The Class 10 Maths Chapter 10 Circles contains details like the properties of circle and the existence of the tangents to a circle. In this chapter, taken from the Unit 4 Geometry of CBSE Syllabus 201920, the 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. Also, keeping in mind the fact that the number of MCQs are likely to increase for the upcoming board exams, we have compiled here the CBSE Class 10 Maths Chapter 10Circles Objective Questions.
These CBSE Class 10 Maths objective questions are most likely categorised topic wise.
SubTopics Discussed in Chapter 10 Circles
Find here some of the subtopics that are discussed via these objective questions we have given below:
10.1 Introduction to Circles (5 MCQs From This Topic)
10.2 Tangent to the Circle (8 MCQs From The Given Topic)
10.3 Theorems (7 MCQs Listed From Topic)
Download CBSE Class 10 Maths Chapter 10Circles 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 center 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
LetÂ OA = a and OC = b.
SinceÂ OCÂ âŠ¥Â AB, OC bisects AB
[âˆµÂ perpendicular from the centre to a chord bisects the chord].In rightÂ Î”Â 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 theÂ centres of these circles areÂ 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,Â (3) âŸ¹x=7âˆ’z=7âˆ’4=3
AndÂ (2.1) âŸ¹y=6âˆ’z=6âˆ’4=2
Therefore, the radii of the circles are 3Â cm, 2Â cm and 4 cm.

 A point P is 13 cm from the centre of the circle. The length of the tangent drawn from P to the circle is 12cm. Find the radius of the circle.
 5cm
 7cm
 10cm
 12cm
 A point P is 13 cm from the centre of the circle. The length of the tangent drawn from P to the circle is 12cm. Find the radius of the circle.
Answer: (A) 5cm
Solution:
Since,
tangent toÂ 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, radius of the circle is 5 cm

 In the adjoining figure ‘O’ is the center of circle,Â âˆ CAO =Â 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”AOCÂ 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”BOCÂ 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 ACBÂ —(angle at the center is twice the angle at the circumference)
= 2Ã—60Â°
=120Â°

 A: What is a line called, if it meets the circle at only one point?
B: Collection of all points equidistant from a fixed point is ______.
1: Chord
2: Tangent
3: Circle
4: Curve
5:Â Secant
Which is correct matching?


 A2; B4
 A5; B4
 A4; B1
 A2; B3

Answer: (D) A2; B3
Solution: Tangent is a line which touches the circle at only 1 point.
Collection of all points equidistant from a fixed point is called a circle.
Tangent to the Circle

 A point A is 26 cm away from the centre of a circle and the length of 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 haveÂ âˆ OTA = 90Â°. In rightÂ â–³Â OTA, 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.
=>Â Â Â Â Â Â OT = âˆš100 = 10cm
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 =Â 180Â°Â andÂ âˆ B+Â 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 cyclic quadrilateral

 In the given figure, AB is the diameter of the circle. Find the value ofÂ âˆ Â 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)
ThereforeÂ âˆ ACD =Â 45Â°

 Find the value ofÂ âˆ Â DCE:


 80Â°
 75Â°
 90Â°
 100Â°

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

 ABCD is a cyclic quadrilateral PQ is a tangent at B. IfÂ âˆ DBQ =Â 65Â°, thenÂ 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°
InÂ 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 =Â 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
 None of these
 9.6cm
 10.8cm
 4.8cm
 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: (B) 9.6cm
Solution:
Consider the triangles OAB and OAC are congruent as
AB=AC
OAÂ is common
OB = OC = 5cm.
SoÂ âˆ OAB =Â âˆ OAC
Draw OD perpendicular to AB
Hence AD = AB/2 =Â 6/2Â = 3 cm as the perpendicular to the chord from the center 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 Area of OAB =Â 1/2Â AB x OD =Â 1/2Â 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 an 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 =Â 90Â°
ThereforeÂ BE is the altitude of the triangle OAB with AO as base.
AlsoÂ this implies BE =EC or BC =2BE
Therefore the area of theÂ â–³Â 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
 15cm
 30cm
 23cm

Answer: (C) 30cm
Solution:
Given that
OB = OD =17
AB = 16Â â‡’ AE = BE = 8 cm as perpendicular from centre to the chord bisects the chords
EF = 23 cm
ConsiderÂ Â 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


 10cm
 8cm
 6cm
 4cm

Answer: (B) 8cm
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 center 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
 4cm
 5cm
 None of these
 6cm
 A point P is 10 cm from the center 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) 6cm
Solution:
Given that OP = 10 cm, PQ = 8 cm
As, tangent to a circle is perpendicular to the line joining the centre of the circle to the tangent at the point of contact to the circle.
Angle OQP =Â 90^{2}
Applying 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.
Ans: Radius of the circle is 6 cm.

 In fig, O is the centre of the circle, CA is tangent at A and CB is tangent at B drawn to the circle. IfÂ âˆ ACB =Â 75Â°, thenÂ 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. IfÂ âˆ OPA =Â 35Â°, thenÂ 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 common base
ThereforeÂ â–³OAP â‰…Â â–³OBP
âˆ AOP =Â âˆ BOP
Ans:Â âˆ BOP =Â 55Â°

 A tangent PQ at a point P of a circle of radius 5 cm meets a line through the center O at a point Q such that OQ =13Â Â cm. Length PQ is:
 âˆš119
 8.5cm
 13cm
 12cm
 A tangent PQ at a point P of a circle of radius 5 cm meets a line through the center O at a point Q such that OQ =13Â Â cm. Length PQ is:
Answer: (D) 12cm
Solution:
Given that OP = 5 cm, OQ = 13Â cm
To find PQ
Applying 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
QP =12Â cm

 The length of the tangent from a point A to a circle, of radius 3 cm, is 4 cm. The distance of A from the centre of the circle is
 âˆš7
 7cm
 5cm
 25cm
 The length of the tangent from a 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) 5cm
Solution:
Given that AB = 4 cm, OB =3 cm
To find OA
Applying 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 center O such thatÂ âˆ POQ =Â 110Â°, then,Â âˆ PTQ is equal to:


 90Â°
 80Â°
 70Â°
 60Â°

Answer: (C)70Â°
Solution:
We know thatÂ âˆ OQT=âˆ OPT=90Â°.
AlsoÂ 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. BC is the diameter of the circle.Â âˆ BAQ =Â 60Â°, find Â âˆ ABC :


 25Â°
 30Â°
 45Â°
 60Â°

Answer: (B) 30Â°
Solution:
Join OA
As theÂ tangentÂ atÂ anyÂ point ofÂ aÂ circleÂ isÂ perpendicularÂ toÂ the radiusÂ throughÂ theÂ pointÂ ofÂ 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
ThereforeÂ âˆ OBA =Â 30Â°
âˆ ABC =Â 30Â°
These objective questions from the 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 for them to score good marks in the examination.
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