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I'm Up and Down, and Round and Round

NCERT Class 9 · Mathematics Based on NCERT Class 9 Mathematics textbook · Free CBSE study kit

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Chapter Notes

Overview
This chapter introduces the circle as a set of points equidistant from a fixed point, exploring its symmetries, chords, arcs, and angles. It exists in the syllabus to build foundational geometry skills for understanding circular shapes in nature and mathematics. The one big idea is that all circles share universal properties—like equal chords subtending equal angles at the centre—which can be proven using congruence and symmetry.
Key Concepts (15)
Circle
A circle is the set of all points on the plane that are equidistant from a given point on that plane.
e.g. A bangle or a coin has all points on its boundary at the same distance from its centre.
Centre
The given point from which all points on the circle are equidistant is called the centre of the circle.
e.g. In a wheel, the centre is the point where the axle is fixed.
Radius
The distance from the centre to any point on the circle is the radius of the circle.
e.g. If you draw a circle with a compass, the distance between the needle and pencil tip is the radius.
Chord
A line segment joining two points on the circle is called a chord of the circle.
e.g. A tight thread tied at two points on a wheel forms a chord.
Diameter
A chord passing through the centre of a circle is called a diameter.
e.g. The longest chord of a circle, like the width of a circular plate through its centre.
Locus
The set of points that satisfy a given condition is also called the locus of points that satisfy the condition.
e.g. The locus of points equidistant from a fixed point is a circle.
Arc
An arc of a circle is a connected portion of the circle, defined by two points on the circle and the curve connecting them along the circle’s edge.
e.g. The curved part of a slice of pizza from crust to crust is an arc.
Minor Arc
The smaller of the two arcs between two points on a circle is called the minor arc.
e.g. If you cut a circle at two points, the shorter curved path is the minor arc.
Major Arc
The bigger of the two arcs between two points on a circle is called the major arc.
e.g. The longer curved path around a circle between two points is the major arc.
Circumcentre
The centre O of the circle that passes through the vertices A, B and C of triangle ABC is called the circumcentre of triangle ABC.
e.g. For an acute triangle, the circumcentre lies inside the triangle.
Circumcircle
The circle that passes through the vertices of a triangle is called the circumcircle and is said to circumscribe the triangle.
e.g. Drawing a circle through the three corners of a triangular park.
Cyclic Quadrilateral
When the vertices of a quadrilateral are concyclic, the quadrilateral is called a cyclic quadrilateral.
e.g. A rectangle inscribed in a circle is a cyclic quadrilateral.
Concyclic
Points that lie on the same circle are called concyclic.
e.g. The four corners of a square are concyclic because they lie on the circumcircle of the square.
Angle Subtended by an Arc at the Centre
The angle subtended by the arc AB at the centre is the measure of the angle AOB, as we sweep along the arc from OA to OB.
e.g. If you stand at the centre and look at two points on the circle, the angle between your lines of sight is the angle subtended by the arc.
Angle Subtended by an Arc at a Point on the Circle
The angle ACB where C is any point on the circle but not on the arc is the angle subtended by the arc at a point on the circle outside the arc.
e.g. If you stand on the circumference and look at two other points on the circle, the angle you see is the angle subtended by the arc.
Formulas & Equations (4)
Chord length = 2√(r² - d²)
Reads as: Chord length equals two times the square root of r squared minus d squared
Variables: r = radius of the circle, d = perpendicular distance from the centre to the chord
Use when: When given the radius and perpendicular distance from centre to chord, find the chord length.
⚠ Students forget to multiply by 2 after taking the square root, giving only half the chord length.
∠ at centre = 2 × ∠ at any point on the remaining part of the circle
Reads as: Angle at the centre equals twice the angle at any point on the remaining part of the circle
Variables: ∠ at centre = angle subtended by an arc at the centre, ∠ at point = angle subtended by the same arc at any point on the circle outside the arc
Use when: To find the angle at the circumference given the central angle, or vice versa.
⚠ Students apply this theorem when the point is on the same arc as the endpoints, but it only holds for points on the remaining part of the circle.
Sum of opposite angles of a cyclic quadrilateral = 180°
Reads as: Sum of opposite angles of a cyclic quadrilateral equals 180 degrees
Variables: ∠A + ∠C = 180°, ∠B + ∠D = 180° for cyclic quadrilateral ABCD
Use when: To find missing angles in a cyclic quadrilateral or to prove a quadrilateral is cyclic.
⚠ Students add adjacent angles instead of opposite angles.
Angle in a semicircle = 90°
Reads as: Angle in a semicircle equals 90 degrees
Variables: If AB is a diameter and C is any point on the circle, then ∠ACB = 90°
Use when: To prove a triangle is right-angled when its hypotenuse is a diameter of the circumcircle.
⚠ Students think any angle in a circle is 90°, but it only applies when the chord is a diameter.
How It Works
Construction of a circle through three non-collinear points
  1. Step 1: Draw the perpendicular bisector of AB.
  2. Step 2: Draw the perpendicular bisector of AC (or BC).
  3. Step 3: The intersection point O of the two perpendicular bisectors is the circumcentre.
  4. Step 4: With centre O and radius OA (or OB or OC), draw the circle.
💡 The perpendicular bisectors of any two sides of the triangle always intersect at a single point, which is the circumcentre.
Proving equal chords subtend equal angles at the centre
  1. Step 1: Draw two equal chords AB and DE in a circle with centre C.
  2. Step 2: Note that CA = CB = CD = CE = radius.
  3. Step 3: Given AB = DE, by SSS congruence, triangle CAB is congruent to triangle CDE.
  4. Step 4: Therefore, ∠ACB = ∠DCE.
💡 Always mention SSS congruence and that radii are equal.
Proving the perpendicular from centre bisects the chord
  1. Step 1: Draw chord AB with midpoint M and centre C.
  2. Step 2: Triangle CAB is isosceles with CA = CB.
  3. Step 3: M is midpoint, so AM = BM.
  4. Step 4: By SAS congruence, triangle CMA is congruent to triangle CMB.
  5. Step 5: Therefore, ∠CMA = ∠CMB = 90°, so CM ⟂ AB.
💡 The converse also holds: the perpendicular from centre to chord bisects the chord.
Proving the angle subtended by an arc at the centre is double the angle at any point on the circle
  1. Step 1: Draw arc AFB with centre C and point D on the circle outside the arc.
  2. Step 2: Join D to C and extend to meet the circle at E on arc AFB.
  3. Step 3: In triangle DCB, CB = CD, so ∠CBD = ∠CDB. Exterior angle ∠BCE = 2∠BDC.
  4. Step 4: Similarly, in triangle ADC, ∠ACE = 2∠CDA.
  5. Step 5: Add angles: ∠BCA = ∠BCE + ∠ECA = 2(∠BDC + ∠CDA) = 2∠BDA.
💡 There are two cases depending on where the extended line from centre meets the circle; both yield the same result.
Proving a quadrilateral is cyclic using equal angles
  1. Step 1: Given segment AB subtends equal angles at C and D on the same side of AB.
  2. Step 2: Draw the circle through A, B, C (non-collinear).
  3. Step 3: Assume D is outside the circle; then AD meets circle at E. ∠AEB = ∠ACB (angles in same segment).
  4. Step 4: But ∠AEB > ∠ADB (exterior angle theorem), contradicting ∠ACB = ∠ADB.
  5. Step 5: Similarly, D cannot be inside. Hence D lies on the circle.
💡 Use proof by contradiction — assume D is not on the circle and show it leads to an impossible conclusion.
Exam Traps — Don't Make These Mistakes
⚠ Confusing chord and diameter
✗ A chord is a line segment from the centre to the circle.
→ A chord joins two points on the circle; a radius joins the centre to a point on the circle.
✓ A chord is a line segment joining any two points on the circle. A diameter is a special chord that passes through the centre.
⚠ Misapplying the angle at centre theorem
✗ The angle subtended by an arc at any point on the circle is half the angle at the centre.
→ This is only true for points on the remaining part of the circle (outside the arc), not for points on the arc itself.
✓ The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
⚠ Assuming all chords equidistant from centre are equal
✗ If two chords are at the same distance from the centre, they are equal in length.
→ This is actually true (Theorem 7), but students often forget to mention that the chords must be in the same circle.
✓ In a given circle, chords that are equidistant from the centre have equal length.
⚠ Incorrectly applying cyclic quadrilateral property
✗ In a cyclic quadrilateral, the sum of all angles is 180°.
→ The sum of all four angles of any quadrilateral is 360°. Only opposite angles sum to 180°.
✓ In a cyclic quadrilateral, the sum of a pair of opposite angles is 180°.
⚠ Forgetting the 'same side' condition for concyclicity
✗ If AB subtends equal angles at C and D, then A, B, C, D are concyclic.
→ The theorem requires that C and D lie on the same side of AB. If they are on opposite sides, the points may not be concyclic.
✓ If a line segment AB subtends equal angles at two other points C and D that lie on the same side of AB, then the four points are concyclic.
Compare & Contrast
Definition and relation to centre
Chord
Diameter
A chord is any line segment joining two points on the circle. A diameter is a chord that passes through the centre, and it is the longest chord.
Size and central angle
Minor Arc
Major Arc
The minor arc subtends an angle less than 180° at the centre, while the major arc subtends an angle greater than 180° at the centre.
Position relative to the triangle
Circumcentre of Acute Triangle
Circumcentre of Obtuse Triangle
For an acute-angled triangle, the circumcentre lies inside the triangle. For an obtuse-angled triangle, the circumcentre lies outside the triangle.
Exam Focus
Marks pattern: 2 MCQs (1 mark each), 1 SA-I (2 marks), 1 SA-II (3 marks), 1 LA (4 marks) — total 10 marks
Most asked: Angle subtended by an arc at the centre and at the circumference · Properties of chords (equal chords, perpendicular from centre) · Cyclic quadrilateral properties and proofs · Circumcircle construction and circumcentre location
Don't skip: Proof-based questions on Theorem 9 (angle at centre is double angle at circumference); Numerical problems on chord length using Pythagoras theorem; Finding angles in cyclic quadrilaterals
One-Line Revision (read aloud before the exam)
  1. A circle is the set of all points in a plane equidistant from a fixed point called the centre.
  2. The radius is the distance from the centre to any point on the circle.
  3. A chord is a line segment joining two points on the circle.
  4. A diameter is a chord passing through the centre and is the longest chord.
  5. Equal chords of a circle subtend equal angles at the centre.
  6. The perpendicular from the centre to a chord bisects the chord.
  7. Chords equidistant from the centre are equal in length.
  8. The longer chord is closer to the centre.
  9. The angle subtended by an arc at the centre is double the angle subtended at any point on the remaining part of the circle.
  10. The angle in a semicircle is 90°.
  11. A cyclic quadrilateral has opposite angles summing to 180°.
  12. If a line segment subtends equal angles at two points on the same side, the four points are concyclic.
  13. The circumcentre of an acute triangle lies inside the triangle; for an obtuse triangle, it lies outside.
  14. For a right-angled triangle, the circumcentre is at the midpoint of the hypotenuse.
  15. Infinitely many circles pass through two given points; their centres lie on the perpendicular bisector of the segment joining them.
  16. A unique circle passes through three non-collinear points.

MCQs — 10 Questions with Answers

Q1. In a circle with centre O, what is the relationship between a chord AB and the radius?

  • A. The chord must always be equal to the radius
  • B. The chord can be any length from 0 to the diameter ✓
  • C. The chord is always less than the radius
  • D. The chord can never equal the diameter

Answer: B — Any chord can have length ranging from a point (0) up to the diameter (2r), depending on where its endpoints lie on the circle.

Q2. Which of the following is NOT a property of a circle?

  • A. All radii of a circle are equal in length
  • B. A circle has only one line of symmetry ✓
  • C. The diameter is the longest chord
  • D. The centre of a circle lies on every diameter

Answer: B — A circle has infinite lines of symmetry (every diameter is a line of symmetry), not just one.

Q3. How many circles can pass through points A and B if they lie 4 cm apart on a plane?

  • A. Exactly one circle with radius 2 cm
  • B. Exactly two circles
  • C. Infinitely many circles ✓
  • D. No circle at all

Answer: C — All centres of circles through A and B lie on the perpendicular bisector of AB, so infinitely many circles can be drawn with different radii.

Q4. Ramesh draws a circle passing through three points P, Q and R. He finds that P, Q and R are collinear. What can you conclude?

  • A. Ramesh has made an error; no circle can pass through three collinear points ✓
  • B. The circle must have its centre on the line through P, Q and R
  • C. Ramesh can draw multiple circles through these three points
  • D. The radius of the circle is half the distance PR

Answer: A — By Theorem 1, a unique circle can only pass through three non-collinear points; collinear points cannot lie on any circle.

Q5. If the perpendicular bisectors of sides AB and AC of triangle ABC intersect at point O, what does point O represent?

  • A. The centroid of the triangle
  • B. The centre of the circumcircle of triangle ABC ✓
  • C. The orthocentre of the triangle
  • D. The incentre of the triangle

Answer: B — O is equidistant from A and B (on perpendicular bisector of AB) and equidistant from A and C (on perpendicular bisector of AC), making it the circumcentre.

Q6. A circle has radius 7 cm. What is the maximum length of a chord in this circle?

  • A. 7 cm
  • B. 14 cm ✓
  • C. 49 cm
  • D. 21 cm

Answer: B — The longest chord is the diameter, which equals 2 × radius = 2 × 7 = 14 cm.

Q7. Two perpendicular bisectors of a triangle's sides meet at point O. How many perpendicular bisectors of the triangle's sides will pass through point O?

  • A. Only two perpendicular bisectors
  • B. All three perpendicular bisectors ✓
  • C. Only one perpendicular bisector
  • D. No perpendicular bisectors

Answer: B — The circumcentre O lies on the perpendicular bisector of each side of the triangle, so all three perpendicular bisectors intersect at O.

Q8. If point P is equidistant from two given points A and B on a plane, on which geometric figure must P lie?

  • A. The line segment AB
  • B. A circle with diameter AB
  • C. The perpendicular bisector of AB ✓
  • D. A line parallel to AB

Answer: C — The perpendicular bisector of AB is precisely the locus of all points equidistant from A and B.

Q9. A circle is drawn with centre O and radius 5 cm. Points M and N lie on the circle such that angle MON = 60°. What can you say about chord MN compared to the radius?

  • A. Chord MN is equal to the radius (5 cm) ✓
  • B. Chord MN is less than the radius
  • C. Chord MN is greater than the radius but less than the diameter
  • D. Chord MN equals the diameter

Answer: A — In triangle OMN, OM = ON = 5 cm and angle MON = 60°, making it equilateral; therefore MN = 5 cm (equal to the radius).

Q10. Three distinct non-collinear points A, B and C are marked on a plane. You are asked to find a circle passing through all three. Which two perpendicular bisectors must you draw to locate the centre uniquely?

  • A. Perpendicular bisectors of any two sides of triangle ABC will work ✓
  • B. Only the perpendicular bisector of side AB will determine the centre
  • C. All three perpendicular bisectors must be drawn to locate the centre
  • D. The perpendicular bisectors of sides AB and BC must be drawn

Answer: A — Any two perpendicular bisectors of the triangle's sides will intersect at the unique circumcentre; using perpendicular bisectors of AB and AC, or AB and BC, or AC and BC all work equivalently.

Flashcards

What is a circle?

A circle is the set of all points on a plane that are equidistant from a fixed point called the centre.

Define the radius of a circle.

The radius is the distance from the centre of the circle to any point on the circle.

What is a chord?

A chord is a line segment joining any two points on the circle.

What is a diameter?

A diameter is a chord that passes through the centre of the circle, and it is the longest chord.

How many circles pass through two given points on a plane?

Infinitely many circles pass through two points, with their centres lying on the perpendicular bisector of the line segment joining those two points.

State the relationship between diameter and radius.

The diameter is twice the radius: d = 2r.

Theorem: How many circles pass through three non-collinear points?

There is a unique circle passing through any three non-collinear points, called the circumcircle of the triangle.

Why does a circle have no straight line of symmetry?

A circle has infinite lines of symmetry; every diameter is a line of reflection symmetry.

What is the locus of points equidistant from two given points A and B?

The perpendicular bisector of the line segment AB is the locus of all points equidistant from A and B.

Can three collinear points determine a circle?

No, three collinear points cannot determine a circle because a circle cannot pass through three points that lie on a straight line.

Important Board Questions

Define a circle. What is the relationship between radius and diameter? [2 marks]

State the definition as the set of equidistant points from centre. Then state that diameter d = 2r, where r is radius.

Explain why infinitely many circles can pass through two given points on a plane. On which line do the centres of all such circles lie? [3 marks]

Use the property that if a circle passes through two points, its centre is equidistant from both points. State that all such equidistant points lie on the perpendicular bisector of the line segment joining the two points.

State and prove Theorem 1: There is a unique circle passing through three non-collinear points. Explain why collinear points cannot lie on a circle and construct the circumcircle of a given triangle. [5 marks]

Prove by showing the centre O must lie on perpendicular bisectors of AB and AC simultaneously, intersecting at unique point. Explain collinear points would require perpendicular bisectors to be parallel (never meeting). Construction: draw perpendicular bisectors of any two sides; intersection point is circumcentre O; radius is distance OA.

Assertion-Reason Questions for I'm Up and Down, and Round and Round

Q1. Assertion (A):

A circle can be defined as the locus of points equidistant from a given point called the centre.

Reason (R):

The distance from the centre to any point on the circle is constant and is called the radius.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both the assertion and reason are true, and the reason correctly explains why the assertion is valid. The definition of a circle relies on all points being equidistant from the centre, and this constant distance is the radius. The reason directly supports the assertion.

Q2. Assertion (A):

For any two given points A and B, there are infinitely many circles passing through both points.

Reason (R):

The centres of all circles passing through A and B lie on the perpendicular bisector of segment AB.

Answer: B — Both A and R are true, but R is NOT the correct explanation of A — Both statements are true. Every point on the perpendicular bisector of AB is equidistant from A and B, so infinitely many circles can be drawn with centres on this line. However, the reason does not explain WHY there are infinitely many circles—it only describes WHERE the centres are located. The reason is true but not the correct explanation of the assertion itself; the true reason is that the perpendicular bisector is an infinite line with infinitely many points.

Q3. Assertion (A):

Three collinear points cannot be the vertices of a circle.

Reason (R):

If three points are collinear, the perpendicular bisectors of the segments formed cannot intersect at a single point to define a unique centre.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both the assertion and reason are true. When three points are collinear, the perpendicular bisectors of AB and AC are parallel lines (both perpendicular to the same line), so they never intersect. Without a unique intersection point, there is no unique centre, and thus no circle can pass through all three collinear points. The reason correctly explains the assertion.

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Case-Based Questions (CBQ) for I'm Up and Down, and Round and Round

Case Study:

Raj is planning to construct a circular garden that must be accessible from three existing pathways at points A, B, and C on the ground. The three pathways are arranged such that no two are parallel and they do not all meet at a single point. Raj needs to find the exact location of the garden's centre so that the circular fence is equidistant from starting points of all three pathways.

Q1. What is the first step Raj must take to find the centre of the circular garden?

  • A. Draw the perpendicular bisectors of segments AB and AC ✓
  • B. Find the midpoint of triangle ABC
  • C. Draw a circle with AB as diameter
  • D. Locate the average distance from all three points

Answer: A

Q2. Why must Raj find the perpendicular bisectors of the segments between the three pathways?

  • A. To determine the angle at each pathway
  • B. Because every point on the perpendicular bisector is equidistant from the two endpoints ✓
  • C. To measure the total perimeter of the garden
  • D. To find the shortest path between pathways

Answer: B

Q3. After drawing perpendicular bisectors of AB and AC, where will the centre of the circular garden be located?

  • A. At any point on either perpendicular bisector
  • B. At the intersection of the two perpendicular bisectors ✓
  • C. At the midpoint of the longest pathway
  • D. At the centroid of triangle ABC

Answer: B

Q4. Why is it guaranteed that the perpendicular bisectors of AB and AC will intersect at exactly one point?

  • A. Because A, B, and C are not collinear, so the perpendicular bisectors are not parallel ✓
  • B. Because the perpendicular bisectors are always shorter than the pathways
  • C. Because Raj has only three pathways, not four
  • D. Because the circle must have a positive radius

Answer: A

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