CBSE Class 10 Maths Chapter 10 Circles Objective Questions are based on concepts, such as the properties of a circle and the existence of the tangents to a circle. In this chapter, taken from the Unit 4 Geometry of CBSE Class 10 Maths Syllabus, students are introduced to some complex terms such as tangents, tangents to a circle, number of tangents from a point on the circle, and so on. We have compiled here the objective questions from CBSE Class 10 Maths Chapter 10 – Circles, keeping in mind that the number of MCQs is likely to increase for the upcoming board exams.
Also, these CBSE Class 10 Maths Objective Questions are mostly categorised topicwise.
SubTopics Covered in Chapter 10 – Circles
Find below some of the subtopics that are included in the given objective questions:
 10.1 Introduction to Circles (5 MCQs from the Topic)
 10.2 Tangent to the Circle (8 MCQs from the Topic)
 10.3 Theorems (7 MCQs from the Topic)
Download CBSE Class 10 Maths Chapter 10 – Circles Objective Questions PDF Free
Introduction to Circles

 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 OCÂ âŠ¥Â AB
Let OA = a, and OC = b.
Since OC âŠ¥ AB, and OC bisects AB,
It is perpendicular from the centre to the chord and bisects the chord.
In the right Î” ACO, we have
OA^{2}=OC^{2}+AC^{2}Â Â [by Pythagoras’ theorem]

 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.
 2 cm, 3 cm, 4 cm
 1 cm, 2 cm, 4 cm
 1 cm, 2.5 cm, 3.5 cm
 3 cm, 4 cm, 1 cm
 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.
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 = 5Â Â Â Â â€¦..(1)
y+z = 6Â Â Â …… (2) (â‡’Â y=6z)…Â Â (2.1)
x+z = 7Â Â Â Â â€¦..(3)
Adding (1), (2) and (3), we haveÂ 2(x+y+z) =5+6+7=18
âŸ¹x+y+z=9…. (4)
Using (1) in (4), we haveÂ 5+z=9âŸ¹z=4
Now, using (3) âŸ¹x=7âˆ’z=7âˆ’4=3
AndÂ (2.1) âŸ¹y=6âˆ’z=6âˆ’4=2
Therefore, the radii of the circles are 2 cm, 3 cm and 4 cm.

 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.
 5 cm
 7 cm
 10 cm
 12 cm
 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.
Answer: (A) 5 cm
Solution:
Since tangent to a circle is perpendicular to the radius through the point of contact,
So,Â âˆ OTP=90Â°
So, in triangle OTP
(OP)^{2}=(OT)^{2}+(PT)^{2}
13^{2}=(OT)^{2}+12^{2}
(OT)^{2}=13^{2}âˆ’12^{2}
OT^{2}=25
OT= âˆš25
OT = 5
So, the radius of the circle is 5 cm.

 In the adjoining figure, ‘O’ is the centre of the circle, âˆ CAO = 25Â° Â andÂ âˆ CBO =Â 35Â°. What is the value ofÂ âˆ AOB?


 120Â°
 110Â°
 55Â°
 Data insufficient

Answer: (A) 120Â°
Solution:
InÂ Î”AOC,
OA=OCÂ Â Â ——–(radii of the same circle)
âˆ´ Î”AOC is an isosceles triangle.
â†’âˆ OAC=âˆ OCA=25Â°—– (base angles of an isosceles triangle )
InÂ Î”BOC,
OB=OCÂ Â Â ——–(radii of the same circle)
âˆ´ Î”BOC is an isosceles triangle.
â†’âˆ OBC=âˆ OCB=35Â°Â —–(base angles of an isosceles triangle )
âˆ ACB=25Â°+35Â°=60Â°
âˆ AOB=2Ã—âˆ ACB —(angle at the centre is twice the angle at the circumference)
= 2Ã—60Â°
= 120Â°
Therefore, the value of âˆ AOB is 120Â°.

 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:Â 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

 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.


 2âˆš313
 12
 7
 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 âˆ OTA = 90Â°. In the right â–³OTA, we have
OT^{2}Â =Â OA^{2}Â â€“Â Â AT^{2}
= [(26)^{2}Â â€“Â (24)^{2}] = (26 + 24) (26 â€“ 24) = 100.
=> OT = âˆš100 = 10 cm
Hence, the radius of the circle is 10 cm.

 The quadrilateral formed by joining the angle bisectors of a cyclic quadrilateral is a
 cyclic quadrilateral
 parallelogram
 square
 Rectangle
 The quadrilateral formed by joining the angle bisectors of a cyclic quadrilateral is a
Answer: (A) cyclic quadrilateral
Solution:
ABCD is a cyclic quadrilateral
âˆ´ âˆ A +âˆ C = 180Â° and âˆ B+ âˆ D =Â Â 180Â°
1/2âˆ A+1/2Â âˆ C =Â 90Â°Â andÂ 1/2Â âˆ B+1/2Â âˆ D =Â Â 90Â°
x + z =Â 90Â°Â and y + w =Â Â 90Â°
InÂ â–³ARB andÂ â–³CPD, x+y +Â âˆ ARB =Â 180Â°Â and z+w+Â âˆ CPD =Â Â 180Â°
âˆ ARB =Â 180Â°Â â€“ (x+y) andÂ âˆ CPD =Â 180Â°Â â€“ (z+w)
âˆ ARB+âˆ CPD =Â 360Â°Â â€“ (x+y+z+w) =Â 360Â°Â â€“ (90+90)
=Â 360Â°Â â€“Â 180Â°Â âˆ ARB+âˆ CPD =Â Â 180Â°
âˆ SRQ+âˆ QPS =Â 180Â°
The sum of a pair of opposite angles of a quadrilateral PQRS isÂ 180âˆ˜.
Hence, PQRS is a cyclic quadrilateral.

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


 25Â°
 45Â°
 60Â°
 30Â°

Answer: (B) 45Â°
Solution: OB = OD (radius)
âˆ Â ODB =Â âˆ Â OBD
âˆ Â ODB +Â âˆ OBD +Â âˆ BOD =Â 180Â°
2âˆ ODB +Â 90Â°Â =Â 180Â°
âˆ ODB =Â 45Â°
âˆ OBD =Â âˆ ACD (Angle subtended by the common chord AD)
Therefore, âˆ ACD = 45Â°

 Find the value of âˆ DCE:


 80Â°
 75Â°
 90Â°
 100Â°

Answer: (A) 80Â°
Solution: âˆ Â BAD =1/2Â BOD
âˆ BAD =1/2(160Â°)
âˆ BAD = 80Â°
ABCD is a cyclic quadrilateral
âˆ BAD + âˆ BCD = 180Â°
âˆ BCD = 100Â°
âˆ DCE = 180Â° âˆ BCD
âˆ DCE = 180Â°â€“ 100Â° = 80Â°
Therefore, âˆ DCE = 80Â°.

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


 35Â°
 85Â°
 90Â°
 115Â°

Answer: (D) 115Â°
Solution:
Join OB and OD
We know that OB is perpendicular to PQ
âˆ OBD =Â âˆ OBQ –Â âˆ DBQ
âˆ OBD =Â 90Â°Â â€“Â 65Â°
âˆ OBD =Â 25Â°
OB = OD (radius)
âˆ OBD =Â âˆ ODB =Â 25Â°
In â–³ODB,
âˆ OBD +Â âˆ ODB +Â âˆ BOD =Â 180Â°
25Â°Â +Â 25Â°Â +Â âˆ BOD =Â 180Â°
âˆ BOD =Â 130Â°
âˆ BAD =Â 1/2Â âˆ BOD
(Angle subtended by a chord on the centre is double the angle subtended on the circle)
âˆ BAD =Â 1/2Â (130Â°)
âˆ BAD =Â 65Â°
ABCD is a cyclic quadrilateral
âˆ BCD +Â âˆ BAD =Â 180Â°
âˆ BCD +Â 65Â°Â =Â 180Â°
âˆ BCD = 180Â° – 65Â° = 115Â°
Therefore, âˆ BCD = 115Â°.

 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.
 9.6 cm
 10.8 cm
 4.8 cm
 None of these
 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.
Answer: (A) 9.6 cm
Solution:
Consider the triangles OAB and OAC are congruent as
AB=AC
OAÂ is common
OB = OC = 5cm.
So, âˆ OAB = âˆ 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.
InÂ â–³ADO
OD^{2}=Â AO^{2}Â â€“Â AD^{2}
OD^{2} =Â 5^{2}Â â€“Â 3^{2}
OD = 4 cm
So, the area of OAB = 1/2 AB x OD = 1/2 6 x 4 = 12 sq. cm.Â Â Â Â Â Â Â â€¦.. (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,
âˆ OAB = âˆ OAC.
Therefore, triangles being congruent, âˆ AEB = âˆ AEC = 90Â°
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 â–³ OAB =Â Â½Ã—AOÃ—BE =Â Â½ Ã— 5Ã—BE = 12 sq. cm as arrived in eq (i).
BE = 12Â Ã—Â 2/5 = 4.8cm
Therefore, BC = 2BE = 2Ã—4.8 cm = 9.6 cm.
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


 None of these
 15 cm
 30 cm
 23 cm

Answer: (C) 30 cm
Solution:
Given that
OB = OD =17
AB = 16 â‡’ AE = BE = 8 cm, as perpendicular to the chord from the centre bisects the chords
EF = 23 cm
ConsiderÂ â–³OEB
OE^{2}Â =Â Â OB^{2}Â –Â EB^{2}
OE^{2}Â =Â Â 17^{2}Â –Â Â 8^{2}
OE = 15 CM
OF = EF – OE
OF = 23 – 15
OF = 8 cm
FD^{2}Â =Â OD2Â –Â OF^{2}
FD^{2}Â =Â 17^{2}Â –Â Â 8^{2}
FD = 15
Therefore, CD = 2FD = 30 cm

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


 10 cm
 8 cm
 6 cm
 4 cm

Answer: (B) 8 cm
Solution: âˆ OAC = âˆ CBP =Â 90Â°
âˆ OCA = âˆ PCB (Vertically opposite angle)
Triangle OAC is similar to PBC
OA/PBÂ =Â OC/PC
3/3Â =Â OC/PC
OC = PC
But, PO = 10 cm
Therefore, OC = PC = 5cm
AC^{2Â }=Â OC^{2}Â â€“Â Â OA^{2}
AC^{2Â }=Â 5^{2}Â â€“Â Â 3^{2}
AC = 4 cm
Similarly, BC = 4 cm
Therefore, AB = 8 cm
Theorems

 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
 4 cm
 5 cm
 None of these
 6 cm
 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
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 =Â 90^{2}
Applying the Pythagoras theorem to triangle OPQ,
OQ^{2}Â +Â QP^{2}Â =Â OP^{2}
OQ^{2Â }+Â 8^{2Â }=Â 10^{2}
OQ^{2Â }= 10064
=36
OQ = 6 cm.
Hence, the radius of the circle is 6 cm.

 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 âˆ ACB = 75Â°, then âˆ AOB =


 75Â°
 85Â°
 95Â°
 105Â°

Answer: (D) 105Â°
Solution: âˆ OAC =Â âˆ OBC =Â 90Â°â€‹
âˆ OAC +Â âˆ OBC +Â âˆ ACB +Â âˆ AOB =Â 360Â°â€‹Â ….. (sum of angles of a quadrilateral)
90Â°â€‹â€‹ +Â 90Â°â€‹â€‹ +Â 75Â°â€‹â€‹ +Â âˆ AOB = 360Â°
âˆ AOB =Â 105Â°

 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 âˆ OPA = 35Â°, then âˆ POB =


 55Â°
 65Â°
 85Â°
 75Â°

Answer: (A) 55Â°
Solution: âˆ OAP =âˆ OBP =Â 90Â°
âˆ AOP =Â 180Â°Â 35Â°Â 90Â°
âˆ AOP =Â 55Â°
OA = OB
AP = PB
OP is a common base
Therefore, â–³OAP â‰… â–³OBP
âˆ AOP =Â âˆ BOP
Therefore, âˆ BOP = 55Â°

 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.
 âˆš119
 8.5 cm
 13 cm
 12 cm
 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.
Answer: (D) 12 cm
Solution:
Given that OP = 5 cm, OQ = 13Â cm
To find PQ
Apply the Pythagoras theorem to triangle OPQ,
OP^{2}Â +Â QP^{2}Â =Â OQ^{2}
5^{2Â }+Â QP^{2}Â =Â 13^{2}
QP^{2}Â = 169Â â€“ 25= 144
QP = âˆš144cm
Thus, QP = 12 cm.

 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
 âˆš7
 7 cm
 5 cm
 25 cm
 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
Answer: (C) 5 cm
Solution:
Given that AB = 4 cm, OB =3 cm
To find OA,
Apply the Pythagoras theorem to triangle OAB
OB^{2Â }+Â AB^{2}Â =Â OA^{2}
3^{2}Â +Â 4^{2Â }= OA
OA^{2Â }= 25
OA = 5 cm
Therefore, the distance of A from the centre of the circle is 5 cm.

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


 90Â°
 80Â°
 70Â°
 60Â°

Answer: (C) 70Â°
Solution:
We know thatÂ âˆ OQT=âˆ OPT=90Â°.
Also, âˆ OQT+âˆ OPT+âˆ POQ+âˆ PTQ=360Â°.
âˆ PTQ=360Â°â€“90Â°â€“90Â°â€“110Â°
=Â 70Â°
âˆ´ âˆ PTQ = 70Â°

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


 25Â°
 30Â°
 45Â°
 60Â°

Answer: (B) 30Â°
Solution:
Join OA
As the tangent at any point of a circle is perpendicular to the radius through the point of contact,
âˆ OAQ =Â 90Â°
âˆ OAB =Â âˆ OAQ –Â âˆ BAQ
âˆ OAB =Â 90Â°Â â€“Â 60Â°
âˆ OAB =Â 30Â°
OA = OB (radius)
âˆ OAB =Â âˆ OBA
Therefore, âˆ OBA = 30Â°
âˆ ABC =Â 30Â°
These objective questions from Chapter 10 of CBSE Class 10 textbooks are framed based on the concepts involved in circles. This will help the students to understand the concept better and also to score good marks in the Class 10 board examination.
Also, find below some additional questions for the students to practise.
CBSE Class 10 Maths Chapter 10 Extra Questions
Â 1. What is the number of tangents drawn at a point of the circle?
(a) one
(b) two
(c) none
(d) infinite
2. If the length of a tangent drawn from a point at a distance of 10 cm from the circle is 8 cm., what is the radius of the circle?Â
(a) 4 cm
(b) 5 cm
(c) 6 cm
(d) 7 cm
3. What is the distance between two parallel tangents of a circle of radius 4 cm?
(a) 2 cm
(b) 4 cm
(c) 6 cm
(d) 8 cm
To access more study resources of CBSE Class 10, visit BYJUâ€™S website or download BYJUâ€™S – The Learning App.
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