I'm Up and Down, and Round and Round Class 9: Circles Notes

I'm Up and Down, and Round and Round is the Class 9 chapter on circles. This guide covers it in order: parts of a circle, symmetry, circles through points, the chord theorems, arcs and angles, and cyclic quadrilaterals. Each theorem comes with a worked example, and the important questions at the end have full model answers.

The circle and its parts

A circle is the set of all points in a plane that are the same distance from a fixed point. That fixed point is the centre, and the fixed distance is the radius. Because every point meets one condition, a circle is also called a locus of points (a set of points that satisfy a given geometric condition).

• Chord: a line segment joining any two points on the circle.
• Diameter: a chord through the centre. It is the longest chord.
• Angle subtended by a chord at the centre: join both ends of the chord to the centre. If A is the centre and BC is a chord, ∠BAC is that angle.

Example: A circle has radius 8 cm. Its longest chord is the diameter = 2 × 8 cm = 16 cm.

Rotation and reflection symmetry

Turn a circle about its centre by any angle (1°, 90°, 180°) and it looks exactly the same. This is complete rotational symmetry, which is why a turning wheel keeps its shape.

A circle also has reflection symmetry across every line through its centre, that is, across every diameter. A square has 4 lines of symmetry and a regular hexagon has 6, but a circle has infinitely many, because infinitely many diameters can be drawn.

How many circles pass through given points?

• One point: infinitely many circles.
• Two points A and B: still infinitely many. All their centres lie on the perpendicular bisector of AB, because every point on it is equidistant from A and B.
• Three non-collinear points: exactly one circle (Theorem 1). It is the circumcircle of the triangle, and its centre is the circumcentre.
• Three collinear points: no circle, because the perpendicular bisectors are parallel and never meet.

Constructing the circumcircle:

1. Take three points A, B and C that are not on one line.
2. Draw the perpendicular bisectors of AB and BC.
3. Their meeting point O is the centre.
4. Draw the circle with centre O and radius OA. It passes through A, B and C.

For a right-angled triangle, the circumcentre is the midpoint of the hypotenuse.

Equal chords and angles at the centre

• Theorem 2: Equal chords subtend equal angles at the centre. If AB = DE (centre C), then ∠ACB = ∠DCE. Proof idea: CA, CB, CD and CE are all radii, so ΔCAB ≅ ΔCDE by SSS.
• Theorem 3 (converse): If two chords subtend equal angles at the centre, the chords are equal. Proof idea: SAS congruence.

Worked example: ∠ACB = 54°, ∠DCE = 54° and AB = 7 cm. The central angles are equal, so by Theorem 3, DE = AB = 7 cm.

The perpendicular from the centre

• Theorem 4: The line from the centre to the midpoint of a chord is perpendicular to the chord. The triangle formed by the centre and the chord is isosceles (two sides are radii), and the median to the base of an isosceles triangle is also its altitude.
• Theorem 5 (converse): The perpendicular from the centre to a chord bisects the chord.

These give a right triangle, so the Baudhāyana–Pythagoras theorem applies. With radius r, distance d from the centre and chord length L:

(L/2)² + d² = r²

Worked example 1: r = 5 cm, d = 3 cm. Let half the chord be x. x² + 3² = 5², so x² + 9 = 25, x² = 16, x = 4 cm. Chord = 2 × 4 = 8 cm.

Worked example 2: A chord of 16 cm lies in a circle of radius 10 cm. How far is it from the centre? Half chord = 8 cm. d² = 10² − 8² = 100 − 64 = 36, so d = 6 cm.

The distance d must be less than the radius for a chord to exist.

Distance of chords from the centre

• Theorem 6: Equal chords are equidistant from the centre.
• Theorem 7 (converse): Chords at the same distance from the centre are equal.
• Theorem 8: The longer the chord, the closer it is to the centre.

The diameter is the longest chord, and its distance from the centre is zero. As a chord moves away from the centre it gets shorter, until at a distance equal to the radius it shrinks to a single point. That is where a tangent would touch the circle.

Example: Chord P is 4 cm from the centre and chord Q is 2 cm from it. Q is closer, so Q is longer.

Angles subtended by arcs

An arc is a connected part of a circle. A minor arc has a central angle less than 180°; a major arc has one greater than 180°.

• Theorem 9: The angle an arc subtends at the centre is double the angle it subtends at any point on the remaining part of the circle.
• Angle in a semicircle: a diameter subtends 180° at the centre, so it subtends 180° ÷ 2 = 90° at any point on the circle.
• Same segment: angles subtended by the same arc at points on the same segment of the circle are equal.

Examples: If ∠AOB = 70° at the centre, the angle at the circle is 70° ÷ 2 = 35°. If an arc makes 45° at the circle, it makes 2 × 45° = 90° at the centre.

Cyclic quadrilaterals

Points on the same circle are concyclic. A quadrilateral with all four vertices on one circle is a cyclic quadrilateral.

• Theorem 10: If a segment joining two points subtends equal angles at two other points on the same side of it, all four points are concyclic.
• Theorem 11: Opposite angles of a cyclic quadrilateral add up to 180°: ∠A + ∠C = 180° and ∠B + ∠D = 180°.
• Theorem 12 (converse): If opposite angles of a quadrilateral add up to 180°, it is cyclic.

Rectangles and squares can be cyclic; a general rhombus is not.

Remember this

• Diameter = 2 × radius, and it is the longest chord.
• Three non-collinear points fix exactly one circle.
• Equal chords ⇔ equal central angles ⇔ equal distances from the centre.
• (L/2)² + d² = r² links chord, distance and radius.
• Angle at centre = 2 × angle at the circle; angle in a semicircle = 90°.
• Opposite angles of a cyclic quadrilateral add up to 180°.

Important questions with answers

1. Why do all circles through A and B have centres on the perpendicular bisector of AB? Each centre must be equidistant from A and B, and every such point lies on the perpendicular bisector of AB.

2. Why can no circle pass through three collinear points? The perpendicular bisectors of the two segments are parallel, so they never meet to give a centre.

3. A circle has radius 13 cm and a chord is 5 cm from the centre. Find the chord. Half chord = √(13² − 5²) = √(169 − 25) = √144 = 12 cm. Chord = 24 cm.

4. A perpendicular from the centre meets a 14 cm chord. Find each part. The perpendicular bisects the chord, so each part is 14 ÷ 2 = 7 cm.

5. Two chords subtend 54° each at the centre and one is 7 cm. Find the other. Equal central angles give equal chords, so it is also 7 cm.

6. In a cyclic quadrilateral ABCD, ∠A = 75°. Find ∠C. ∠C = 180° − 75° = 105°.

7. Can a quadrilateral with opposite angles 100° and 70° be cyclic? No. 100° + 70° = 170°, not 180°.

8. A right-angled triangle has hypotenuse 10 cm. Find the radius of its circumcircle. The circumcentre is the midpoint of the hypotenuse, so the radius is 10 ÷ 2 = 5 cm.

Common mistakes to avoid

• Doubling when you should halve: the angle at the centre is the bigger one.
• Using the full chord in (L/2)² + d² = r² instead of half the chord.
• Assuming any three points fix a circle; they must be non-collinear.
• Calling every parallelogram cyclic; only those with opposite angles adding to 180° are.

When you want to practise these theorems step by step, study this chapter with Joy on Learnijoy.