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MathematicsClass 9I’m Up and Down, and

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

By ConceptScroll Team · Published on 17 July 2026 · 3 min read

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

I’m Up and Down, and – this guide gives you a concise, exam-ready overview of I’m Up and Down, and 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 given points. For two points A and B, infinitely many circles can pass through them. The centres of all such circles lie on the perpendicular bisector of the segment AB. The smallest circle passing through A and B has its centre at the midpoint of AB, making AB the diameter. For three points, the situation is different. If the three points are collinear, no circle passes through all three. However, if the points are non-collinear, there exists a unique circle passing through them, called the circumcircle. Its centre, the circumcentre, is the intersection of the perpendicular bisectors of the sides of the triangle formed by the three points. The circumcentre's position varies with the type of triangle: inside for acute, outside for obtuse, and at the midpoint of the hypotenuse for right-angled triangles.

📊 Diagram: Fig. 5.4: Circles through two points; Fig. 5.5: The circumcircle of triangle ABC; Fig. 5.6: Obtuse-angled triangle: Circumcentre O is outside the triangle; Fig. 5.7: Right-angled triangle: Circumcentre O is at the midpoint of the hypotenuse

🔗 Connection: This understanding of circles through points sets the stage for studying chords and the angles they subtend.

Frequently asked questions

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 and is isosceles.

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

Let two isosceles triangles OAB and OCD be formed by chords AB and CD respectively, with OA = OB and OC = OD (radii). If AB = CD (equal bases), then by the Side-Side-Side (SSS) congruence criterion, triangles OAB and OCD are congruent because OA = OC (radii), OB = OD (radii), and AB = CD (given).

1. Can you explain why the converse to Theorem 4 is true, i.e., why does the perpendicular from the centre of a circle to a chord of the circle bisect the chord? (Hint: Use Fig. 5.12. You are told that ∠CMA = ∠CMB = 90°. You need to show that AM = BM.)

Solution:

Given: A chord AB of a circle with centre C, and a perpendicular CM from C to AB.

To prove: CM bisects AB, i.e., AM = BM.

Proof: 1. Since CM is perpendicular to AB, ∠CMA = ∠CMB = 90°. 2. In triangles CMA and CMB:

  • CM = CM (common side)
  • CA = CB (radii of the circle)
  • ∠CMA = ∠CMB = 90°

3. By RHS congruence criterion, ΔCMA ≅ ΔCMB. 4. Therefore, corresponding parts are equal, so AM = BM.

Hence, the perpendicular from the centre of a circle to a chord bisects the chord.

2. 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.

Solution:

Given: Triangle ABC is isosceles with AB = AC, inscribed in a circle with centre O.

To prove: The altitude from A to BC passes through the centre O.

Proof: 1. Since AB = AC, triangle ABC is isosceles with vertex A. 2. The altitude from A to BC is also the median and angle bisector of angle A. 3. Let the altitude from A meet BC at D. 4. Since ABC is inscribed in the circle, points A, B, C lie on the circle. 5. The perpendicular bisector of chord BC passes through the centre O. 6. Sin

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