Circles Class 9: Chords, Arcs, Angles and Cyclic Quadrilaterals

What Circles is about

Weightage: This chapter is part of Unit IV, Geometry, which carries 25 of the 80 marks in the school annual exam, shared with Introduction to Euclid's Geometry: Axioms and Postulates, Lines and Angles, Triangles – Congruence Theorems and 4-gons (Quadrilaterals).Source: CBSE Mathematics syllabus, Class IX 2026-27, course structure, Unit IV Geometry

Key concepts

What a circle is: centre, radius, chord, diameter and arcs

NCERT section 5.1 Definitions

A circle is the set of all points in a plane at the same distance from a fixed point in that plane. Such a set of points is called a locus. The fixed point is the centre; the fixed distance is the radius.

• A chord joins two points of the circle. The angle it makes at the centre is the angle it subtends there.
• A chord through the centre is a diameter, the longest chord, of length $2r$.
• An arc is a connected part of the circle between two points. Of the two arcs, the smaller is the minor arc (central angle below 180°) and the bigger is the major arc (central angle above 180°).

Worked example

A circle has radius 6.5 cm. (i) What is the length of its longest chord? (ii) A chord subtends 140° at the centre. Is its minor arc or major arc subtending 220°?

Try it first, then show the solutionSolution

1. (i) The longest chord is a diameter: $2\times 6.5=13$ cm.
2. (ii) The two arcs cut off by the chord subtend $140°$ and $360°-140°=220°$ at the centre.
3. The one subtending more than 180° is the major arc.

Answer: (i) 13 cm (ii) the major arc subtends 220°; the minor arc subtends 140°.

Symmetries of a circle

NCERT section 5.2 Symmetries of a Circle

A circle has complete rotational symmetry: turn it about its centre through any angle and it looks exactly the same (like a turning wheel). It also has reflection symmetry in every diameter: fold a paper circle so the edges meet and the crease is a diameter.

So a circle has infinitely many lines of symmetry, all passing through the centre. Compare: a square has 4, a regular pentagon 5, a regular hexagon 6.

This symmetry is the reason behind many results: if you rotate a chord around the centre, its length, the angle it subtends and its distance from the centre all stay the same.

Worked example

How many lines of reflection symmetry do a square, a regular hexagon and a circle have? Through which point do all the circle's lines of symmetry pass?

Try it first, then show the solutionSolution

1. Square: 2 diagonals and 2 lines joining midpoints of opposite sides, so 4.
2. Regular hexagon: 3 through opposite vertices and 3 through midpoints of opposite sides, so 6.
3. Circle: every diameter is a line of symmetry, so infinitely many.
4. Every diameter passes through the centre.

Answer: 4, 6 and infinitely many; all pass through the centre.

How many circles through given points? The circumcircle

NCERT section 5.3 How Many Circles?

• Through one point: infinitely many circles.
• Through two points A and B: infinitely many, and their centres all lie on the perpendicular bisector of AB (the locus of points equidistant from A and B). The smallest has AB as diameter.
• Through three non-collinear points: exactly one circle. Its centre is where the perpendicular bisectors of AB and AC meet (NCERT Theorem 1).
• Through three collinear points: none, because the perpendicular bisectors are parallel and never meet.

The circle through the vertices of a triangle is its circumcircle and its centre the circumcentre. It lies inside an acute-angled triangle, outside an obtuse-angled triangle, and at the midpoint of the hypotenuse of a right-angled triangle.

Worked example

A right-angled triangle has legs 15 cm and 20 cm. Where is its circumcentre, and what is the radius of its circumcircle?

Try it first, then show the solutionSolution

1. Hypotenuse $=\sqrt{15^2+20^2}=\sqrt{225+400}=\sqrt{625}=25$ cm.
2. For a right-angled triangle, the circumcentre is the midpoint of the hypotenuse.
3. Radius $=$ half the hypotenuse $=12.5$ cm.

Answer: At the midpoint of the 25 cm hypotenuse; radius 12.5 cm.

Equal chords and equal angles at the centre

NCERT section 5.4 Chords and the Angles They Subtend

NCERT Theorem 2: equal chords of a circle subtend equal angles at the centre.
*Why:* if $AB=DE$ in a circle with centre C, then $CA=CD$ and $CB=CE$ (radii) and $AB=DE$, so $△CAB\cong △CDE$ by SSS, giving $\angle ACB=\angle DCE$.

NCERT Theorem 3 (converse): chords that subtend equal angles at the centre are equal.
*Why:* two radii and the included angle match, so the triangles are congruent by SAS.

Useful fact: the triangle formed by a chord and the centre is isosceles (two sides are radii), so its base angles are equal.

Worked example

In a circle with centre O, chords AB and CD are equal and $\angle AOB=70°$. Find $\angle COD$ and $\angle OCD$.

Try it first, then show the solutionSolution

1. Equal chords subtend equal angles at the centre, so $\angle COD=70°$.
2. In $△OCD$, $OC=OD$ (radii), so $\angle OCD=\angle ODC$.
3. $\angle OCD=\frac{180°-70°}{2}=55°$.

Answer: $\angle COD=70°$ and $\angle OCD=55°$.

The perpendicular from the centre bisects a chord

NCERT section 5.5 Midpoints and Perpendicular Bisectors of Chords

NCERT Theorem 4: the line joining the centre to the midpoint of a chord is perpendicular to the chord.
*Why:* with M the midpoint of AB and C the centre, $CA=CB$, $AM=BM$ and CM is common, so $△CMA\cong △CMB$ (SSS). Then $\angle CMA=\angle CMB$, and since they add to 180°, each is 90°.

NCERT Theorem 5 (converse): the perpendicular from the centre to a chord bisects the chord (RHS congruence).

Consequence: the perpendicular bisector of any chord passes through the centre. That is how you find the centre of a circle you only partly have: draw two chords and their perpendicular bisectors.

In calculations, radius r, half-chord $\frac{c}{2}$ and distance d from the centre form a right triangle: $r^2=d^2+{\left(\frac{c}{2}\right)}^2$.

Worked example

A chord of length 24 cm is drawn in a circle of radius 13 cm. How far is the chord from the centre?

Try it first, then show the solutionSolution

1. The perpendicular from the centre bisects the chord, so half the chord is 12 cm.
2. $d^2=r^2-12^2=169-144=25$.
3. $d=5$ cm.

Answer: 5 cm

Distance of chords from the centre

NCERT section 5.6 Distance of Chords from the Centre

The distance of a chord from the centre means the perpendicular distance.

• NCERT Theorem 6: equal chords are equidistant from the centre (congruent triangles, or the Baudhāyana–Pythagoras theorem).
• NCERT Theorem 7 (converse): chords equidistant from the centre are equal.
• NCERT Theorem 8: of two unequal chords, the longer one is nearer the centre. Since $r^2=d^2+(\text{half-chord}{)}^2$ is fixed, a bigger half-chord means a smaller d.

Formula: chord length $=2\sqrt{r^2-d^2}$. The diameter ($d=0$) is the longest chord; as d grows to r, the chord shrinks to a point.

Worked example

In a circle of radius 10 cm, chords PQ and RS have lengths 16 cm and 12 cm. Find their distances from the centre and check that the longer chord is nearer.

Try it first, then show the solutionSolution

1. PQ: half-chord 8, $d=\sqrt{100-64}=6$ cm.
2. RS: half-chord 6, $d=\sqrt{100-36}=8$ cm.
3. $16>12$ and $6<8$: the longer chord is nearer the centre.

Answer: PQ is 6 cm and RS is 8 cm from the centre.

Angle at the centre is double the angle on the circle

NCERT section 5.7 Angles Subtended by an Arc

NCERT Theorem 9: the angle subtended by an arc at the centre is twice the angle the same arc makes at any point on the rest of the circle.

*Why (one case):* join the point D on the circle to the centre C and extend to E. Triangles DCA and DCB are isosceles (radii), so each exterior angle at C is twice the base angle at D: $\angle ACE=2\angle ADC$ and $\angle BCE=2\angle BDC$. Adding gives $\angle ACB=2\angle ADB$. (When E falls outside the arc, subtract instead of adding.)

For a major arc, the angle at the centre is the reflex angle, so the angle on the circle can be more than 90°.

Worked example

O is the centre and $\angle AOB=130°$ (minor arc AB). C is a point on the major arc and D a point on the minor arc. Find $\angle ACB$ and $\angle ADB$.

Try it first, then show the solutionSolution

1. Minor arc AB subtends 130° at the centre, so at C (outside the minor arc) it subtends $\frac{130°}{2}=65°$.
2. Major arc AB subtends the reflex angle $360°-130°=230°$ at the centre.
3. D is outside the major arc, so $\angle ADB=\frac{230°}{2}=115°$.
4. Check: $65°+115°=180°$, as expected since ACBD is a cyclic quadrilateral.

Answer: $\angle ACB=65°$ and $\angle ADB=115°$.

Angles in the same segment and the angle in a semicircle

NCERT section 5.7 Angles Subtended by an Arc

From the last theorem, every point on the circle outside a given arc sees that arc at the same angle (half the central angle). So angles in the same segment are equal: $\angle AEB=\angle ADB=\angle AFB$.

Corollary: the angle subtended by a diameter at any point of the circle is 90° (the angle in a semicircle), because the central angle is a straight angle, 180°.

This property belongs to circles alone: points inside the circle see the chord at a bigger angle and points outside see it at a smaller angle.

Worked example

AB is a diameter of a circle and C is a point on the circle with $\angle CAB=35°$. Find $\angle ACB$ and $\angle ABC$.

Try it first, then show the solutionSolution

1. The angle in a semicircle is a right angle, so $\angle ACB=90°$.
2. Angle sum of $△ABC$: $\angle ABC=180°-90°-35°=55°$.

Answer: $\angle ACB=90°$ and $\angle ABC=55°$.

Concyclic points and cyclic quadrilaterals

NCERT section 5.8 Concyclicity of Points

Points on one circle are concyclic.

• NCERT Theorem 10: if segment AB subtends equal angles at two points C and D on the same side of AB, then A, B, C, D are concyclic.
• A quadrilateral (4-gon) whose vertices lie on a circle is cyclic.
• NCERT Theorem 11: the opposite angles of a cyclic quadrilateral add up to 180°. (Each is half of the two central angles that together make 360°.)
• NCERT Theorem 12 (converse): if a pair of opposite angles of a quadrilateral adds to 180°, it is cyclic.

A consequence: the exterior angle of a cyclic quadrilateral equals the interior opposite angle.

Worked example

In a cyclic quadrilateral ABCD, $\angle A=(3x+5)°$ and $\angle C=(2x+25)°$. Find x, $\angle A$ and $\angle C$.

Try it first, then show the solutionSolution

1. Opposite angles of a cyclic quadrilateral add to 180°: $(3x+5)+(2x+25)=180$.
2. $5x+30=180$, so $x=30$.
3. $\angle A=95°$ and $\angle C=85°$; check: $95+85=180$.

Answer: $x=30$; $\angle A=95°$, $\angle C=85°$.

Maths in real life: Circles

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Angles in a Circle and Cyclic Quadrilaterals
• Chord Theorems and Distance from the Centre

Circles formula sheet

Definition

$\text{circle}=\{P:OP=r\}$

All points in the plane at distance r from the centre O.

Diameter

$d=2r$

The longest chord.

Radius, half-chord and distance

$r^2=d^2+{\left(\frac{c}{2}\right)}^2$

d is the perpendicular distance of a chord of length c from the centre.

Chord length

$c=2\sqrt{r^2-d^2}$

Equal d means equal chords; smaller d means a longer chord.

Distance from the centre

$d=\sqrt{r^2-{\left(\frac{c}{2}\right)}^2}$

Central angle and inscribed angle

$\angle AOB=2\angle ACB$

C on the circle, outside the arc AB.

Angle in a semicircle

$\angle ACB=90^{\circ }\text{ if AB is a diameter}$

Angles in the same segment

$\angle ACB=\angle ADB$

C and D on the same side of chord AB, on the circle.

Cyclic quadrilateral

$\angle A+\angle C=180^{\circ },\angle B+\angle D=180^{\circ }$

And the converse.

Exterior angle of a cyclic quadrilateral

$\text{exterior }\angle \text{ at D}=\text{interior }\angle B$

Because both are supplementary to $\angle ADC$.

Base angles in a chord triangle

$\angle OAB=\angle OBA=\frac{180^{\circ }-\angle AOB}{2}$

OA = OB are radii.

Circumradius of a right triangle

$R=\frac{\text{hypotenuse}}{2}$

Circumcentre at the midpoint of the hypotenuse.

Chord with a 60° central angle

$\angle AOB=60^{\circ }\Rightarrow AB=r$

Triangle AOB is equilateral.

Number of circles

$\text{2 points}:\infty ;\text{3 non-collinear}:1;\text{3 collinear}:0$

How the school exam asks Circles

The question patterns to expect, and the method that works for each.

Tricks to remember Circles

Short memory aids. Say each one aloud once, then use it on a question the same day.

• Every chord hides a right triangle

  Radius is the hypotenuse; half the chord and the distance from the centre are the two legs.

  Use it when: All chord length and distance calculations.

• Centre sees double

  The centre is closer to the action, so it sees the arc at double the angle a point on the circle sees.

  Use it when: Central angle = 2 × inscribed angle.

• Opposite friends share 180

  In a cyclic quadrilateral, opposite angles are friends who split ₹180 between them.

  Use it when: Cyclic quadrilateral angle questions.

• Longer chord, closer to home

  The diameter lives at the centre (distance 0) and is the longest; shorter chords move out towards the edge.

  Use it when: Comparing distances of unequal chords.

• Acute in, obtuse out, right on the edge

  Circumcentre: inside an acute triangle, outside an obtuse one, on the hypotenuse of a right one.

  Use it when: MCQs on the position of the circumcentre.

Circles: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section E style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 4-mark question split into 1 + 1 + 2 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Circles

The slips examiners see most often, and what to write instead.

• ✗Mistake: Using the whole chord instead of half

  The bisecting step is skipped.

  Wrong working

  $r=13$, chord 24: $d=\sqrt{169-576}$

  Right working

  Half-chord 12: $d=\sqrt{169-144}=5$ cm

  ✓Do this: Always write 'the perpendicular from the centre bisects the chord' and halve the chord first.

  Usually costs: 1 to 2 marks.

• ✗Mistake: Halving the wrong angle for a major arc

  The non-reflex central angle is used for the angle on the minor arc.

  Wrong working

  $\angle AOB=130°$, D on the minor arc, so $\angle ADB=65°$.

  Right working

  D on the minor arc sees the major arc: $\angle ADB=\frac{360°-130°}{2}=115°$.

  ✓Do this: Ask which arc the angle stands on; a point on one arc sees the other arc.

  Usually costs: 1 mark per angle.

• ✗Mistake: Adding distances for chords on the same side

  The position of the chords is not drawn.

  Wrong working

  Chords at 12 cm and 5 cm on the same side are 17 cm apart.

  Right working

  Same side: $12-5=7$ cm apart; opposite sides: $12+5=17$ cm.

  ✓Do this: Draw both chords before calculating.

  Usually costs: The final answer mark.

• ✗Mistake: Adding adjacent angles of a cyclic quadrilateral to 180°

  Opposite and adjacent are confused.

  Wrong working

  ABCD cyclic, $\angle A=75°$, so $\angle B=105°$.

  Right working

  $\angle A+\angle C=180°$, so $\angle C=105°$; $\angle B$ is not fixed by $\angle A$.

  ✓Do this: Opposite means across the diagonal: A with C, B with D.

  Usually costs: 1 mark.

• ✗Mistake: Proofs without reasons

  Statements are written but the congruence rule or the reason is left out.

  Wrong working

  $△OAM\cong △OBM$, so $AM=BM$.

  Right working

  $OA=OB$ (radii), $\angle OMA=\angle OMB=90°$, OM common, so $△OAM\cong △OBM$ (RHS) and $AM=BM$ (CPCT).

  ✓Do this: Write the reason in brackets after every step.

  Usually costs: 1 to 2 marks in a proof.

• ✗Mistake: Assuming the circumcentre is always inside the triangle

  Most diagrams show acute triangles.

  Wrong working

  For an obtuse triangle, the circumcentre is inside.

  Right working

  For an obtuse triangle, the circumcentre is outside; for a right triangle it is the midpoint of the hypotenuse.

  ✓Do this: Check the largest angle first.

  Usually costs: 1-mark MCQ.

NCERT exercise guide: Circles

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Practice questions on Circles

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQExam level

   The longest chord of a circle of radius 7 cm is
   (a) 7 cm
   (b) 14 cm
   (c) 21 cm
   (d) 49 cm

   Hint

   The longest chord passes through the centre.

   Answer

   (b) 14 cm

2. Q2 · 1 mark · MCQExam level

   The circumcentre of a right-angled triangle lies
   (a) inside the triangle
   (b) outside the triangle
   (c) at the vertex with the right angle
   (d) at the midpoint of the hypotenuse

   Hint

   The hypotenuse is a diameter of the circumcircle.

   Answer

   (d)

3. Q3 · 1 mark · Assertion-reasonExam level

   Assertion (A): The angle subtended by a diameter at any point of the circle is 90°.
   Reason (R): An arc makes an angle at the centre that is twice the angle it makes at any point on the rest of the 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.

   Hint

   What angle does a diameter make at the centre?

   Answer

   (a). The diameter's arc subtends 180° at the centre, so half of it, 90°, at the circle.

4. Q4 · 2 marks · Very short answerExam level

   A chord is 7 cm from the centre of a circle of radius 25 cm. Find its length.

   Hint

   $c=2\sqrt{r^2-d^2}$

   Answer

   $2\sqrt{625-49}=2\times 24=48$ cm.

5. Q5 · 2 marks · Very short answerHarder exam level

   O is the centre of a circle and $\angle BOC=140°$ (minor arc BC). A is a point on the major arc and D a point on the minor arc. Find $\angle BAC$ and $\angle BDC$.

   Hint

   D sees the major arc.

   Answer

   $\angle BAC=70°$; $\angle BDC=\frac{360°-140°}{2}=110°$.

6. Q6 · 3 marks · Short answerHarder exam level

   ABCD is a cyclic quadrilateral with $\angle DBC=55°$ and $\angle BAC=45°$. Find $\angle BDC$, $\angle CAD$ and $\angle BCD$.

   Hint

   Use angles in the same segment, then the cyclic quadrilateral property.

   Answer

   $\angle BDC=\angle BAC=45°$ (same segment on BC). $\angle CAD=\angle CBD=55°$ (same segment on CD). $\angle BAD=100°$, so $\angle BCD=80°$.

7. Q7 · 3 marks · Short answerHarder exam level

   In a circle of radius 17 cm, chords AB = 30 cm and CD = 16 cm. Find their distances from the centre and state which is nearer.

   Hint

   Half-chords 15 and 8.

   Answer

   AB: $\sqrt{289-225}=8$ cm. CD: $\sqrt{289-64}=15$ cm. AB, the longer chord, is nearer.

8. Q8 · 3 marks · Short answerHarder exam level

   ABCD is a cyclic quadrilateral and side CD is extended to E. Prove that $\angle BCE=\angle BAD$.

   Hint

   Angles on a line and opposite angles.

   Answer

   $\angle BCE+\angle BCD=180°$ (linear pair) and $\angle BAD+\angle BCD=180°$ (cyclic quadrilateral). So $\angle BCE=\angle BAD$.

9. Q9 · 5 marks · Long answerHarder exam level

   Prove that the angle subtended by an arc at the centre is double the angle subtended at a point on the remaining part of the circle (take the case where the line from the point through the centre meets the arc).

   Hint

   Join the point to the centre and extend; use isosceles triangles and the exterior angle property.

   Answer

   Join D to centre C and extend to E on the arc. $CA=CD$, so $\angle CAD=\angle CDA$ and exterior $\angle ACE=2\angle CDA$. Similarly $\angle BCE=2\angle CDB$. Adding, $\angle ACB=2(\angle CDA+\angle CDB)=2\angle ADB$.

10. Q10 · 3 marks · Short answerHarder exam level

    PQRS is a cyclic quadrilateral with $\angle Q=(4y-10)°$ and $\angle S=(2y+40)°$. Find y. What does your answer tell you about the diagonal PR?

    Hint

    Q and S are opposite.

    Answer

    $6y+30=180$, so $y=25$ and $\angle Q=\angle S=90°$. PR subtends 90° on the circle, so PR is a diameter.

11. Q11 · 4 marks · Case-basedHOTS / competency

    A broken circular plate shows only an arc. A chord AB across it is 24 cm long, and the arc rises 8 cm above the middle of AB. Find the radius of the plate, and say how you could find its centre by construction.

    Hint

    The centre lies on the perpendicular bisector of AB, at distance $r-8$ from AB.

    Answer

    $r^2=12^2+(r-8{)}^2$ gives $16r=208$, so $r=13$ cm. Construction: draw two chords and their perpendicular bisectors; they meet at the centre.

12. Q12 · 3 marks · Short answerHOTS / competency

    Point A is 6 cm from the centre O of a circle of radius 10 cm. Find the lengths of the shortest and the longest chords through A, and explain why.

    Hint

    The chord farthest from O is the shortest.

    Answer

    Longest: the diameter through A, 20 cm. Shortest: the chord perpendicular to OA, since its distance from O (6 cm) is the greatest possible; length $2\sqrt{100-36}=16$ cm.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.Define a circle as a locus.Show answer

   The set of all points in a plane at a fixed distance (the radius) from a fixed point (the centre).

2. 2.How many circles pass through three collinear points?Show answer

   None.

3. 3.Where is the circumcentre of a right-angled triangle?Show answer

   At the midpoint of the hypotenuse.

4. 4.What congruence rule shows that equal chords subtend equal angles at the centre?Show answer

   SSS (two radii and the equal chords).

5. 5.What does the perpendicular from the centre do to a chord?Show answer

   It bisects the chord.

6. 6.Which of two unequal chords is nearer the centre?Show answer

   The longer one.

7. 7.State the relation between the angle at the centre and the angle on the circle.Show answer

   The angle at the centre is double the angle subtended at any point on the remaining part of the circle.

8. 8.What is the angle in a semicircle?Show answer

   90°.

9. 9.What is special about opposite angles of a cyclic quadrilateral?Show answer

   They add up to 180°.

10. 10.When are four points concyclic (one test)?Show answer

    When the segment joining two of them subtends equal angles at the other two, on the same side.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you define a circle using the word 'locus'?
• ☐Can you say how many circles pass through 1, 2 and 3 points?
• ☐Can you construct the circumcircle of a triangle?
• ☐Can you prove that equal chords subtend equal angles at the centre?
• ☐Can you prove that the perpendicular from the centre bisects a chord?
• ☐Can you find a chord's distance from the centre given r and the chord?
• ☐Can you handle two parallel chords on the same and on opposite sides?
• ☐Can you find an angle on the circle from a central angle, including a reflex one?
• ☐Can you explain why the angle in a semicircle is 90°?
• ☐Can you solve for x in a cyclic quadrilateral and test whether four angles can be cyclic?

Circles: frequently asked questions