Class 10 Maths Chapter 10 Circle NCERT Solutions is an important chapter for students preparing for the Class 10 CBSE Board Examination. The NCERT textbook should be at the centre of your Mathematics preparation because it helps you understand the fundamental concepts, theorems, diagrams and methods that form the foundation of many examination questions. Simply completing the NCERT exercises once is not enough. Students should revise the chapter repeatedly and make NCERT Mathematics a part of their regular revision cycle. A good revision cycle can be: learn the concept → solve NCERT examples → attempt NCERT exercise questions without seeing the solution → check your mistakes → revise the concept → solve the question again → move towards PYQs and sample papers. When you follow this cycle regularly, the concepts become stronger and you become more confident in solving questions independently.
The Class 10 Maths NCERT Solutions provided on this page are meant to support your learning when you are unable to understand a question or want to check your method after making your own attempt. Do not make the mistake of reading every solution directly. First take your notebook, try the question yourself, and only then use the solution for verification and learning. Repeated revision of NCERT questions can help you improve your accuracy, understanding and ability to present mathematical solutions properly in the examination. Therefore, make a habit of returning to NCERT again and again instead of studying the chapter only once.
Connecting Class 10 Circles with What You Studied in Class 9
Before beginning Class 10 Maths Chapter 10 Circles, stop for a moment and remember what you studied about circles in Class 9 Mathematics. You have already developed an important foundation in the previous class. In Class 9, you studied concepts related to chords of a circle and the angle subtended by a chord at the centre, along with other basic properties of circles. Those ideas are not left behind; they become useful as you move into the more advanced concepts of Class 10.
In Class 10, you will take your understanding of circles one step further. You will study important concepts and theorems related to tangents to a circle, the relationship between a tangent and the radius at the point of contact, properties involving chords and tangents, and circles circumscribed around quadrilaterals. You will also learn how these properties are used to solve mathematical problems and prove geometrical statements.
So, do not approach this chapter as something completely new. Think of it as a continuation of your Class 9 learning. The concepts you studied earlier about chords and angles at the centre will help you connect with the new ideas introduced in Class 10. Whenever you come across a theorem or a diagram, try to recall what you already know about circles. This connection between Class 9 and Class 10 concepts can make the chapter easier to understand and remember.
While studying the chapter, pay special attention to the figures, given information, construction of the argument, theorem being applied and final conclusion. In geometry, understanding the diagram and the relationship between different parts of the figure is often just as important as knowing the theorem.
Use these Class 10 Maths Chapter 10 Circles NCERT Solutions CBSE Board as a learning and revision tool—not as a shortcut. First attempt every question yourself, then check the solution, identify your mistakes and revise the related concept. With repeated NCERT practice and proper revision, you can build a strong foundation for solving CBSE Board examination questions, previous-year questions and sample papers with confidence.
Watch Class 10 Maths Chapter 10 Circles NCERT Solutions
Now it is time to put your understanding into practice. Before watching the video below, take your Class 10 Maths NCERT textbook and notebook and try to solve the questions from Chapter 10 – Circles yourself. If you are watching the video on a particular exercise, pause the video when the question appears and solve it independently before continuing.
After making your attempt, watch the complete solution carefully. Compare your method with the explained solution and pay attention to the theorem used, logical steps, calculations, diagrams and final answer. If you make a mistake, don’t worry—identify it, understand why it happened, and solve the question again.
The video below provides a step-by-step explanation of the Class 10 Maths Chapter 10 Circles NCERT Solutions, helping you revise the chapter, strengthen your concepts and prepare more confidently for the CBSE Board Examination.
Remember: Don’t just watch the solution—pause, solve, compare and learn.
Exercise 10.1
Q1. How many tangents can a circle have?
Ans.A circle has infinite points and through each point a tangents can be drawn so a circle can have infinite tangents.
tangents at various pointsQ2.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) A tangent to a circle intersects it in one point.
(ii)A line intersecting a circle in two points is called a secant.
(iii)A circle can have two parallel tangents at the most.
(iv)The common point of a tangent to a circle and the circle is called the contact point.
Q3.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) cmAns.
As we know OP⊥ PQ, because P is the contact point of the circle , PQ is the tangent on the circle and OP is the radius of circle.
ΔOPQ is the right triangle
So, OQ² = OP² + PQ²
PQ² = OQ² – OP²
PQ² = 12² – 5²
PQ² = 144 – 25 =119
PQ =√119
Hence the length of PQ is√119 cm
Class 10 Maths NCERT Solutions of Chapter 10 Circle
Exercise 10.2
In question number 1 to 3, choose the correct option and give justification.
Q1.From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the center is 25 cm. The radius of the circle is.
(A) 7 cm (B)12 cm (C)15 cm (D)24.5Ans.
Let O is the centre of circle ,we are given that OQ = 25 cm, PQ = 24 cm
O P Q 25 cm24 cmAccording to theorem OP ⊥ PQ, where OP is the radius and PQ is the tangent
Therefore ΔOPQ is the right triangle
Applying Pythagoras theorem
OQ² = PQ² + OP²
OP² = OQ² − PQ²
OP² = 25² − 24²
OP² = 625 − 576
OP² = 49
OP = 7
Therefore the length of the radius OP is 7 cm, so (A) is the right option.
Q2.In the given fig. 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°Ans.
O P Q T 110°As we are given TP and TQ are tangents of the circle ,therefore P and Q will be contact points of tangents.
O is the centre of circle , therefore OP and OQ will be the radii of the given circle
∵OP⊥ TP and OQ ⊥ TQ ( The radius is perpendicular to the tangent)
∴∠OQT = 90° and ∠ OPT = 90°
∠POQ + ∠PTQ + ∠OQT + ∠ OPT = 360° (Angle sum property of quadrilateral)
110° + ∠PTQ + 90° + 90° = 360°
∠PTQ = 360° − 290°
∠PTQ = 70°
Therefore right option is (B).
Q3. If tangents PA and PB from a point P to a circle with centre O are inclined to each other at an angle of 80° then ∠POA is equal to
(A) 50°(B) 60°(C) 70°(D) 80°Ans.
O A B P 80°∠PAO = ∠PBO = 90° ( The radius is perpendicular to the tangent)
∠PAO + ∠PBO + ∠AOB + ∠APB = 360° (Angle sum property of quadrilateral)
90° + 90° + 80° +∠AOB = 360°
∠AOB = 360° − 260°
∠AOB = 100°
Considering ΔPOA and ΔPOB
OP = OP (Common)
OA = OB (Radii of same circle)
∠PAO = ∠PBO = 90° ( The radius is perpendicular to the tangent)
ΔPOA ≅ ΔPOB (RHS rule )
∠POA = ∠POB (By CPCT)
∠POA +∠POB = 100°
2∠POA = 100°
∠POA = 50°
Q4. Prove that the tangents drawn at the ends of diameter of a circle are parallel to each other.
Ans. Let PQ and RS be the tangents drawn on both ends of the diameter MN.
OPQRSMNMO and NO are the radii of the circle.
PM ⊥ MO and RN ⊥ NO (Radius is perpendicular to the tangent)
∴ ∠PMO = 90° and ∠RNO = 90°
∠PMO + ∠RNO = 90° + 90° = 180°
Since the angles, ∠PMO and ∠RNO are co-interior angles with MN as the transversal:
Here the sum of co-interior angles is 180°, therefore RS ∥ PQ.
Q5. Prove that perpendicular at the point of contact to the tangent to a circle passes through the centre.
Ans. Let PQ be the tangent to a circle with centre O, where T is the point of contact of the tangent.
ORTPQLet TR ⊥ PQ, and suppose it does not pass through the centre O.
∠OTQ = 90° (The angle between tangent and radius)
∠RTQ = 90° (We have supposed)
∴ ∠OTQ = ∠RTQ
This is only possible if line segments TR and TO coincide with each other. Therefore, TR must pass through the centre of the circle.
Q6. The length of tangent AB from a point A at a distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.
Ans. Let O be the centre of the circle, B be the point of contact, and AB be the tangent.
Given: OA = 5 cm, AB = 4 cm
O B A r4 cm5 cm∠ABO = 90° (Radius OB ⊥ Tangent AB)
∴ ΔABO is a right-angled triangle.
By Pythagoras Theorem in ΔABO:
OA2 = OB2 + AB2
(5)2 = OB2 + (4)2
25 = OB2 + 16
OB2 = 25 – 16
OB2 = 9
OB = √9 = 3 cm
Therefore, the radius of the circle is 3 cm.
Q7. 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.
Ans. Let O be the common centre of the concentric circles, and let AB be the chord of the larger circle which touches the smaller circle at point C.
O A C B 3 cm5 cmOC ⊥ AB (The radius is perpendicular to the tangent at the point of contact)
∴ ΔOCB is a right-angled triangle.
By Pythagoras Theorem in ΔOCB:
OB2 = OC2 + BC2
BC2 = OB2 – OC2
BC2 = 52 – 32 = 25 – 9 = 16
BC = √16 = 4 cm
AB = 2 × BC (Perpendicular from the centre to a chord bisects the chord)
AB = 2 × 4 = 8 cm
Therefore, the length of the chord of the larger circle which touches the smaller circle is 8 cm.
Q8. A quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB + CD = AD + BC
Ans. Let the circle touch the sides of quadrilateral ABCD at points P, Q, R, and S respectively.
O A B C D PQRSWe know that the lengths of tangents drawn from an external point to a circle are equal.
From point A:
AP = AS …(i)
From point B:
BP = BQ …(ii)
From point C:
CR = CQ …(iii)
From point D:
DR = DS …(iv)
Adding equations (i), (ii), (iii), and (iv):
(AP + BP) + (CR + DR) = (AS + BQ) + (CQ + DS)
(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
∴ AB + CD = AD + BC
Q9. In fig. XY and X’Y’ are two parallel tangents to a circle with centre O and another tangent with point of contact C intersecting XY at A and X’Y’ at B. Prove that ∠AOB = 90°.
Ans. In the given fig. joining the points A to O, B to O and O to C.
Let ∠PAC = 2x
∠OAC = x (The line joining an external point and centre of the circle bisects the angle between the tangents drawn from the external point)
∠PAC + ∠QBC = 180° (Sum of co-interior angles)
∠QBC = 180 − 2x
∠OBC = 90 − x
Considering ΔAOB
∠AOB + ∠OBC + ∠OAC = 180° (Angle sum property of triangle)
∠AOB + 90 − x + x = 180
∠AOB = 90
Hence ∠AOB = 90°
Q10. Prove that 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 point of contact at the centre.
Ans.
GIVEN. Two tangents AB and BC inclined with angle ∠ABC and their contact points on the circle are A and C, AC subtends ∠AOC to the centre.
TO PROVE. ∠AOC + ∠ABC = 180°
CONSTRUCTION. Join A to C
PROOF. ∠OAB = 90° and ∠OCB = 90° ( The radius is perpendicular to tangent)
In quadrilateral OABC
∠OAB + ∠OCB + ∠AOC + ∠ABC = 360° ( Angle sum property of quadrilateral)
90 + 90 + ∠AOC + ∠ABC = 360°
∠AOC + ∠ABC = 360° − 180 = 180
Therefore the angle between two tangents is supplementary to the angle subtended by the line segment joining the point of contact at the centre.
Q11. Prove that the parallelogram circumscribing a circle is a rhombus.
Answer.
Given → Parallelogram ABCD circumscribing a circle with center O which is touching it at P, Q, R and S points.
To Prove → The given parallelogram ABCD is a rhombus
Proof → AB = DC ………(i) (Opposite side of the parallelogram)
AD = BC ………(ii) (Opposite side of the parallelogram)
AP = AS ………(iii) [Tangents drown on a circle from a same external point are equal]
BP = BQ ………(iv) [Tangents drown on a circle from a same external point are equal]
DR = DS ………(v) [Tangents drown on a circle from a same external point are equal]
CR = CQ ………(vi) [Tangents drown on a circle from a same external point are equal]
HINT: Write the tangents corresponding to opposite sides same side as written above.
Adding (iii) to (vi) all equation we get
AP + BP + DR + CR = AS + BQ + DS + CQ
Arranging these line segments in order to get the side of parellelogram
(AP + BP) + (DR + CR) = (AS + DS )+ (BQ + CQ)
AB + DC = AD + BC
From equation number (i) and (ii)
DC + DC = AD + AD
2DC = 2AD
DC = AD
It is clear that AB = BC = DC= AD
All sides of the given parallelogram are equal therefore it is a rhombus.
Q12. A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segment BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see fig.). Find the sides AB and AC.
Ans.
Let AF = AE = x (Tangents drawn from an external point to the circle)
Drawing OF ⊥ AC, OE ⊥ AB and OD ⊥ BC
Area of ΔABC = Sum of area of all Δ’s inside it
arΔABC = 4x + 24 + 32 = 4x + 56…………(i)
AC = (6 + x) cm, BC = 8 + 6 = 14 cm, AB = (8 + x) cm
s = 6 + x + 8 + x + 142 = 28 + 2x2 = 14 + x
∵ arΔ = √s(s − a)(s − b)(s − c)
= √(x + 14)(x + 14 − x − 6)(x + 14 − x − 8)(x + 14 − 14)
= √48x (x + 14)
From (i)
= √48x (x + 14) = 56 + 4x
Squaring both sides
48x(x + 14) = (4x + 56)2
48 x2 + 672 x = 16 x2 + 3136 + 448 x
48 x2 – 16x2 + 672x – 448 x – 3136 = 0
32 x2 + 224 x – 3136 = 0
x2 + 7 x – 98 = 0
x2 + 14 x – 7x – 98 = 0
x (x + 14) – 7( x + 14) = 0
(x + 14)(x – 7) = 0
x = 7, x = -14, neglecting x = -14, since length of a line segment can’t be negative
Hence AC = AF + CF = 7 + 6 = 13 cm and AB = AE + BE = 7 + 8 = 15 cm
Q13. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplimentry angles at the centre of the circle.
Ans.
GIVEN. ABCD is a quadrilateral
TO PROVE. ∠AOB + ∠DOC = 180°
∠AOD + ∠BOC = 180°
PROOF. Considering the Δ’s AOC and DOC
The line joining to the centre from the point tangents drawn bisect the angle between the tangents.
Therefore we have
∠AOB = 180° – (∠A + ∠B)/2…………(i)
∠DOC = 180° – (∠D + ∠C)/2…………(ii)
Adding both the equation
∠AOB + ∠DOC = 360° -(∠A + ∠B + ∠D + ∠C)/2
∠AOB + ∠DOC = 360° – 360°/2 [since ∠A + ∠B + ∠D + ∠C=360°
∠AOB + ∠DOC = 180°
Similarly
∠AOD + ∠BOC = 180°
Hence opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Quick Revision of Class 10 Maths Chapter 10 Circles
After completing the NCERT Solutions of Class 10 Maths Chapter 10 Circles, students should revise all the important concepts, definitions and theorems of the chapter.
Important points to remember:
- A tangent to a circle touches the circle at only one point.
- The tangent at any point of a circle is perpendicular to the radius through the point of contact.
- The lengths of two tangents drawn from an external point to a circle are equal.
- Students should practise questions based on tangents and their properties.
- While solving geometry questions, always draw the figure carefully and write the steps of the proof in a proper order.
How to Prepare Chapter 10 Circles
First, understand all the concepts and theorems given in the NCERT textbook. Then try to solve all the NCERT questions yourself. Use the solutions given above to check your answers and understand the steps wherever you have difficulty.
After completing the NCERT exercises, practise previous-year questions, sample-paper questions and other important questions from Circles. This will help you become more confident before the CBSE Board Examination.
Watch the Video Solutions
Students can also watch the complete video solutions of Class 10 Maths Chapter 10 Circles for a step-by-step explanation of the questions. Video explanations can make it easier to understand the diagrams, theorems and methods used in solving the questions.
Conclusion
The Class 10 Maths Chapter 10 Circles NCERT Solutions given above provide step-by-step solutions to help students understand the chapter and check their answers. Students should not simply memorise the solutions. They should first try each question themselves and then use the solutions for checking and learning.
For better CBSE Board Exam preparation, revise the important theorems regularly and practise NCERT, previous-year and sample-paper questions. A proper revision cycle can help you remember the concepts and improve your performance in the examination.
Class 10 Maths Chapter-wise NCERT Solutions
Chapter 3 – Pair of Linear Equations in Two Variables
Chapter 4 – Quadratic Equations
Chapter 5 – Arithmetic Progressions
Chapter 7 – Coordinate Geometry
