In-Depth Class 10 Circles Notes for Exam Preparation
Class 10 Circles Notes provide a comprehensive understanding of the properties, theorems, and practical applications of circles. These notes cover tangents, secants, chords, angle subtended by arcs, and cyclic quadrilaterals with detailed explanations and solved examples. Understanding circles is vital for geometry, trigonometry, and problem-solving in competitive exams. The notes include step-by-step methods to solve problems, shortcuts, and tips for remembering key properties. With structured practice exercises, students can improve their calculation speed, accuracy, and conceptual clarity. Circles appear frequently in board exams, so these notes help students revise effectively and tackle questions confidently. By practicing these notes, students gain a strong foundation in geometric reasoning, visualization, and analytical skills. Whether it is solving tangent problems or applying circle theorems, these notes are an essential resource for Class 10 students aiming for academic success. After reading the note, solving NCERT questions with the help of the NCERT Solutions for class 10 Maths will help in building the foundation.
Class 10 Circles Notes, Solved example & Questions
History of Circles
The word "circle" derives from the Greek, kirkos "a circle," from the base ker- which means to turn or bend. The origins of the words "circus" and "circuit" are closely related. The circle has been known since before the beginning of recorded history. Natural circles would have been observed, such as the Moon, Sun, and a short plant stalk blowing in the wind on sand, which forms a circle shape in the sand. The circle is the basis for the wheel, which, with related inventions such as gears, makes much of modern civilization possible.
In mathematics, the study of the circle has helped inspire the development of geometry, astronomy, and calculus. Early science, particularly geometry and astrology and astronomy, was connected to the divine for most medieval scholars, and many believed that there was something intrinsically "divine" or "perfect" that could be found in circles.
A circle is the locus of a point which moves in a plane in such a way that its distance from a fixed point remains constant.
Secant
A line, which intersects a circle in two distinct points, is called a secant.
Tangent
A line meeting a circle only in one point is called a tangent to the circle at that point.
The point at which the tangent line meets the circle is called the point of contact.
Length of Tangent
The length of the line segment of the tangent between a given point and the given point of contact with the circle is called the length of the tangent from the point to the circle.
- There is no tangent passing through a point lying inside the circle.
- There is one and only one tangent passing through a point lying on a circle.
- There are exactly two tangents through a point lying outside a circle.
Theorem
A tangent to a circle i perpendicular to the radius through the point of contact.
Given: A circle C (O, r) and a tangent AB at a point P.
To prove: OP AB
Construction: Take any points Q, other than P on the tangent AB. Join OQ. Suppose OQ meets the circle at R.
Proof: Among all line segments joining the point O to a point on AB, the shorted one is perpendicular to
AB. So, to prove that OP AB, it is sufficient to prove that OP is shorter than any other segment joining O to any point of AB.
Clearly OP = OR.
Now, OQ OR + RQ
OQ > OR
OQ > OP ( OP = OR)
Thus, OP is shorter than any other segment joining O to any point of AB.
Hence, OP AB.
Theorem
Lengths of two tangents drawn from an external point to a circle are equal.
[Diagram shows a circle with center O, and point A external to the circle. Two tangent lines are drawn from A to the circle, touching at points P and Q]
Given: AP and AQ are two tangents drawn from a point A to a circle C (O, r).
To prove: AP = AQ.
Construction: Join OP, OQ and OA.
Proof: In △AOQ and △APO
∠OQA = ∠OPA [Tangent at any point of a circle is perp. to radius through the point of contact]
AO = AO [Common]
OQ = OP [Radius]
So by R.H.S. criterion of congruency △AOQ ≅ △AOP.
∴ AQ = AP. [By CPCT] Hence Proved
Result:
(i) If two tangents are drawn to a circle from an external point, then they subtend equal angles at the centre. ∠OAQ = ∠OAP [By CPCT].
(ii) If two tangents are drawn to a circle from an external point, they are equally inclined to the segment, joining the centre to that point ∠OAQ = ∠OAP [By CPCT].
1. If all the sides of a parallelogram touches a circle, show that the parallelogram is a rhombus.
Sol.
Given: Sides AB, BC, CD and DA of a ||gm ABCD touch a circle at P, Q, R and S respectively.
To prove: ||gm ABCD is a rhombus.
Here's the text conversion with all mathematical symbols and expressions:
Proof:
AP = AS ......(i)
BP = BQ ......(ii)
CR = CQ ......(iii)
DR = DS ......(iv)
[Tangents drawn from an external point to a circle are equal]
Adding (1), (2), (3) and (4), we get
⇒ AP + BP + CR + DR = AS + BQ + CQ + DS
⇒ (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
⇒ AB + CD = AD + BC
⇒ AB + AB = AD + AD [In a ‖gm ABCD, opposite side are equal]
⇒ 2AB = 2AD or AB = AD.
But AB = CD AND AD = BC [Opposite sides of a ‖ gem]
∴ AB = BC = CD = DA.
Hence, ‖gm ABCD is a rhombus.
2. From a point P, 10 cm away from the centre of a circle, a tangent PT of length 8 cm is drawn. Find the radius of the circle.
Sol. Let O be the centre of the given circle and let P be a point such that OP = 10 cm.
Let PT be the tangent such that PT = 8 cm.
Join OT.
Now PT is a tangent at T and OT is the radius through T.
[THIS IS FIGURE: A circle with center O, point P outside the circle, tangent PT touching the circle at point T, with a right angle marked at T between the radius OT and tangent PT]
∴ OT ⊥ PT [Radius is ⊥ to tangent at the point of contact]
In the right ΔOTP, we have
OP² = OT² + PT² [by Pythagoras' theorem]
⇒ OT = √(OP² - PT²) = √((10)² - (8)²) cm = √36 cm = 6 cm
Hence, the radius of the circle is 6 cm.
3. In the given figure, O is the centre of a circle; PA and PB, pair of tangents drawn to the circle from point P outside the circle. If ∠AOB = 117°, find θ.
[FIGURE: A diagram showing a circle with center O. Point P is outside the circle. Two tangent lines PA and PB are drawn from P to the circle, touching at points A and B respectively. The angle at P between the two tangents is marked as θ. The angle ∠AOB at the center is marked as 117°. The radius OA and OB are shown perpendicular to the tangents at points A and B.]
Sol.
In quad. AOBP, ∠A = ∠B = 90°. [Radius is ⊥ to tangent at the point of contact]
∴ ∠AOB + ∠APB = 180° [∵ sum of angles of quad. is 360°]
⇒ 117° + θ = 180°
⇒ θ = 180° - 117° = 63°
∴ θ = 63°.
4. In the given figure, PA and PB are tangents to the circle drawn from an external point P. CD is another tangent touching the circle at Q. If PA = PB = 12 cm and QD = 3 cm, find the length of PD.
[FIGURE: A diagram showing a circle with center (not explicitly marked). Point P is external to the circle on the left side. Two tangent lines PA and PB are drawn from P to the circle, touching at points A (top) and B (bottom). A third tangent line CD passes through point C (above A) and point D (below B), touching the circle at point Q. The circle's center appears to be between points A and B.]
Sol.
Point D outside the circle. DQ and DB is the pair of tangents drawn to the circle from the point D.
⇒ DB = DQ
⇒ DB = 3 cm. (∵ QD = DQ = 3 cm is given)
Now, PD + DB = PB
⇒ PD + 3 cm = 12 cm (∵ PB = 12 cm is given)
⇒ PD = 9 cm.
5. PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q intersect at a point T. Find the length of TP.
[Diagram shows a circle with center O and radius 5 cm, chord PQ of length 8 cm, and tangent lines at P and Q intersecting at external point T]
Sol.
Let TP = x and TR = y.
ΔTPQ is isosceles Δ,
∴ TR is the ⊥ bisector of PQ,
∴ PR = RQ = 4 cm
Also, OR = √(OP² - PR²) = √(5² - 4²) = 3 cm.
In right Δ PRT, we have
x² = y² + 16. ... (i)
In right Δ OPT, we have
⇒ x² + 5² = (y + 3)² ... (ii)
Subtracting (i) from (ii), we have
25 = 6y - 7 or y = 32/6 = 16/3.
Therefore, x² = (16/3)² + 16 = 16/9(16 + 9) = (16 × 25)/9. [from equation (i)]
Or x = 20/3.
6. A circle touches the BC of a △ABC at P and touches AB and AC when produced at Q and R respectively as shown in the figure. Show that = 1/2 (Perimeter of △ABC).
Sol. Given: A circle is touching side BC of △ABC at P and touching AB and AC when produced at Q and R respectively.
[THIS IS FIGURE: A triangle ABC with a circle inscribed that touches BC at point P, and touches the extended lines AB and AC at points Q and R respectively]
To prove: AQ = 1/2 (perimeter of △ABC).
Proof: AQ = AR ......(i) BQ = BP ......(ii) CP = CR ........(iii)
[Tangents drawn from and external point to a circle are equal]
Now, perimeter of △ABC = AB + BC + CA = AB + BP + PC + CA = (AB + BQ) + (CR + CA) [From (ii) and (iii)] = AQ + AR = AQ + AQ [From (i)]
AQ = 1/2 (perimeter of △ABC).
7. Prove that the tangents at the extremities of any chord make equal angles with the chord.
Sol.
Let AB be a chord of a circle with centre O, and let AP and BP be the tangents at A and B respectively. Suppose, the tangents meet at point P. Join OP. Suppose OP meets AB at C.
[DIAGRAM: A circle with center O, chord AB, tangents at A and B meeting at point P, with OP intersecting AB at C]
We have to prove that ∠PAC = ∠PBC.
In triangles PCA and PCB
PA = PB [∵ Tangent from an external point are equal]
∠APC = ∠BPC. [∵ PA and PB are equally inclined to OP]
And PC = PC. [Common]
So, by SAS criteria of congruence
ΔPAC ≅ ΔBPC
⇒ ∠PAC = ∠PBC. [By CPCT]
8. Prove that the segment joining the points of contact of two parallel tangents passes through the centre.
Sol.
Let PAQ and RBS be two parallel tangents to a circle with centre O. Join OA and OB. Draw OC||PQ. Now, PA||CO
[THIS IS FIGURE: A circle with center O, two parallel tangents PAQ (top) and RBS (bottom), points A and B as points of contact, point C to the left of O, and a vertical dashed line through A, O, and B]
⇒ ∠PAO + ∠COA = 180° [Sum of co-interior angle is 180°]
⇒ 90° + ∠COA = 180° [∵ ∠PAO = 90]
⇒ ∠COA = 90°.
Similarly, ∠COB = 90°.
∴ ∠COA + ∠COB = 90° + 90° = 180°.
Hence AOB is a straight line passing through O.
9. In the given figure, PA and PB is a pair of tangents drawn to a circle having its centre at O. If ∠APB = 52°, find the ∠PAB and ∠PBA.
[FIGURE: A diagram showing a circle with center O. Point P is external to the circle. Two tangent lines are drawn from P to the circle, touching at points A and B. The angle at P (∠APB) is marked as 52°. Lines PA and PB extend beyond the points of tangency.]
Sol. In ΔPAB,
PA = PB [Tangents from common point are equal in length]
⇒ ∠PAB = ∠PBA [Angles opposite to equal sides]
= 1/2 × (180° - 52°)
= 1/2 × 128° (∵ ∠PAB + ∠PBA + 52° = 180°)
⇒ ∠PAB = ∠PBA = 64°.
10. A point P is 15 cm from the centre of a circle. The radius of the circle is 5 cm. Find the length of the tangent drawn to the circle from the point P.
Sol. Let PA = x cm be the length of the tangent to the circle. OA is perpendicular to PA at A.
OA = 5 cm (radius)
OP = 15 cm (given).
[There is a diagram showing a circle with center O, radius 5 cm to point A on the circle, point P outside the circle at distance 15 cm from O, and a tangent line PA of length x cm from P to A]
By Pythagoras theorem,
OP² = PA² + OA²
⇒ (15)² = x² + (5)²
⇒ x² = 225 - 25 = 200
⇒ x = √200 = 10√2.
∴ Length of the tangent = 10√2 cm.
11. In the given figure O is the centre of the circle and PT is the tangent drawn from the point P to the circle. Secant PAB passes through the centre O of the circle. If PT = 6 cm and PA = 3 cm, then find the radius of the circle.
[FIGURE: A circle with center O. Point P is outside the circle. PT is a tangent to the circle at point T, with PT = 6 cm marked. A secant line PAB passes through P, A, and B, where AB is a diameter passing through center O. PA = 3 cm is marked. Angle OTP = 90° is shown.]
Sol.
In figure AB is a diameter of the circle and OA = r (radius of the circle). Join O & T. ∠OTP = 90°. [Radius is ⊥ to tangent at the point of contact]
In right angled ΔOTP, OP = OA + PA = (r + 3) cm.
By Pythagoras theorem,
OP² = OT² + PT²
⇒ (r + 3)² = r² + (6)²
⇒ r² + 6r + 9 = r² + 36
⇒ 6r = 27
⇒ r = 9/2 = 4.5
Hence, the radius of the circle = 4.5 cm
12. The incircle of ΔABC touches the sides BC, CA and AB at D, E and F, respectively. If AB = AC, prove that BD = CD.
Sol. Since tangents drawn from an external point to a circle are equal in length.
∴ AF = AE [Tangents from A] ..... (i)
BF = BD [Tangents from B] ..... (ii)
CD = CE [Tangents from C] ..... (iii)
[THIS IS FIGURE: A triangle ABC with an incircle. The incircle touches side BC at point D, side CA at point E, and side AB at point F. Points B, D, and C are shown on the bottom, with A at the top vertex.]
Adding (i), (ii) and (iii), we get
AF + BF + CD = AE + BD + CE
⇒ AB + CD = AC + BD
But AB = AC. [Given]
∴ CD = BD.
13. A circle is touching the side BC of a △ABC at P and touching AB and AC produced at Q and R respectively. Prove that AQ = 1/2 (Perimeter of △ABC).
Solution:
2AQ = AQ + AR (∵ AQ = AR) [Tangent from a common point]
= (AB + BQ) + (AC + CR) = AB + AC + (BQ + CR)
= AB + AC + (BP + CP) [∵ BQ = BP and CR = CP]
= AB + AC + BC = (Perimeter of △ABC)
[THIS IS FIGURE: A triangle ABC with a circle inscribed. The circle touches side BC at point P, and touches the extensions of sides AB and AC at points Q and R respectively. Points B, P, and C are marked along the bottom, with Q to the left and R to the right of the circle.]
⇒ AQ = 1/2 (Perimeter of △ABC).
14. 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 segments joining the points of contact to the centre.
Sol.
Given: PA and PB are the tangents drawn from a point P to a circle with centre O. Also, the line segments OA and OB are drawn.
To prove: ∠APB + ∠AOB = 180°.
Proof: We know that the tangent to a circle is perpendicular to the radius through the point of contact.
[THIS IS FIGURE: A diagram showing a circle with center O, external point P, two tangent lines PA and PB touching the circle at points A and B respectively, with radii OA and OB drawn to the points of contact]
∴ PA ⊥ OA ⇒ ∠OAP = 90°, and PB ⊥ OB ⇒ ∠OBP = 90°.
∴ ∠OAP + ∠OBP = (90° + 90°) = 180°.
Hence, ∠APB + ∠AOB = 180°. [∵ Sum of all the ∠s of a quadrilateral is 360°]
15. Prove that in two concentric circles, the chord of the larger circle, which touches the smaller circle, is bisected at the point of contact.
Sol. Let O be the common centre of two concentric circles C₁ and C₂ and a chord AB of the larger circle C₁ touching the smaller circle C₂ at the point P.
Join OP.
[DIAGRAM: Shows two concentric circles with center O, where C₂ is the smaller inner circle and C₁ is the larger outer circle. A horizontal chord AB of the larger circle touches the smaller circle at point P. A dashed line connects O to P vertically.]
Since OP is the radius of the smaller circle and AB is a tangent to this circle at a point P.
∴ OP ⊥ AB
The perpendicular drawn from the centre of a circle to any chord of the circle bisects the chord.
∴ AP = BP
Hence AB is bisected at the point of contact.
Frequently Asked Questions
The circumference of a circle is given by the formula C = 2πr, where r is the radius of the circle, and C = πd, where d is the diameter. This formula is important because the circumference measures the perimeter or boundary length of a circle, which has countless real-world applications. For example, calculating the boundary length of wheels, circular tracks, circular gardens, or pipes relies on this formula.
Understanding circumference also provides a bridge to other topics such as trigonometry, coordinate geometry, and areas of circular figures. The constant π (pi ≈ 3.14159) represents the universal ratio between circumference and diameter, highlighting a deep mathematical relationship observed across all circles.
In exams, questions on circumference may appear as direct formula-based calculations, word problems, or combined with other shapes in geometry. To master this concept, practice applying the formula to different contexts, such as finding the distance a wheel covers in one rotation or determining material required to fence a circular field. This formula is foundational for both practical problem-solving and advanced mathematics, making it essential knowledge for students.
The use of π in mathematics is rooted in history and standardization. Pi (π) is defined as the ratio of a circle’s circumference to its diameter, a constant that has been studied for thousands of years. Since ancient civilizations used diameter as the standard measure of circles, π became universally adopted. Tau (τ), defined as the ratio of circumference to radius, is simply equal to 2π. Advocates argue tau makes formulas more intuitive since many formulas involve 2π, such as those in trigonometry and radians. For example, one complete revolution of a circle is 2π radians, which could be simplified as τ radians.
However, π remains dominant because it is deeply ingrained in mathematical literature, education, and applications across disciplines. Switching to τ would create confusion and inconsistency with centuries of mathematical knowledge. Understanding both constants can deepen comprehension of circle relationships, but for academic and practical purposes, π is essential. Students should focus on mastering π-based formulas, as they are the standard in CBSE, competitive exams, and professional mathematics.
Several circle formulas are essential for exams, as they are frequently applied in problems. These include:
- Circumference: C = 2πr or πd
- Area of a circle: A = πr²
- Arc length: (θ/360) × 2πr, where θ is the central angle
- Area of a sector: (θ/360) × πr²
- Equation of a circle (coordinate geometry): (x – h)² + (y – k)² = r², where (h, k) is the center
- Length of tangent from an external point: √(x² + y² – r²), given a circle centered at origin with radius r
These formulas are critical not just for direct exam questions but also for applied problems involving geometry, mensuration, and coordinate geometry. To retain them, students should create a formula sheet and revise it daily. Practice questions from previous year papers to understand how formulas are applied in different scenarios. Visual aids such as diagrams can make learning easier. Mastery of these formulas ensures strong performance in board exams, competitive exams, and real-world applications of geometry.
One of the most fascinating and unusual circle facts is the property of the circle being the shape with the maximum area for a given perimeter. Teachers often illustrate this by comparing a circle to polygons with the same perimeter, showing how the circle encloses the most space. Another striking concept is the power of a point theorem, which connects tangents, secants, and chords from an external point, revealing hidden relationships in circle geometry. Teachers sometimes demonstrate this using simple paper folds or dynamic geometry software, which makes the theorem less abstract.
Another unusual insight is how circles relate to trigonometric functions: the unit circle provides a visual foundation for sine, cosine, and tangent. Beyond pure mathematics, teachers may link circles to planetary orbits, architecture, and engineering designs, making the concept both surprising and practical. These unique insights inspire curiosity and help students view mathematics not as rote memorization but as a subject full of interconnected wonders. Presenting such unusual facts is highly engaging for learners and helps cement long-term understanding.
Circles are fundamental because they combine geometry, algebra, and trigonometry in a single figure. The circle’s properties, such as symmetry, tangents, and chords, lead to many essential theorems used in both school-level and advanced mathematics. Circles also form the foundation for coordinate geometry equations, where their algebraic representation provides practice in analytical problem-solving. In calculus, circles lead to trigonometric functions, derivatives, and integrals involving circular motion. Practically, circles appear in mechanics, engineering, navigation, and even biology think of gears, wheels, ripples in water, and planetary orbits. Circles also introduce students to π (pi), one of the most famous mathematical constants, linking infinite series, geometry, and probability.
The circle’s central role across diverse fields makes it indispensable in mathematics. By understanding circle-related concepts, students gain skills that prepare them for higher-level math, competitive exams, and problem-solving in real life. Teachers emphasize circles not just for exam preparation but also to develop a comprehensive understanding of mathematical relationships.
Some essential circle theorems that every student should master include:
- Angle subtended by the same arc at the circumference is equal.
- Angle at the center is twice the angle at the circumference from the same arc.
- Tangent to a circle is perpendicular to the radius at the point of contact.
- The lengths of tangents drawn from an external point are equal.
- The opposite angles of a cyclic quadrilateral are supplementary.
- Equal chords subtend equal angles at the center.
These theorems are not only important for CBSE and board exams but also form the basis for higher-level competitive exams. Learning them with diagrams, proofs, and real-world analogies—like ladders leaning on walls or bicycle wheels can make them more intuitive. Many problems in geometry integrate these theorems with algebra or trigonometry, so consistent practice is vital. A strong grasp of these theorems also improves logical reasoning and proof-writing skills, which are crucial in advanced mathematics. Students should revise them regularly, practice proof-based questions, and attempt application-based problems to gain mastery.
The best way to prepare for circle-related mathematics topics is to start with NCERT textbooks, as they provide the foundational concepts tested in most exams. Once the basics are clear, move on to sample papers, question banks, and practice worksheets that include a mix of direct formula applications and theorem-based questions. Online educational platforms and school board websites often publish official sample papers and solved solutions, which are valuable for exam practice. Students should also use curated resource hubs that categorize questions chapter-wise for quick access.
For competitive exams like JEE advanced question banks focusing on problem-solving and analytical skills are recommended. Search-optimized FAQs, study guides, and exam-focused blogs also serve as quick navigation points. Organizing your resources by topic formulas, theorems, application problems will make revision efficient. Bookmarking or downloading PDFs ensures offline access. Always prefer resources aligned with your syllabus to avoid confusion. By using structured navigation, students save time and focus their preparation more effectively.
Efficient navigation starts with understanding exactly what you need: formulas, theorems, or practice questions. Use specific search queries like “CBSE Class 10 circle important questions with solutions” or “circle theorems for JEE with practice problems.” Many educational sites categorize their content using headings and tags, which helps filter out unrelated material.
Use resources that provide chapter-wise filters or exam-year filters, allowing you to target circle-related content directly. Avoid vague searches, as they often lead to irrelevant results. Bookmark trusted websites that consistently provide accurate, exam-oriented resources. Creating a personal digital index (like folders on your device for formula sheets, solved examples, and past papers) ensures that once you find reliable content, you won’t need to search repeatedly. Smart navigation is about minimizing distractions, so rely on structured resources that emphasize syllabus-specific preparation. This approach makes exam revision faster and more effective.
The most useful online content is structured, visual, and exam-oriented. Step-by-step solutions with diagrams make circle problems easier to understand than plain text. Infographics and animations that show how formulas like circumference or arc length are applied help students visualize concepts. Short explanatory videos are excellent for breaking down theorems or problem-solving techniques. For exam preparation, downloadable PDFs containing formula sheets, question banks, and solved papers provide offline access and reduce reliance on internet searches.
FAQ-style articles are helpful for quick answers, especially if optimized with common queries like “what is the circumference formula” or “important circle theorems for Class 10.” Students should also look for resources that combine text with practice exercises, ensuring both explanation and application. High-quality online content saves time, clarifies concepts, and builds exam confidence. Prioritize reliable and syllabus-aligned resources for the best results.
Crop circles are large patterns created by flattening crops in fields. While they are often associated with folklore or unexplained phenomena, they also involve geometric patterns, many of which are circular. From a mathematical perspective, crop circles can demonstrate concepts of symmetry, concentric circles, and radial geometry. However, they are not typically relevant to school-level mathematics examinations or formal study of circles. They are best viewed as examples of how geometry can appear in art, culture, and unexplained natural or human-made phenomena. For students focusing on mathematics preparation, crop circles may serve as interesting visuals but are not academically essential. They should be considered a curiosity rather than a study priority.
Students should learn to filter irrelevant material by focusing on syllabus-specific keywords. For example, instead of general searches like “circle facts,” use queries like “circle formulas Class 10 CBSE” or “circle theorems with proofs.” If unclear or distracting content appears, it should be ignored to save time. Creating a structured study plan with reliable textbooks, official sample papers, and curated online resources ensures you stay on track.
Maintaining a personal library of vetted notes, solved examples, and formula sheets reduces the need to rely on random online searches. Developing digital literacy knowing how to identify high-quality, syllabus-aligned resources is key to avoiding confusion. Ultimately, focus on relevance, clarity, and exam preparation to handle noise effectively.