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MathematicsClass 9I’m Up and Down, and Round and Round

I’m Up and Down, and Round and Round | Class 9 Mathematics Notes

By ConceptScroll Team · Published on 17 July 2026 · 4 min read

I’m Up and Down, and Round and Round – this guide gives you a concise, exam-ready overview of I’m Up and Down, and Round and Round from Class 9 Mathematics, written by ConceptScroll editors and reviewed against the latest NCERT textbook.

5.3 HOW MANY CIRCLES?

This section investigates the number of circles passing through two or three given points on a plane. For two points A and B, infinitely many circles can pass through them because the centres of such circles lie on the perpendicular bisector of segment AB. The smallest such circle has the midpoint of AB as its centre, with radius half the length of AB, making AB the diameter. For three points A, B, and C, if they are collinear (lie on a straight line), no circle passes through all three. However, if they are non-collinear, there exists a unique circle passing through them, called the circumcircle. The centre of this circle, the circumcentre, is the intersection point of the perpendicular bisectors of AB and AC. The circumcentre's position varies with the triangle type: inside for acute, outside for obtuse, and at the midpoint of the hypotenuse for right-angled triangles.

📊 Diagram: Fig. 5.4 illustrates circles passing through two points with centres on the perpendicular bisector. Fig. 5.5 shows the circumcircle of triangle ABC with circumcentre O inside the triangle. Additional figures (5.6 and 5.7) depict circumcentres outside obtuse triangles and at the hypotenuse midpoint in right-angled triangles.

🧪 Activity: Students are encouraged to explore how many circles pass through two points and to consider the radius variations as the centre moves along the perpendicular bisector.

🔗 Connection: This understanding of circle construction sets the stage for exploring properties of chords and angles subtended by them.

Frequently asked questions

1. Draw ΔABC with AB = 5 cm, ∠A = 70° and ∠B = 60°. Draw the circumcircle of ΔABC. Is the centre inside or outside the triangle? 2. Draw ΔABC with AB = 5 cm, ∠A = 100°, AC = 4 cm. Draw the circumcircle of ΔABC. Is the centre inside or outside the triangle? 3. Draw ΔABC, with AB = 6 cm, BC = 7 cm and CA = 7 cm. Draw the circumcircle of ΔABC. Let the circumcentre be O. Measure OA, OB, OC. 4. What is the least possible radius of a circle through two points A and B?

1. To draw ΔABC with AB = 5 cm, ∠A = 70°, and ∠B = 60°:

  • Draw a line segment AB = 5 cm.
  • At point A, construct an angle of 70°.
  • At point B, construct an angle of 60°.
  • The intersection of these two rays is point C.
  • Draw the circumcircle by finding the perpendicular bisectors of at least two sides and locating their intersection (circumcentre).
  • Since both angles are acute and sum less than 180°, the circumcentre lies inside the triangle.

2. To draw ΔABC with AB = 5 cm, ∠A = 100°, and AC

Think, Draw and Infer 1. A, B and C are three collinear points. Can you find a point P such that PA = PB = PC? What can you say about the perpendicular bisectors of AB and BC? Draw and check. Can you show that for three collinear points A, B and C, the perpendicular bisector of AB and BC are parallel? Is it possible for a circle to pass through collinear points? Can you draw a line that cuts a given circle in three distinct points? 2. The circumcircle of a given ΔABC is drawn. Can there be other triangles congruent to ΔABC that share the same circumcircle?

1. For three collinear points A, B, and C:

  • It is not possible to find a point P such that PA = PB = PC unless all three points coincide.
  • The perpendicular bisectors of AB and BC are lines perpendicular to AB and BC at their midpoints.
  • Since A, B, and C are collinear, these perpendicular bisectors are parallel lines.
  • Because the perpendicular bisectors do not intersect, there is no unique circumcentre.
  • Hence, no circle can pass through three collinear points.
  • A line cannot cut a circl
1. Show that the triangle formed by a chord and the centre of the circle is isosceles.

Consider a circle with centre O and a chord AB. Join OA and OB. Since OA and OB are radii of the circle, OA = OB. Therefore, triangle OAB has two equal sides OA and OB, making it an isosceles triangle.

2. Show that if two such isosceles triangles (occurring in the previous question) have equal base length, they are congruent to each other.

Let the two isosceles triangles be OAB and OCD, where O is the centre of the circle, and AB and CD are chords with equal length (AB = CD). Since OA = OB and OC = OD (radii of the circle), and AB = CD (given), by the SSS congruence criterion, triangle OAB is congruent to triangle OCD.

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