Circles defines a circle as the locus of points equidistant from a centre and studies its perfect symmetry. It shows that infinitely many circles pass through two points but exactly one passes through three non-collinear points, the circumcircle. It then proves twelve theorems on chords, their distances from the centre, angles subtended by arcs and cyclic quadrilaterals. Each theorem is explained with congruent triangles or the Baudhayana-Pythagoras theorem, so the chapter trains proof writing as well as chord and angle numericals for school exams.
Key Concepts
1. What a Circle Is: Centre, Radius, Chord and Diameter
Every circle, big or small, has a centre, and every point on the circle is at the same distance from that centre. NCERT turns this observation into the definition. All figures in this chapter are drawn on a flat sheet, a two-dimensional plane.
Circle: the set of all points on the plane that are equidistant from a given point on that plane.
A set of points that satisfies a given condition is called a locus. So a circle is also the locus of points equidistant from a given point. The given point is the centre, and the distance from the centre to any point on the circle is the radius.
- Chord: the line segment joining two points B and C on the circle.
- Angle subtended by a chord at the centre: if A is the centre, chord BC subtends ∠BAC at the centre.
- Diameter: a chord that passes through the centre. Its length is 2r.
- Arc: a connected portion of the circle, fixed by two end points and the curve joining them along the circle.
- Minor and major arc: two points A and B split the circle into two arcs. The smaller one is the minor arc, the bigger one is the major arc.
- Semicircle: each of the two equal arcs cut off by a diameter.
The CBSE course structure also lists two region words, segment and sector. A segment is the region between a chord and one of its arcs (minor segment with the minor arc, major segment with the major arc). A sector is the region between two radii and the arc joining their ends.
Illustration: in a circle of radius 5 units the longest chord is a diameter, 10 units. There is no smallest chord, since a chord can be made as short as you like.
2. Symmetries of a Circle
A circle is perfectly symmetric in two ways.
- Rotational symmetry: rotate a circle about its centre by any angle and it looks exactly the same. That is why you cannot tell which point of a rotating wheel is touching the ground. NCERT calls this complete rotational symmetry.
- Reflection symmetry: fold a paper circle so that the boundaries overlap. The crease passes through the centre, so it is a diameter. Every diameter is a line of reflection symmetry, so a circle has infinitely many lines of symmetry.
By comparison, a square has only 4 lines of symmetry, a regular pentagon 5 and a regular hexagon 6.
Rotating a circle carries a chord to another chord of the same length, with the same central angle. The theorems below prove with congruent triangles what this rotation suggests.
3. Circles Through Two Points and Through Three Points
Key fact: the perpendicular bisector of a segment AB is the locus of points equidistant from A and B. Every point on it is equidistant from A and B, and every point equidistant from A and B lies on it.
Two points. The centre O of any circle through A and B satisfies OA = OB, so O lies on the perpendicular bisector of AB. Every point of that bisector works as a centre. So infinitely many circles pass through two given points, and their centres all lie on the perpendicular bisector of AB.
- The smallest such circle has its centre at the midpoint of AB; its radius is AB/2 and AB is a diameter.
- Moving the centre away from AB increases the radius; there is no largest circle.
Theorem 1: There is a unique circle passing through three non-collinear points.
Why: the centre O of such a circle has OA = OB, so it lies on the perpendicular bisector of AB. It also has OA = OC, so it lies on the perpendicular bisector of AC. Since A, B and C are not collinear, these two bisectors are not parallel and meet at exactly one point. That point is O, and the circle with centre O and radius OA passes through A, B and C.
If A, B, C are collinear, the perpendicular bisectors of AB and BC are both perpendicular to the same line, so they are parallel and never meet. No point is equidistant from all three, and no circle passes through three collinear points. For the same reason, a line cannot cut a circle in three distinct points.
Circumcircle and circumcentre. Three non-collinear points form a triangle. The circle through its vertices is the circumcircle; it circumscribes the triangle and the triangle is inscribed in it. Its centre, where the perpendicular bisectors of the sides meet, is the circumcentre.
| Type of triangle | Position of the circumcentre |
|---|---|
| Acute-angled | Inside the triangle |
| Right-angled | At the midpoint of the hypotenuse |
| Obtuse-angled | Outside the triangle |
Illustration: a right triangle has legs 6 cm and 8 cm. Its hypotenuse is √(36 + 64) = 10 cm, so the circumcentre is the midpoint of the hypotenuse and the circumradius is 5 cm.
Construction: draw the perpendicular bisectors of two sides, mark their meeting point O and draw the circle with centre O through A.
4. Equal Chords and the Angles They Subtend at the Centre
Tie a thread tightly between two points of a wheel and rotate the wheel. The thread becomes a new chord of the same length, and the angle it makes at the centre does not change. Theorems 2 and 3 prove this.
Theorem 2: Equal chords of a circle subtend equal angles at the centre of the circle.
Given: chords AB and DE of a circle with centre C, AB = DE. To show: ∠ACB = ∠DCE.
Why: CA = CB = CD = CE = r. So in ΔCAB and ΔCDE: CA = CD, CB = CE and AB = DE. By SSS, ΔCAB ≅ ΔCDE, so ∠ACB = ∠DCE.
Theorem 3 (converse): Chords of a circle that subtend equal angles at the centre are equal.
Given: ∠ACB = ∠DCE. To show: AB = ED.
Why: AC = DC and BC = EC (radii), and the included angles are equal. By SAS, ΔACB ≅ ΔDCE, so AB = ED.
A useful fact: the triangle formed by a chord and the centre is always isosceles, because its two sides from the centre are radii. Its base angles are equal. If the chord subtends 60° at the centre, all three angles are 60° and the chord equals the radius.
Illustration: in a circle of radius 12 cm, chord AB subtends 60° at the centre O. ΔOAB has OA = OB, so ∠OAB = ∠OBA = (180° − 60°)/2 = 60°. The triangle is equilateral and AB = 12 cm.
5. The Perpendicular from the Centre to a Chord
Theorem 4: The line joining the centre of a circle and the midpoint of a chord of the circle is perpendicular to the chord.
Given: chord AB of a circle with centre C, M the midpoint of AB. To show: CM ⊥ AB.
Why: CA = CB, so ΔCAB is isosceles and ∠A = ∠B. Also AM = BM. By SAS, ΔCMA ≅ ΔCMB, so ∠CMA = ∠CMB. These two angles form a linear pair, so ∠CMA + ∠CMB = 180°, and each is 90°.
Theorem 5 (converse): The perpendicular from the centre of a circle to a chord of the circle bisects the chord.
Why: now ∠CMA = ∠CMB = 90°. In right triangles CMA and CMB, the hypotenuses CA and CB are radii and CM is common. By RHS, ΔCMA ≅ ΔCMB, so AM = BM.
Consequence: the perpendicular bisector of any chord passes through the centre, since the centre is equidistant from the chord’s ends.
The key right triangle. With radius r, distance d from the centre to the chord, and half-chord AM, the Baudhāyana-Pythagoras theorem in ΔCMA gives
r2 = d2 + (AM)2, so chord AB = 2 × AM = 2√(r2 − d2).
6. Distance of Chords from the Centre
The distance of a chord from the centre always means the perpendicular distance. By Theorem 5, the foot of that perpendicular is the midpoint of the chord.
Theorem 6: Chords of a circle having the same length are all at the same distance from the centre of the circle.
Given: AB = FG in a circle with centre C; E and H are the midpoints of AB and FG. To show: CE = CH. NCERT gives two explanations.
- Method 1: CA = CF, CB = CG (radii) and AB = FG, so ΔCAB ≅ ΔCFG by SSS. Corresponding altitudes of congruent triangles are equal, so CE = CH.
- Method 2: AE = FH (halves of equal chords), ∠CEA = ∠CHF = 90° and CA = CF (radii). By RHS, ΔCEA ≅ ΔCHF, so CE = CH.
Theorem 7 (converse): Chords of a circle that are equidistant from the centre have equal length. If CE = CH, then AE2 = r2 − CE2 = r2 − CH2 = FH2, so AE = FH and the full chords are equal.
Theorem 8: Let AB and DE be two chords of a circle with centre C. Suppose AB > DE. Then the distance from C to AB is less than the distance from C to DE.
Why: drop perpendiculars CF to AB and CG to DE. AC = CD (radii), so CF2 + AF2 = AC2 = CD2 = CG2 + GD2. F and G are midpoints, so AB > DE gives AF > GD, so AF2 > GD2. Then CF2 < CG2, so CF < CG.
In short, the longer the chord, the closer it is to the centre.
- The chord nearest the centre is the one through the centre (distance 0). That is the diameter, the longest chord.
- As a chord is pushed away from the centre it shrinks. When the distance reaches r, the chord becomes a single point of length 0.
| Distance d from centre (r = 5 cm) | Half-chord √(25 − d2) | Chord length |
|---|---|---|
| 0 cm | 5 cm | 10 cm (diameter) |
| 3 cm | 4 cm | 8 cm |
| 4 cm | 3 cm | 6 cm |
| 5 cm | 0 cm | 0 (a single point) |
7. Angles Subtended by an Arc
Angle at the centre. The angle subtended by arc AB at the centre O is the angle swept as we move from OA to OB along that arc. A minor arc subtends an angle less than 180°. A major arc subtends a reflex angle, more than 180°. A semicircle subtends exactly 180°. The central angles of the minor arc and the major arc on the same two end points add up to 360°.
Angle at a point on the circle. The angle subtended by arc AXB at a point on the circle outside the arc is ∠ACB, where C is any point of the circle not on arc AXB. Its measure does not depend on which such point C you choose.
Theorem 9: The angle subtended by an arc at the centre of the circle is double the angle subtended by the arc at any point on the circle outside the arc.
Proof (case where DC extended meets the arc): let C be the centre, AFB the arc and D a point on the circle outside it. Join DC and extend it to meet arc AFB at E.
- ΔDCB is isosceles (CB = CD), so ∠CBD = ∠CDB. ∠BCE is an exterior angle of ΔBCD, so ∠BCE = ∠CBD + ∠CDB = 2∠BDC.
- ΔADC is isosceles (CA = CD), so ∠CAD = ∠CDA. ∠ACE is an exterior angle of ΔADC, so ∠ACE = 2∠CDA.
- Adding: ∠BCA = ∠BCE + ∠ECA = 2(∠BDC + ∠CDA) = 2∠BDA.
Other case: if DC extended meets the circle at a point E outside arc AFB, the same two exterior-angle steps give ∠ACE = 2∠ADC and ∠BCE = 2∠BDC, and this time you subtract: ∠ACB = ∠ACE − ∠BCE = 2(∠ADC − ∠BDC) = 2∠ADB.
Angles in the same segment are equal. Every point D on the circle outside arc AB gives ∠ADB = ½ (angle of the arc at the centre). So all such angles are equal. This property belongs to the circle alone: for points inside or outside the circle, the angle subtended by AB is not the same for all such points.
Corollary: The angle subtended by a diameter at any point on the circle is 90°. If AB is a diameter, the arc not containing D subtends a straight angle, 180°, at the centre. So ∠ADB = ½ × 180° = 90°. This is the angle in a semicircle. It also explains why the circumcentre of a right triangle is the midpoint of the hypotenuse.
Illustrations:
- An arc subtends 70° at the centre. At any point on the remaining part of the circle it subtends 35°.
- A major arc subtends a reflex angle of 250° at the centre. At a point on the minor arc it subtends 125°.
8. Concyclic Points and Cyclic Quadrilaterals
Points that lie on the same circle are called concyclic. Any three non-collinear points are concyclic (Theorem 1). Four points need a test.
Theorem 10: If a line segment AB joining two points A, B subtends equal angles at two other points C, D that lie on the same side of AB, then the four points lie on a circle.
Why (by contradiction): draw the circle through A, B, C. If D is outside it, AD meets the circle at E, and ∠ADB < ∠AEB = ∠ACB (exterior angle of ΔBED, then same segment), which contradicts ∠ACB = ∠ADB. If D is inside, extending AD gives the reverse inequality and the same contradiction. So D lies on the circle.
A quadrilateral whose four vertices are concyclic is a cyclic quadrilateral.
Theorem 11: The sum of two opposite angles of a cyclic quadrilateral is 180°.
Why: let ABCD be cyclic with centre O. A is outside arc BCD, so ∠BAD = ½ (reflex ∠BOD). C is outside arc BAD, so ∠BCD = ½ (∠BOD). The reflex angle and the ordinary angle BOD together make the complete angle at O. So ∠BAD + ∠BCD = ½ × 360° = 180°. The same argument gives ∠B + ∠D = 180°.
Theorem 12 (converse): If two opposite angles of a quadrilateral add up to 180°, then the vertices of the quadrilateral lie on a circle, i.e., they are concyclic.
Why (by contradiction): take the circle through A, B, D. If C were outside it, let E be where CD meets the circle. ABED is cyclic, so ∠BAD + ∠BED = 180°. With ∠BAD + ∠BCD = 180°, this gives ∠BED = ∠BCD. But ∠BED is an exterior angle of ΔBEC, so it is greater than ∠BCE, which is ∠BCD. This contradiction (and a similar one when C is inside) shows C lies on the circle.
Consequences you should know:
- Exterior angle of a cyclic quadrilateral: the exterior angle at any vertex equals the interior opposite angle. If E is on CD extended, ∠CDE = 180° − ∠ADC = ∠ABC.
- Rectangle: the only parallelogram that can be inscribed in a circle. A parallelogram has equal opposite angles; cyclic means they add to 180°, so each is 90°.
Formula and Theorem Sheet
| No. | Result (NCERT numbering) | Proved using |
|---|---|---|
| 1 | A unique circle passes through three non-collinear points | Perpendicular bisectors of two sides meet at one point |
| 2 | Equal chords subtend equal angles at the centre | SSS congruence |
| 3 | Chords subtending equal angles at the centre are equal | SAS congruence |
| 4 | Line from centre to midpoint of a chord is perpendicular to it | SAS, linear pair |
| 5 | Perpendicular from centre to a chord bisects it | RHS congruence |
| 6 | Equal chords are equidistant from the centre | SSS (altitudes) or RHS |
| 7 | Chords equidistant from the centre are equal | RHS or Pythagoras |
| 8 | Of two unequal chords, the longer is nearer the centre | Baudhāyana-Pythagoras theorem |
| 9 | Angle at centre = 2 × angle at a point on the circle outside the arc | Isosceles triangles, exterior angle theorem |
| Corollary | Angle subtended by a diameter at the circle = 90° | Theorem 9 with a 180° central angle |
| 10 | Equal angles on the same side of AB make A, B, C, D concyclic | Contradiction, exterior angle |
| 11 | Opposite angles of a cyclic quadrilateral add to 180° | Theorem 9, complete angle 360° |
| 12 | Opposite angles adding to 180° make a quadrilateral cyclic | Contradiction, Theorem 11 |
| Formula | Chord = 2√(r2 − d2); r2 = d2 + (chord/2)2 | Theorem 5 and Pythagoras |
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Important Definitions
| Term | Definition |
|---|---|
| Circle | The set of all points on a plane that are equidistant from a given point on that plane |
| Locus | The set of points that satisfy a given condition |
| Chord | A line segment joining two points on the circle |
| Diameter | A chord passing through the centre; length 2r; the longest chord |
| Arc | A connected portion of the circle defined by two end points and the curve joining them along the circle |
| Minor arc, major arc | Of the two arcs between A and B, the smaller is the minor arc and the bigger is the major arc |
| Segment | The region between a chord and one of the arcs it cuts off |
| Sector | The region between two radii and the arc joining their end points |
| Circumcircle, circumcentre | The circle through the three vertices of a triangle, and its centre, where the perpendicular bisectors of the sides meet |
| Concyclic points | Points that lie on the same circle |
| Cyclic quadrilateral | A quadrilateral (4-gon) whose four vertices lie on a circle |
Solved Examples (NCERT-Based)
Example 1: Chord from radius and distance (NCERT End-of-Chapter Exercise 1)
In a circle, a chord is 5 cm away from the centre. If the radius of the circle is 13 cm, what is the length of the chord?
Solution: let O be the centre, AB the chord and M the foot of the perpendicular from O. By Theorem 5, M is the midpoint of AB. In right ΔOMA, OA = 13 cm and OM = 5 cm.
AM2 = OA2 − OM2 = 169 − 25 = 144, so AM = 12 cm.
AB = 2 × AM = 24 cm.
Example 2: Distance of a chord (NCERT End-of-Chapter Exercise 3)
The diameter of a circle is 26 cm. A chord of length 24 cm is drawn in the circle. Find the distance from the centre of the circle to the chord.
Solution: radius r = 26/2 = 13 cm. Half the chord = 24/2 = 12 cm, since the perpendicular from the centre bisects the chord.
d2 = r2 − 122 = 169 − 144 = 25, so d = 5 cm.
Example 3: Radius from a chord (NCERT End-of-Chapter Exercise 9)
The distance of a chord of length 16 cm from the centre of a circle is 6 cm. Find the radius of the circle.
Solution: half-chord = 8 cm and d = 6 cm. r2 = 82 + 62 = 64 + 36 = 100, so r = 10 cm.
Example 4: Parallel chords on opposite sides (NCERT Exercise Set 5.3, Q3)
Two parallel chords of lengths 6 cm and 8 cm are on opposite sides of the centre of a circle. If the radius of the circle is 5 cm, find the distance between the midpoints of the chords.
Solution: the perpendicular from the centre O to each chord passes through its midpoint. Because the chords are parallel, both perpendiculars lie along one line through O, so the two midpoints and O are collinear.
Chord of 6 cm: d1 = √(25 − 9) = 4 cm. Chord of 8 cm: d2 = √(25 − 16) = 3 cm.
The chords are on opposite sides of O, so the distance between the midpoints = 4 + 3 = 7 cm. (On the same side it would be 4 − 3 = 1 cm.)
Example 5: Parallel chords on the same side (NCERT End-of-Chapter Exercise 18)
Two parallel chords of lengths 10 cm and 24 cm are on the same side of the centre of a circle. The distance between the chords is 7 cm. Find the radius of the circle.
Solution: the longer chord is nearer the centre (Theorem 8). Let its distance from the centre be x cm; the 10 cm chord is then (x + 7) cm away.
From the 24 cm chord: r2 = 122 + x2 = 144 + x2.
From the 10 cm chord: r2 = 52 + (x + 7)2 = 25 + x2 + 14x + 49.
Equating: 144 + x2 = 74 + x2 + 14x, so 14x = 70 and x = 5.
r2 = 144 + 25 = 169, so r = 13 cm. Check with the other chord: 25 + 122 = 169.
Example 6: Chord from a central angle (NCERT Exercise Set 5.6, Q1)
In a circle with centre O, the central angle AOB is 60°. If the radius of the circle is 12 cm, what is the length of the chord AB?
Solution: OA = OB = 12 cm, so ΔOAB is isosceles and ∠OAB = ∠OBA.
∠OAB + ∠OBA = 180° − 60° = 120°, so each is 60°.
All angles are 60°, so ΔOAB is equilateral and AB = 12 cm.
Example 7: Angle at the circle (NCERT End-of-Chapter Exercise 2)
An arc of a circle subtends an angle of 70° at the centre. What is the measure of the angle subtended by the arc at a point on the circle?
Solution: by Theorem 9, the angle at a point on the circle outside the arc is half the central angle: ½ × 70° = 35°.
Example 8: Opposite angles of a cyclic quadrilateral (NCERT End-of-Chapter Exercise 7)
ABCD is a cyclic quadrilateral inscribed in a circle. If ∠A measures 75°, what is the measure of ∠C? If ∠B measures 110°, what is the measure of ∠D?
Solution: by Theorem 11, opposite angles add to 180°.
∠C = 180° − 75° = 105°. ∠D = 180° − 110° = 70°.
Check: 75° + 110° + 105° + 70° = 360°.
Example 9: Cyclic quadrilateral with algebra (NCERT End-of-Chapter Exercise 8)
Quadrilateral PQRS is inscribed in a circle. If ∠P = (2x + 10)° and ∠R = (3x − 20)°, find the value of x and the measures of ∠P and ∠R.
Solution: P and R are opposite vertices, so ∠P + ∠R = 180°.
(2x + 10) + (3x − 20) = 180, so 5x − 10 = 180, 5x = 190 and x = 38.
∠P = 2(38) + 10 = 86°; ∠R = 3(38) − 20 = 94°. Check: 86° + 94° = 180°.
Example 10: Can the quadrilateral be drawn? (NCERT in-text exercise after Theorem 11)
A cyclic quadrilateral has angles measuring ∠A = 80°, ∠B = 110°, ∠C = 100°, and ∠D = 70°. Can such a quadrilateral be drawn? Explain why or why not.
Solution: the angles add to 80 + 110 + 100 + 70 = 360°, so a quadrilateral with these angles exists.
Opposite pairs: ∠A + ∠C = 80° + 100° = 180° and ∠B + ∠D = 110° + 70° = 180°.
By Theorem 12, a quadrilateral whose opposite angles add to 180° is cyclic. So yes, such a cyclic quadrilateral can be drawn.
Example 11: Area of a cyclic quadrilateral (NCERT End-of-Chapter Exercise 10)
A cyclic quadrilateral has sides 5, 5, 12, 12 units. Find its area.
Solution: take the sides in order: AB = BC = 5 and CD = DA = 12. Join BD.
ΔBAD ≅ ΔBCD by SSS (BA = BC, AD = CD, BD common), so ∠A = ∠C.
The quadrilateral is cyclic, so ∠A + ∠C = 180°. Hence ∠A = ∠C = 90°.
Area = area of ΔBAD + area of ΔBCD = ½ × 5 × 12 + ½ × 5 × 12 = 30 + 30 = 60 square units.
If the sides are taken in the order 5, 12, 5, 12, the quadrilateral is a parallelogram inscribed in a circle, so it is a 5 by 12 rectangle, and the area is again 60 square units.
Example 12: Regular hexagon in a circle (NCERT End-of-Chapter Exercise 19)
A regular hexagon is inscribed in a circle of radius r. Find the length of the sides of the hexagon and the distance of each side from the centre of the circle.
Solution: the six sides are equal chords, so they subtend equal angles at the centre (Theorem 2). Each angle is 360°/6 = 60°.
Each side and two radii form an isosceles triangle with apex angle 60°, so it is equilateral (as in Example 6). Side = r.
The perpendicular from the centre bisects the side, so distance = √(r2 − (r/2)2) = √(3r2/4) = (√3/2) r.
Competency-Based Questions (with answers)
1. Case-based: Finding the centre of a broken plate
An archaeologist finds a piece of a circular plate. She marks two non-parallel chords AB and PQ on the curved edge and draws the perpendicular bisector of each. The two bisectors meet at point O. She measures one chord as 30 cm and its distance from O as 8 cm.
(i) Why is O the centre of the plate? (ii) Find the radius of the plate.
Answer: (i) The perpendicular bisector of every chord passes through the centre. O lies on both bisectors, and two non-parallel lines meet in only one point, so O is the centre. (ii) Half-chord = 15 cm, d = 8 cm, r = √(225 + 64) = √289 = 17 cm.
2. Case-based: A well for three villages
Three villages P, Q and R are not on one straight line. The panchayat wants to dig a well at a point equally far from all three.
(i) Where should the well be? (ii) If the angle at Q in ΔPQR is 90° and PR = 4 km, where is the well and how far is it from each village? (iii) Is there such a point if the three villages lie on one straight road?
Answer: (i) At the circumcentre of ΔPQR, where the perpendicular bisectors of PQ and QR meet (Theorem 1). (ii) For a right triangle the circumcentre is the midpoint of the hypotenuse PR. The well is at the midpoint of PR, 2 km from each village. (iii) No. For collinear points the perpendicular bisectors are parallel and never meet.
3. Assertion-Reason
Assertion (A): Infinitely many circles pass through two given points A and B.
Reason (R): The centre of any circle through A and B lies on the perpendicular bisector of AB, and every point of that bisector is equidistant from A and B.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer: (a). Each point of the bisector is the centre of a circle through A and B, and the bisector has infinitely many points.
4. Assertion-Reason
Assertion (A): Three collinear points A, B, C can lie on one circle.
Reason (R): For collinear points A, B, C, the perpendicular bisectors of AB and BC are parallel.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer: (d). Both bisectors are perpendicular to the same line, so they are parallel and have no common point. No point is equidistant from A, B and C, so no circle passes through them.
5. Assertion-Reason
Assertion (A): Two chords of lengths 7 cm each in a circle subtend equal angles at the centre.
Reason (R): The diameter is the longest chord of a circle.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer: (b). A is Theorem 2 (equal chords, equal central angles, by SSS). R is true, but it compares chord lengths and says nothing about central angles, so it does not explain A.
6. Source-based: Completing a proof
A student writes the proof of Theorem 9 with the reasons missing. C is the centre, D is on the circle and DC extended meets arc AFB at E.
Step 1: ∠CBD = ∠CDB. Step 2: ∠BCE = ∠CBD + ∠CDB. Step 3: ∠BCE = 2∠BDC. Step 4: in the same way ∠ACE = 2∠CDA. Step 5: ∠BCA = 2∠BDA.
(i) Give the reason for Step 1. (ii) Give the reason for Step 2. (iii) What do you do with Steps 3 and 4 to reach Step 5?
Answer: (i) CB = CD (radii), so ΔDCB is isosceles and its base angles are equal. (ii) ∠BCE is an exterior angle of ΔBCD, and an exterior angle equals the sum of the two interior opposite angles. (iii) Add them: ∠BCE + ∠ECA = ∠BCA and ∠BDC + ∠CDA = ∠BDA, so ∠BCA = 2∠BDA.
7. Error analysis: Chord length versus distance (NCERT Exercise Set 5.5, Q3)
In a circle, if the distance of chord AB from the centre is twice the distance of another chord CD from the centre, then can we conclude that CD = 2 AB? Give reasons for your answer.
Answer: No. Chord length is 2√(r2 − d2), which is not proportional to d. Take r = 5 cm, CD at 2 cm and AB at 4 cm. CD = 2√(25 − 4) = 2√21 ≈ 9.2 cm and AB = 2√(25 − 16) = 6 cm. Then 2AB = 12 cm, which is not equal to CD. What we can conclude is CD > AB, since the chord nearer the centre is longer (Theorem 8).
8. Reasoning: The shortest chord through a point (NCERT End-of-Chapter Exercise 23)
A point A is 6 cm from the centre O of a circle of radius 10 cm. Which chord through A is the shortest, and how long is it? Which chord through A is the longest?
Answer: for any chord through A, its perpendicular distance from O is at most OA, because OA is the hypotenuse of the right triangle formed with the foot of the perpendicular. The distance equals OA only when the chord is perpendicular to OA. The farthest chord is the shortest (Theorem 8), so the shortest chord is the one perpendicular to OA. Its length is 2√(100 − 36) = 2 × 8 = 16 cm. The longest chord through A is the diameter through A, 20 cm.
Important Questions for Exams
1-Mark Questions
Q1. What is the length of the longest chord in a circle of radius 5 units?
Answer: 10 units, the diameter.
Q2. How many circles pass through three collinear points?
Answer: None.
Q3. What is the least possible radius of a circle through two points A and B?
Answer: AB/2, with the midpoint of AB as centre.
Q4. The diameter of a circle is AB. Point C is on the circumference. What is the measure of ∠ACB?
Answer: 90°, the angle in a semicircle.
Q5. Where does the circumcentre of an obtuse-angled triangle lie?
Answer: Outside the triangle.
2-Mark Questions
Q6. Show that the triangle formed by a chord and the centre of the circle is isosceles.
Answer: Let AB be a chord of a circle with centre O. OA and OB are both radii, so OA = OB. A triangle with two equal sides is isosceles, so ΔOAB is isosceles with ∠OAB = ∠OBA.
Q7. A circle has a radius of 15 cm. A chord is drawn. The distance from the centre of the circle to the chord is 9 cm. What is the length of the chord?
Answer: Half-chord = √(152 − 92) = √(225 − 81) = √144 = 12 cm. Chord = 24 cm.
Q8. Explain why the perpendicular from the centre of a circle to a chord bisects the chord.
Answer: Let OM ⊥ AB. In right triangles OMA and OMB, hypotenuse OA = OB (radii) and OM is common. By RHS, ΔOMA ≅ ΔOMB, so AM = BM.
Q9. Let ABCD be a cyclic quadrilateral and E a point on CD extended. Explain why ∠CDE = ∠ABC.
Answer: ∠CDE + ∠ADC = 180° (linear pair). ∠ABC + ∠ADC = 180° (opposite angles of a cyclic quadrilateral). So ∠CDE = ∠ABC.
3-Mark Questions
Q10. Prove that the perpendicular bisector of a chord passes through the centre of the circle.
Answer: Let AB be a chord of a circle with centre O. OA = OB (radii), so O is equidistant from A and B. The perpendicular bisector of AB is the locus of all points equidistant from A and B. So O lies on it. (Equivalently: if M is the midpoint of AB, OM ⊥ AB by Theorem 4, so line OM is the perpendicular bisector, and it contains O.)
Q11. An isosceles triangle ABC is inscribed in a circle, with AB = AC. Show that the altitude from A to BC passes through the centre of the circle.
Answer: Let AD be the altitude, D on BC. In right triangles ADB and ADC, AB = AC and AD is common, so ΔADB ≅ ΔADC (RHS) and BD = DC. So AD is perpendicular to BC and passes through its midpoint: AD is the perpendicular bisector of chord BC. The perpendicular bisector of a chord passes through the centre, so the altitude from A passes through the centre.
Q12. Show that rectangle is the only parallelogram that can be inscribed in a circle.
Answer: Let parallelogram ABCD be inscribed in a circle. In a parallelogram, opposite angles are equal: ∠A = ∠C. The quadrilateral is cyclic, so ∠A + ∠C = 180°. Hence 2∠A = 180° and ∠A = 90°. A parallelogram with a right angle is a rectangle.
Q13. In a circle with centre O, chords AB and AC are congruent. Explain why this statement is true: “The centre of the circle lies on the angle bisector of ∠BAC”.
Answer: Join OA, OB, OC. In ΔOAB and ΔOAC: OB = OC (radii), AB = AC (given) and OA is common. By SSS, ΔOAB ≅ ΔOAC, so ∠OAB = ∠OAC. Hence AO bisects ∠BAC and O lies on the bisector.
5-Mark Questions
Q14. Prove that the angle subtended by an arc at the centre of a circle is double the angle it subtends at any point on the remaining part of the circle. Hence show that the angle in a semicircle is 90°.
Answer: Let O be the centre, AB the arc and P a point on the circle outside it. Join PO and extend it to meet the arc at Q (in the other case, subtract instead of adding). OA = OP, so ∠OAP = ∠OPA; ∠AOQ is the exterior angle of ΔAOP, so ∠AOQ = ∠OAP + ∠OPA = 2∠OPA. In the same way, OB = OP gives ∠BOQ = 2∠OPB. Adding, ∠AOB = ∠AOQ + ∠QOB = 2(∠OPA + ∠OPB) = 2∠APB. For a semicircle, AB is a diameter, the central angle is a straight angle, 180°, so ∠APB = ½ × 180° = 90°.
Q15. Prove that the sum of either pair of opposite angles of a cyclic quadrilateral is 180°. In a cyclic quadrilateral PQRS, ∠P is 40° more than ∠R. Find ∠P and ∠R.
Answer: Let ABCD be cyclic with centre O. A lies outside arc BCD, so ∠BAD = ½ (reflex ∠BOD), the angle swept from OB to OD through C. C lies outside arc BAD, so ∠BCD = ½ ∠BOD, swept through A. The two central angles together make one complete turn, 360°. So ∠BAD + ∠BCD = ½ × 360° = 180°. As the angles of a quadrilateral add to 360°, ∠B + ∠D = 180° too. For PQRS: ∠P + ∠R = 180° and ∠P − ∠R = 40°. Adding, 2∠P = 220°, so ∠P = 110° and ∠R = 70°.
Q16. Let AB and DE be two chords of a circle with centre C and AB > DE. Prove that AB is nearer to the centre than DE. Use this to justify: “There is no chord of a circle that is longer than its diameter.”
Answer: Draw CF ⊥ AB and CG ⊥ DE. By Theorem 5, F and G are midpoints, so AF = AB/2 and GD = DE/2, and AF > GD. In right triangles CFA and CGD, CA = CD = r, so CF2 + AF2 = CG2 + GD2. Since AF2 > GD2, CF2 < CG2, so CF < CG. For the second part: every chord has a distance d ≥ 0 from the centre, and chord = 2√(r2 − d2) ≤ 2r, with equality only when d = 0, that is, when the chord passes through the centre. So the diameter, 2r, is the longest chord.
Common Mistakes and Examiner Tips
- Using the diameter as the hypotenuse. In chord numericals the hypotenuse is the radius. Halve a given diameter before using r2 = d2 + (half-chord)2.
- Forgetting to double the half-chord. The right triangle gives half the chord. Write “chord = 2 × AM” as a separate line.
- Writing Theorem 5 without the reason. Every time you use the midpoint of a chord, write “the perpendicular from the centre bisects the chord”. A number without its reason is an incomplete step.
- Mixing up SSS, SAS and RHS. Theorem 2 uses SSS, Theorem 3 and Theorem 4 use SAS, Theorem 5 uses RHS. Name the correct rule and list the three matching parts in order.
- Halving the wrong way. The angle at the centre is double the angle at the circle. If the angle at the circle is 40°, the central angle is 80°, not 20°.
- Choosing a point on the arc itself. Theorem 9 needs the point outside the arc. For a major arc, the central angle is reflex (360° minus the minor one), and the angle at a point on the minor arc is half of that.
- Adding adjacent angles of a cyclic quadrilateral. Only opposite angles (A with C, B with D) add to 180°. Check which vertices are opposite from the order of the letters.
- Calling every quadrilateral in a figure cyclic. A quadrilateral is cyclic only if all four vertices lie on the circle, or if a pair of opposite angles is supplementary.
- Same side or opposite sides. For two parallel chords, add the distances when the chords are on opposite sides of the centre and subtract when they are on the same side. Read the question for these words.
- Claiming a circle through any three points. The points must be non-collinear. Always write the word “non-collinear” in Theorem 1.
Quick Revision Points
- A circle is the set (locus) of all points on a plane equidistant from a fixed point, the centre.
- A circle has complete rotational symmetry about its centre and reflection symmetry in every diameter.
- Infinitely many circles pass through two points; their centres lie on the perpendicular bisector of the segment joining them.
- Theorem 1: exactly one circle passes through three non-collinear points. No circle passes through three collinear points.
- Circumcentre = meeting point of perpendicular bisectors of the sides: inside for acute, midpoint of hypotenuse for right, outside for obtuse triangles.
- Theorem 2 and 3: equal chords subtend equal central angles, and conversely (SSS and SAS).
- Theorem 4: the line from the centre to the midpoint of a chord is perpendicular to it.
- Theorem 5: the perpendicular from the centre to a chord bisects it (RHS).
- Chord = 2√(r2 − d2), from r2 = d2 + (chord/2)2.
- Theorem 6 and 7: equal chords are equidistant from the centre, and conversely.
- Theorem 8: the longer chord is nearer the centre.
- Theorem 9: angle at the centre = 2 × angle at any point on the circle outside the arc.
- The angle in a semicircle is 90°.
- Theorem 10: if AB subtends equal angles at C and D on the same side of AB, then A, B, C, D are concyclic.
- Theorem 11: opposite angles of a cyclic quadrilateral add to 180°.
- Theorem 12: if a pair of opposite angles of a quadrilateral add to 180°, it is cyclic.
- The exterior angle of a cyclic quadrilateral equals the interior opposite angle.
- The rectangle is the only parallelogram that can be inscribed in a circle.
Weightage and Exam Pattern
Class 9 has no board exam. Circles is assessed in school exams: periodic tests, the half-yearly exam and the annual exam. In the CBSE Class IX Mathematics course structure for 2026-27, the chapter is part of Unit IV, Geometry, which carries 25 marks out of the 80-mark theory paper and is allotted 69 teaching periods. The unit has five chapters: Introduction to Euclid’s Geometry, Lines and Angles, Triangles (Congruence Theorems), 4-gons (Quadrilaterals) and Circles. The course structure does not split the 25 marks between these chapters. The other 20 marks are internal assessment.
| Unit | Marks (out of 80) |
|---|---|
| I. Number System | 07 |
| II. Algebra | 20 |
| III. Coordinate Geometry | 04 |
| IV. Geometry | 25 |
| V. Mensuration | 14 |
| VI. Statistics and Probability | 10 |
The CBSE question paper design weights questions by typology: about 54% remembering and understanding, 24% applying, and 22% analysing, evaluating and creating. For this chapter the course structure asks students to state definitions, construct the circumcircle, explain why each chord and angle theorem is true, and decide when four points are concyclic. Expect short recall items, chord and angle numericals, and “explain why” proofs.
The chord, arc and cyclic quadrilateral results here are the base for the further results about circles in Class 10.
Class 9 Maths · Chapter 5 – swipe through all 10 cards to understand the whole chapter.
What a Circle Is
The set of all points on a plane equidistant from a given point, the centre.
A circle is a locus: the set of points satisfying one condition.
- Chord: segment joining two points on the circle.
- Diameter: chord through the centre, the longest chord.
- Arc: connected part of the circle; minor arc is the smaller one.
Perfect Symmetry
A circle looks the same after any rotation about its centre.
Every diameter is a line of reflection symmetry.
- Complete rotational symmetry: any angle of turn works.
- Infinitely many lines of symmetry, all diameters.
- A square has only 4 lines; a regular hexagon has 6.
Circles Through Points
Two points allow infinitely many circles; three non-collinear points allow exactly one.
Centres of circles through A and B lie on the perpendicular bisector of AB.
- Smallest circle through A and B has radius AB/2.
- Three collinear points: bisectors are parallel, no circle.
- Circumcentre: inside acute, outside obtuse, hypotenuse midpoint for right.
Equal Chords, Equal Angles
Equal chords subtend equal angles at the centre, and conversely.
Theorem 2 uses SSS; Theorem 3 uses SAS.
- Chord plus centre always makes an isosceles triangle.
- 60° at the centre makes the chord equal to the radius.
- Rotating a chord keeps its length and central angle.
Perpendicular from the Centre
The perpendicular from the centre bisects a chord, and the line to the midpoint is perpendicular.
So the perpendicular bisector of every chord passes through the centre.
- Theorem 4 uses SAS and a linear pair.
- Theorem 5 uses RHS congruence.
- Two chord bisectors meet at the centre of a paper circle.
Chord and Distance
Radius, distance and half-chord form a right triangle.
Distance from the centre always means perpendicular distance.
- r = 13, d = 5 gives chord 24.
- Chord 16 at distance 6 gives r = 10.
- Halve a given diameter before using the formula.
Nearer Means Longer
Equal chords are equidistant from the centre; the longer chord is nearer.
Theorems 6, 7 and 8; the diameter is the longest chord.
- Equidistant chords are equal (converse, Theorem 7).
- Parallel chords: add distances on opposite sides, subtract on the same side.
- At distance r the chord shrinks to a single point.
Angle at Centre is Double
An arc subtends at the centre double the angle it subtends at any point on the circle outside it.
Proof uses isosceles triangles and the exterior angle theorem.
- 70° at the centre gives 35° at the circle.
- Angles in the same segment are equal.
- Angle in a semicircle = 90°.
Cyclic Quadrilaterals
A quadrilateral with all four vertices on a circle has opposite angles adding to 180°.
Converse (Theorem 12): supplementary opposite angles make it cyclic.
- ∠A = 75° gives ∠C = 105°.
- Exterior angle equals the interior opposite angle.
- Rectangle is the only parallelogram that is cyclic.
When Four Points Are Concyclic
If AB subtends equal angles at C and D on the same side of AB, then A, B, C, D lie on one circle.
Theorem 10, proved by contradiction with the exterior angle theorem.
- Any three non-collinear points are always concyclic.
- Four points need a test: equal angles or supplementary opposite angles.
- C and D must be on the same side of AB.
📝 Practice Circles - 10 questions
Concept-check questions · with answers & explanations
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Why A: the assertion is the wheel example from the chapter and it is true. The reason is also true, and it is exactly why the wheel looks unchanged: whatever angle the wheel turns through, the circle maps onto itself.
Why not B: R is not an unrelated fact here; it directly explains A.
Why not C: R is true; a circle has rotational symmetry for every angle, not just some angles like a square.
Why not D: A is true; a turning wheel really does look identical at all times.
Remember: a circle looks the same after a turn through any angle at all.
Why A: Step 1 (centre on perpendicular bisector): the centre O lies on the perpendicular bisector of AB, so with M the midpoint, triangle OMA is right-angled at M and AM = 3 cm. Step 2 (Baudhāyana-Pythagoras): OM² = OA² – AM² = 25 – 9 = 16, so OM = 4 cm.
Why not B: 3 cm is AM, half of AB, not the distance of the centre from M.
Why not C: √34 comes from adding 25 + 9 instead of subtracting.
Why not D: 2 cm is 5 – 3, subtracting lengths instead of their squares.
Remember: radius is the hypotenuse; distance² = r² – (half of AB)².
Why A: AB = AC means A is equidistant from B and C, so A lies on the perpendicular bisector of chord BC. O also lies on it. The altitude from A is the line through A perpendicular to BC, which is that same perpendicular bisector, so it passes through O.
Why not B: O is on BC only when BC is a diameter, that is when ∠A = 90°, which is not always so.
Why not C: O is the midpoint of AB only if AB is a diameter, which is not given.
Why not D: the altitude from B passes through O only if BA = BC as well, which would make the triangle equilateral.
Remember: the apex of an isosceles triangle, the centre and the base midpoint lie on one line.
Why C: Step 1 (longer chord is closer): let the 24 cm chord be x cm from the centre, so the 10 cm chord is x + 7 cm away. Step 2 (Baudhāyana-Pythagoras for each chord): r² = 12² + x² = 5² + (x + 7)², so 144 + x² = 25 + x² + 14x + 49, giving 14x = 70 and x = 5. Step 3: r² = 144 + 25 = 169, so r = 13 cm.
Why not A: 12 cm is the half-chord of the longer chord.
Why not B: 17 cm is 10 + 7, adding a chord length and the gap.
Why not D: 15 cm does not satisfy either triangle: 12² + x² = 225 gives x = 9, and then the other chord would need 5² + 16² = 225, which is false.
Remember: set r² equal for both chords and solve for the unknown distance.
Why A: Step 1 (isosceles triangle): OA = OB, so ∠OBA = 25° and ∠AOB = 180° – 50° = 130°. Step 2 (Theorem 9): C is on the circle outside the minor arc AB, so ∠ACB = 130° / 2 = 65°.
Why not B: 130° is the centre angle, which must be halved at C.
Why not C: 25° copies the base angle of triangle OAB, which is a different angle.
Why not D: 50° is the sum of the two base angles, not half the centre angle.
Remember: find the centre angle from the isosceles triangle, then halve it.
Why D: Step 1 (Theorem 11): ∠ABC = 180° – ∠ADC = 180° – 130° = 50°. Step 2 (angle in a semicircle): AB is a diameter, so ∠ACB = 90°. Step 3 (angle sum of triangle ABC): ∠BAC = 180° – 90° – 50° = 40°.
Why not A: 50° is ∠ABC, found in the first step, not ∠BAC.
Why not B: 130° makes ∠BAC equal to ∠ADC with no theorem to support it.
Why not C: 90° is ∠ACB, the angle in the semicircle at C.
Remember: find the opposite angle first, then use the right angle on the diameter.
Why D: the longest chord is the one through the centre, the diameter, and diameter = 2 × radius = 2 × 5 = 10 units. Every other chord is closer to the edge and shorter.
Why not A: 5 units is the radius, which is not even a chord because one end is at the centre.
Why not B: 25 is the square of the radius, a calculation slip with no meaning here.
Why not C: chord length is limited; the diameter is the greatest possible chord, though there is no smallest one.
Remember: longest chord = diameter = 2r.
Why B: Step 1 (Baudhāyana-Pythagoras): PR² = 6² + 8² = 36 + 64 = 100, so hypotenuse PR = 10 cm. Step 2 (circumcentre of a right triangle): it lies at the midpoint of the hypotenuse, so the circumradius = 10/2 = 5 cm.
Why not A: Q is a vertex on the circle, it cannot be the centre.
Why not C: 10 cm is the whole hypotenuse, which is the diameter, not the radius.
Why not D: the circumcentre lies inside only for acute-angled triangles, and 7 cm is just the average of 6 and 8.
Remember: right triangle, circumradius = half the hypotenuse.
Why C: Step 1 (isosceles triangle): OA = OB as radii, so the base angles are equal and ∠OBA = ∠OAB = 40°. Step 2 (angle sum of a triangle): ∠AOB = 180° – 40° – 40° = 100°.
Why not A: 40° assumes all three angles are equal, but only the base angles are.
Why not B: 80° is the sum of the two base angles, not the apex angle.
Why not D: 140° is 180° – 40°, forgetting the second base angle.
Remember: two radii make an isosceles triangle; apex = 180° – 2 × base angle.
Why B: Step 1 (Baudhāyana-Pythagoras): the 6 cm chord is √(25 – 9) = 4 cm from the centre and the 8 cm chord is √(25 – 16) = 3 cm from the centre. Step 2 (opposite sides): the distances add, 4 + 3 = 7 cm.
Why not A: 1 cm = 4 – 3 is the answer when both chords are on the same side.
Why not C: 5 cm is the radius, not the gap between the chords.
Why not D: 14 cm is 6 + 8, adding chord lengths.
Remember: opposite sides add distances; same side subtracts them.
Chapter Navigation
Previous: Exploring Algebraic Identities Class 9 Notes
Next: Area and Perimeter Class 9 Notes
Related Chapters in Class 9 Maths
- Coordinate Geometry Class 9 Notes
- Introduction to Polynomials Class 9 Notes
- Number System Class 9 Notes
- Exploring Algebraic Identities Class 9 Notes
- Area and Perimeter Class 9 Notes
- Introduction to Probability Class 9 Notes
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Frequently Asked Questions
A circle is the set of all points on the plane that are equidistant from a given point on that plane. A set of points satisfying a condition is called a locus, so a circle is the locus of points at a fixed distance from a fixed point. The fixed point is the centre and the fixed distance is the radius.
Infinitely many circles pass through two points A and B, because every point on the perpendicular bisector of AB can be a centre. Exactly one circle passes through three non-collinear points, with its centre where the perpendicular bisectors meet. No circle passes through three collinear points, because their perpendicular bisectors are parallel.
The perpendicular from the centre bisects the chord, which makes a right triangle with the radius as hypotenuse. So r² = d² + (half-chord)², and the chord length is 2√(r² − d²). For r = 13 cm and d = 5 cm, the half-chord is 12 cm and the chord is 24 cm.
The longer chord is closer to the centre. Both half-chords and their distances satisfy r² = d² + (half-chord)², with the same r, so a bigger half-chord forces a smaller distance. The diameter passes through the centre at distance zero, which is why it is the longest chord of the circle.
The angle subtended by an arc at the centre is double the angle it subtends at any point on the circle outside the arc. So an arc of 70° at the centre gives 35° at the circle. It follows that angles in the same segment are equal and that the angle in a semicircle is 90°.
A cyclic quadrilateral is a quadrilateral whose four vertices lie on one circle. Its opposite angles add up to 180°. The converse is also true: if a pair of opposite angles of a quadrilateral add to 180°, its vertices are concyclic. For example, if ∠A = 75° in cyclic ABCD, then ∠C = 105°.
The circumcentre is where the perpendicular bisectors of the sides meet. It lies inside an acute-angled triangle, outside an obtuse-angled triangle, and at the midpoint of the hypotenuse of a right-angled triangle. For a right triangle with hypotenuse 10 cm, the circumradius is 5 cm.
Class 9 has no board exam, so Circles is assessed in school exams. In the CBSE 2026-27 course structure it belongs to Unit IV, Geometry, which carries 25 of the 80 theory marks together with four other chapters. The course structure does not give a separate figure for Circles.
