 Circles Objective Questions

 

Introduction to Circles

1. 1. Two concentric circles of radii a and b (a > b) are given. Find the length of the chord of the larger circle, which touches the smaller circle.

Solution: Let O be the common centre of the two circles and AB be the chord of the larger circle, which touches the smaller circle at C.

Join OA and OC.

Then OC ⊥ AB

Let OA = a, and OC = b.

Since OC ⊥ AB, and OC bisects AB,

It is perpendicular from the centre to the chord and bisects the chord.

In the right Δ ACO, we have

OA2=OC2+AC2    [by Pythagoras’ theorem]

2. 2. Three circles touch each other externally. The distance between their centres is 5 cm, 6 cm and 7 cm. Find the radii of the circles.
      1. 2 cm, 3 cm, 4 cm
      2. 1 cm, 2 cm, 4 cm
      3. 1 cm, 2.5 cm, 3.5 cm
      4. 3 cm, 4 cm, 1 cm

Answer: (A) 2 cm, 3 cm, 4 cm

Solution: Consider the below figure wherein three circles touch each other externally.

Since the distances between the centres of these circles are 5 cm, 6 cm and 7 cm, respectively, we have the following set of equations with respect to the above diagram:

x+y = 5     …..(1)

y+z = 6     …… (2) (⇒ y=6-z)…   (2.1)

x+z = 7     …..(3)

Adding (1), (2) and (3), we have 2(x+y+z) =5+6+7=18

⟹x+y+z=9…. (4)

Using (1) in (4), we have 5+z=9⟹z=4

Now, using (3) ⟹x=7−z=7−4=3

And (2.1) ⟹y=6−z=6−4=2

Therefore, the radii of the circles are 2 cm, 3 cm and 4 cm.

3. 3. A point P is 13 cm from the centre of a circle. The length of the tangent drawn from P to the circle is 12 cm. Find the radius of the circle.
      1. 5 cm
      2. 7 cm
      3. 10 cm
      4. 12 cm

Answer: (A) 5 cm

Solution:

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

So, ∠OTP=90°

So, in triangle OTP

(OP)2=(OT)2+(PT)2

132=(OT)2+122

(OT)2=132−122

OT2=25

OT= √25

OT = 5

So, the radius of the circle is 5 cm.

4. 4. In the adjoining figure, ‘O’ is the centre of the circle, ∠CAO = 25°  and ∠CBO = 35°. What is the value of ∠AOB?

4. 4. 1. 120°
      2. 110°
      3. 55°
      4. Data insufficient

Answer: (A) 120°

Solution:

In ΔAOC,

OA=OC      ——–(radii of the same circle)

∴ ΔAOC is an isosceles triangle.

→∠OAC=∠OCA=25°—– (base angles of an isosceles triangle )

In ΔBOC,

OB=OC      ——–(radii of the same circle)

∴ ΔBOC is an isosceles triangle.

→∠OBC=∠OCB=35° —–(base angles of an isosceles triangle )

∠ACB=25°+35°=60°

∠AOB=2×∠ACB —-(angle at the centre is twice the angle at the circumference)

= 2×60°

= 120°

Therefore, the value of ∠AOB is 120°.

5. 5. A. What is a line called if it meets the circle at only one point?

B. The collection of all points equidistant from a fixed point is ______.

1: Chord

2: Tangent

3: Circle

4: Curve

5: Secant

Solution: A. Tangent is a line that touches the circle at only 1 point.

B. The collection of all points equidistant from a fixed point is called a circle.

Tangent to the Circle

6. 6. Point A is 26 cm away from the centre of a circle, and the length of the tangent drawn from A to the circle is 24 cm. Find the radius of the circle.

6. 6. 1. 2√313
      2. 12
      3. 7
      4. 10

Answer: (D) 10

Solution: Let O be the centre of the circle, and let A be a point outside the circle such that OA = 26 cm.

Let AT be the tangent to the circle

Then, AT = 24 cm

Join OT

Since the radius through the point of contact is perpendicular to the tangent, we have ∠OTA = 90°. In the right △OTA, we have

OT2 = OA2 –  AT2

= [(26)2 – (24)2] = (26 + 24) (26 – 24) = 100.

=> OT = √100 = 10 cm

Hence, the radius of the circle is 10 cm.

7. 7. The quadrilateral formed by joining the angle bisectors of a cyclic quadrilateral is a
      1. cyclic quadrilateral
      2. parallelogram
      3. square
      4. Rectangle

Answer: (A) cyclic quadrilateral

Solution:

ABCD is a cyclic quadrilateral

∴ ∠A +∠C = 180° and ∠B+ ∠D =  180°

1/2∠A+1/2 ∠C = 90° and 1/2 ∠B+1/2 ∠D =  90°

x + z = 90° and y + w =  90°

In △ARB and △CPD, x+y + ∠ARB = 180° and z+w+ ∠CPD =  180°

∠ARB = 180° – (x+y) and ∠CPD = 180° – (z+w)

∠ARB+∠CPD = 360° – (x+y+z+w) = 360° – (90+90)

= 360° – 180° ∠ARB+∠CPD =  180°

∠SRQ+∠QPS = 180°

The sum of a pair of opposite angles of a quadrilateral PQRS is 180∘.

Hence, PQRS is a cyclic quadrilateral.

8. 8. In the given figure, AB is the diameter of the circle. Find the value of ∠ ACD.

8. 8. 1. 25°
      2. 45°
      3. 60°
      4. 30°

Answer: (B) 45°

Solution: OB = OD (radius)

∠ ODB = ∠ OBD

∠ ODB + ∠OBD + ∠BOD = 180°

2∠ODB + 90° = 180°

∠ODB = 45°

∠OBD = ∠ACD (Angle subtended by the common chord AD)

Therefore, ∠ACD = 45°

9. 9. Find the value of ∠DCE:

9. 9. 1. 80°
      2. 75°
      3. 90°
      4. 100°

Answer: (A) 80°

Solution: ∠ BAD =1/2 BOD

∠BAD =1/2(160°)

∠BAD = 80°

ABCD is a cyclic quadrilateral

∠BAD + ∠BCD = 180°

∠BCD = 100°

∠DCE = 180°- ∠ BCD

∠DCE = 180°– 100° = 80°

Therefore, ∠DCE = 80°.

10. 10. ABCD is a cyclic quadrilateral, and PQ is a tangent at B. If ∠DBQ = 65°, then ∠BCD is

10. 10. 1. 35°
        2. 85°
        3. 90°
        4. 115°

Answer: (D) 115°

Solution:

Join OB and OD

We know that OB is perpendicular to PQ

∠OBD = ∠OBQ – ∠DBQ

∠OBD = 90° – 65°

∠OBD = 25°

OB = OD (radius)

∠OBD = ∠ODB = 25°

In △ODB,

∠OBD + ∠ODB + ∠BOD = 180°

25° + 25° + ∠BOD = 180°

∠BOD = 130°

∠BAD = 1/2 ∠BOD

(Angle subtended by a chord on the centre is double the angle subtended on the circle)

∠BAD = 1/2 (130°)

∠BAD = 65°

ABCD is a cyclic quadrilateral

∠BCD + ∠BAD = 180°

∠BCD + 65° = 180°

∠BCD = 180° – 65° = 115°

Therefore, ∠BCD = 115°.

11. 11. In a circle of radius 5 cm, AB and AC are the two chords such that AB = AC = 6 cm. Find the length of the chord BC.
        1. 9.6 cm
        2. 10.8 cm
        3. 4.8 cm
        4. None of these

Answer: (A) 9.6 cm

Solution:

Consider the triangles OAB and OAC are congruent as

AB=AC

OA is common

OB = OC = 5cm.

So, ∠OAB = ∠OAC

Draw OD perpendicular to AB

Hence, AD = AB/2 = 6/2 = 3 cm, as the perpendicular to the chord from the centre bisects the chord.

In △ADO

OD2= AO2 – AD2

OD2 = 52 – 32

OD = 4 cm

So, the area of OAB = 1/2 AB x OD = 1/2 6 x 4 = 12 sq. cm.        ….. (i)

Now, AO extended should meet the chord at E, and it is middle of the BC as ABC is isosceles with AB = AC.

Triangles AEB and AEC are congruent, as

AB = AC

AE common,

∠OAB = ∠OAC.

Therefore, triangles being congruent, ∠AEB = ∠AEC = 90°

Therefore, BE is the altitude of the triangle OAB with AO as a base.

Also, this implies BE = EC or BC = 2BE

Therefore, the area of the △ OAB = ½×AO×BE = ½ × 5×BE = 12 sq. cm as arrived in eq (i).

BE = 12 × 2/5 = 4.8cm

Therefore, BC = 2BE = 2×4.8 cm = 9.6 cm.

12. 12. 

In a circle of radius 17 cm, two parallel chords are drawn on opposite sides of a diameter. The distance between the chords is 23 cm. If the length of one chord is 16 cm, then the length of the other is

12. 12. 1. None of these
        2. 15 cm
        3. 30 cm
        4. 23 cm

Answer: (C) 30 cm

Solution:

Given that

OB = OD =17

AB = 16 ⇒ AE = BE = 8 cm, as perpendicular to the chord from the centre bisects the chords

EF = 23 cm

Consider  △OEB

OE2 =  OB2 – EB2

OE2 =  172 –  82

OE = 15 CM

OF = EF – OE

OF = 23 – 15

OF = 8 cm

FD2 = OD2 – OF2

FD2 = 172 –  82

FD = 15

Therefore, CD = 2FD = 30 cm

13. 13. The distance between the centres of equal circles, each of radius 3 cm, is 10 cm. The length of a transverse tangent AB is

13. 13. 1. 10 cm
        2. 8 cm
        3. 6 cm
        4. 4 cm

Answer: (B) 8 cm

Solution: ∠OAC = ∠CBP = 90°

∠OCA = ∠PCB (Vertically opposite angle)

Triangle OAC is similar to PBC

OA/PB = OC/PC

3/3 = OC/PC

OC = PC

But, PO = 10 cm

Therefore, OC = PC = 5cm

AC2 = OC2 –  OA2

AC2 = 52 –  32

AC = 4 cm

Similarly, BC = 4 cm

Therefore, AB = 8 cm

Theorems

14. 14. A point P is 10 cm from the centre of a circle. The length of the tangent drawn from P to the circle is 8 cm. The radius of the circle is equal to
        1. 4 cm
        2. 5 cm
        3. None of these
        4. 6 cm

Answer: (D) 6 cm

Solution:

Given that OP = 10 cm, PQ = 8 cm

A tangent to a circle is perpendicular to the line joining the centre of the circle to the tangent at the point of contact with the circle.

Angle OQP = 902

Applying the Pythagoras theorem to triangle OPQ,

OQ2 + QP2 = OP2

OQ2 + 82 = 102

OQ2 = 100-64

=36

OQ = 6 cm.

Hence, the radius of the circle is 6 cm.

15. 15. In the figure, O is the centre of the circle, CA is tangent at A, and CB is tangent at B drawn to the circle. If ∠ACB = 75°, then ∠AOB =

 

15. 15. 1. 75°
        2. 85°
        3. 95°
        4. 105°

Answer: (D) 105°

Solution: ∠OAC = ∠OBC = 90°​

∠OAC + ∠OBC + ∠ACB + ∠AOB = 360°​  ….. (sum of angles of a quadrilateral)

90°​​ + 90°​​ + 75°​​ + ∠AOB = 360°

∠AOB = 105°

16. 16. PA and PB are the two tangents drawn to the circle. O is the centre of the circle. A and B are the points of contact of the tangents PA and PB with the circle. If ∠OPA = 35°, then ∠POB =

16. 16. 1. 55°
        2. 65°
        3. 85°
        4. 75°

Answer: (A) 55°

Solution: ∠OAP =∠OBP = 90°

∠AOP = 180°- 35°- 90°

∠AOP = 55°

OA = OB

AP = PB

OP is a common base

Therefore, △OAP ≅ △OBP

∠AOP = ∠BOP

Therefore, ∠BOP = 55°

17. 17. 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, such that OQ =13 cm. Find the length of PQ.
        1. √119
        2. 8.5 cm
        3. 13 cm
        4. 12 cm

Answer: (D) 12 cm

Solution:

Given that OP = 5 cm, OQ = 13 cm

To find PQ

Apply the Pythagoras theorem to triangle OPQ,

OP2 + QP2 = OQ2

52 + QP2 = 132

QP2 = 169 – 25= 144

QP = √144cm

Thus, QP = 12 cm.

18. 18. The length of the tangent from point A to a circle of radius 3 cm is 4 cm. The distance of A from the centre of the circle is
        1. √7
        2. 7 cm
        3. 5 cm
        4. 25 cm

Answer: (C) 5 cm

Solution:

Given that AB = 4 cm, OB =3 cm

To find OA,

Apply the Pythagoras theorem to triangle OAB

OB2 + AB2 = OA2

32 + 42 = OA

OA2 = 25

OA = 5 cm

Therefore, the distance of A from the centre of the circle is 5 cm.

19. 19. If TP and TQ are two tangents to a circle with centre O, such that ∠POQ = 110°, then ∠PTQ is equal to:

19. 19. 1. 90°
        2. 80°
        3. 70°
        4. 60°

Answer: (C) 70°

Solution:

We know that ∠OQT=∠OPT=90°.

Also, ∠OQT+∠OPT+∠POQ+∠PTQ=360°.

∠PTQ=360°–90°–90°–110°

= 70°

∴ ∠PTQ = 70°

20. 20. In the given figure, PAQ is the tangent, and BC is the diameter of the circle. If ∠BAQ = 60°, find  ∠ABC.

20. 20. 1. 25°
        2. 30°
        3. 45°
        4. 60°

Answer: (B) 30°

Solution:

Join OA

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

∠OAQ = 90°

∠OAB = ∠OAQ – ∠BAQ

∠OAB = 90° – 60°

∠OAB = 30°

OA = OB (radius)

∠OAB = ∠OBA

Therefore, ∠OBA = 30°

∠ABC = 30°