I’m Up and Down and Round and Round Class 9 Notes and Solutions

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Circles: Complete Visual Guide for Grade 9

Circles: I'm Up and Down, and Round and Round

Have you ever noticed how nature loves round shapes? When raindrops fall on still water, they create perfect circular ripples. The rings inside a tree trunk, the centre of a sunflower, the full moon, and even the sun during an eclipse all look like circles. Humans have been fascinated by this shape for thousands of years, drawing circles in ancient cave paintings long before schools existed.

1. What Is a Circle?

A circle is the set of all points on a flat surface that are the same distance from one fixed point. That fixed point is called the centre, and the fixed distance is called the radius.

O A B Cradius chord diameter
The radius joins the centre to the circle, the chord joins two points on the circle, and the diameter is the longest chord.
  • Chord: A straight line joining any two points on the circle.
  • Diameter: A chord that passes through the centre. It is the longest possible chord.
Imp: Diameter = 2 × Radius

2. Symmetry of a Circle

A circle is perfectly balanced in every direction.

  • Rotational symmetry: Spin the circle around its centre by any angle, and it looks exactly the same.
  • Reflection symmetry: Fold the circle along any diameter, and the two halves match perfectly. This means every diameter acts as a mirror line.

3. How Many Circles Can Pass Through Given Points?

Through Two Points

Pick any two points A and B on a page. You can draw infinitely many circles through them. The centres of all these circles lie on the perpendicular bisector of the line segment AB.

A BK J LPerpendicular bisector The smallest circle occurs when AB is the diameter.
Every possible centre lies on the perpendicular bisector of AB, so infinitely many circles can pass through A and B.

The smallest circle through A and B has AB as its diameter. Its radius is simply AB ÷ 2.

Through Three Points

If the three points lie on a straight line (collinear), no circle can pass through all of them.

If the three points are not collinear, there is exactly one unique circle passing through them. Its centre is the spot where the perpendicular bisectors of the triangle's sides cross.

A B C OPerpendicular bisectors meet at O, the circumcentre.
Three non-collinear points determine exactly one circle. Its centre is the circumcentre of the triangle.

This special circle is called the circumcircle of the triangle, and its centre is the circumcentre.

Where does the circumcentre sit?
→ Acute triangle: Inside the triangle.
→ Obtuse triangle: Outside the triangle.
→ Right triangle: Exactly at the midpoint of the hypotenuse.
Exercise Set 5.1 — Solutions
  1. Solution: Angle C = 180° - 70° - 60° = 50°. All angles are smaller than 90°, so the triangle is acute. The circumcentre lies inside the triangle.
  2. Solution: Angle A = 100°, which is greater than 90°. This is an obtuse triangle, so the circumcentre lies outside the triangle.
  3. Solution: After drawing the perpendicular bisectors, you will find point O. Measuring shows OA = OB = OC because all three are radii of the same circumcircle.
  4. Solution: The least possible radius is when AB acts as the diameter. Radius = AB ÷ 2.
A B C O Acute: circumcentre inside A B C O Obtuse: circumcentre outside A B C O 90° Right: circumcentre on hypotenuse
Answer sketch: where the circumcentre appears for acute, obtuse, and right triangles.

4. Chords and the Angles They Subtend

Theorem 1: Equal chords subtend equal angles at the centre. If two chords in the same circle have the same length, the angles they make at the centre are exactly equal.
Theorem 2: Equal angles at the centre subtend equal chords. If two chords make the same angle at the centre, the chords themselves must be equal in length.
O A B C Dθ θ AB = CD and ∠AOB = ∠COD
Equal chords correspond to equal central angles, and equal central angles correspond to equal chords.
Imp: These two theorems work both ways. Equal chords ↔ Equal central angles.
Exercise Set 5.2 — Solutions
  1. Solution: Join the two ends of any chord to the centre. You get a triangle with two sides as radii. Since radii are equal, the triangle is always isosceles.
  2. Solution: Two isosceles triangles with equal bases (chords) and equal sides (radii) are congruent by the SSS rule.
A B C D O Chord AB Chord CD Each triangle formed by a chord and two radii is isosceles.
Answer sketch: joining chord endpoints to the centre gives isosceles triangles.

5. Perpendicular from the Centre to a Chord

Theorem 3: Line from centre to midpoint of chord is perpendicular to the chord. If you join the centre to the exact middle of any chord, that line will meet the chord at 90°.
Theorem 4 (Converse): Perpendicular from centre bisects the chord. If you drop a perpendicular line from the centre to any chord, it will cut the chord into two equal halves.
O A B MOM ⟂ AB and AM = MB
The line from the centre to the midpoint of a chord is perpendicular to the chord, and the converse is also true.
Exercise Set 5.3 — Solutions
  1. Solution: In triangles OMA and OMB, OA = OB (radii), OM is common, and angles OMA = OMB = 90°. By RHS congruence, the triangles are identical. Hence AM = BM.
  2. Solution: In isosceles triangle ABC with AB = AC, the altitude from A to BC is also the perpendicular bisector of BC. Since the circle's centre lies on the perpendicular bisector of every chord, it must lie on this altitude.
  3. Solution: For the 6 cm chord, half = 3 cm. Distance from centre = √(5² - 3²) = 4 cm. For the 8 cm chord, half = 4 cm. Distance = √(5² - 4²) = 3 cm. Since the chords sit on opposite sides of the centre, total distance between their midpoints = 4 + 3 = 7 cm.
A B C D O M N Chord = 6 cm Chord = 8 cm 4 cm 3 cm Opposite sides of the centre: 4 cm + 3 cm = 7 cm.
Answer sketch for the distance-between-midpoints question in Set 5.3.

6. Distance of Chords from the Centre

Theorem 5: Equal chords are equidistant from the centre. If two chords have the same length, their perpendicular distances from the centre are exactly the same.
Theorem 6: Chords equidistant from the centre are equal. If two chords are the same distance from the centre, they must have the same length.
O A B C Dd d AB = CD ⇒ perpendicular distances from O are equal
Equal chords lie at equal perpendicular distances from the centre. The converse is also true.
Theorem 7: The longer chord is closer to the centre. If one chord is longer than another, then it is nearer to the centre.
Imp: The diameter is the longest chord. Its distance from the centre is zero.
Exercise Set 5.4 — Solutions
  1. Solution: Using Pythagoras: r² = d² + (chord/2)². If two chords are equal, their half-chords are equal. Since r is the same for both, d² must also be the same. So the distances are equal.
  2. Solution: In right triangles CEA and CHF, CE = CH (given), CA = CF (radii), and angles CEA = CHF = 90°. By RHS congruence, AE = FH. Doubling both sides gives AB = GF.
  3. Solution: By Pythagoras: AE² = CA² - CE² and FH² = CF² - CH². Since CA = CF and CE = CH, we get AE = FH, so AB = GF.
A B G F C E H CE CH Equal perpendicular distances from the centre give equal chords.
Answer sketch: CE = CH leads to AB = GF by matching right triangles.
Exercise Set 5.5 — Solutions
  1. Solution: Half chord = √(7² - 6²) = √13. Full chord = 2√13 cm.
  2. Solution: From the right triangle formed by radius r, distance d, and half-chord: (chord/2)² + d² = r². Rearranging gives chord = 2√(r² - d²).
  3. Solution: No. Let the distance of CD be d. Then distance of AB is 2d. Length of CD = 2√(r² - d²) and length of AB = 2√(r² - 4d²). If d > 0, AB is actually shorter than CD, not double. For example, with r = 5 and d = 1, CD = 2√24 and AB = 2√21. Clearly CD ≠ 2AB.
A B M O d r chord/2 (chord/2)² + d² = r²
Answer sketch: the chord formula comes from the right triangle made by radius, distance, and half-chord.

7. Angles Subtended by an Arc

An arc is a curved piece of the circle between two points.

  • Minor arc: The shorter curve (centre angle is less than 180°).
  • Major arc: The longer curve (centre angle is greater than 180°).
Theorem 8: Angle at the centre is twice the angle at the circumference. The angle subtended by an arc at the centre of the circle is double the angle subtended by the same arc at any point on the remaining part of the circle.
O A B D2θ θ Arc AB subtends 2θ at the centre and θ at point D.
The same arc creates a bigger angle at the centre and exactly half of that angle at the circumference.
Example: If arc AB subtends 100° at the centre, it subtends 50° at any point D on the circumference outside the arc.
Corollary 1: Angles in the same segment are equal. All points on the same side of a chord see the chord under exactly the same angle.
Corollary 2: Angle in a semicircle is 90°. Since a semicircle makes an angle of 180° at the centre, the angle at any point on the edge is 180° ÷ 2 = 90°.
Imp: If you stand anywhere on the edge of a semicircle and look at the two ends of the diameter, you always see a perfect right angle.
Exercise Set 5.6 — Solutions
  1. Solution: Triangle AOB has OA = OB = 12 cm and angle AOB = 60°. Since two sides are equal and the angle between them is 60°, all three sides are equal. Therefore, chord AB = 12 cm.
  2. Solution:
    (i) Yes. If X and Y are on opposite sides of chord AB, one angle is θ and the other is 180° - θ.
    (ii) Not always. Equal angles can occur on the same side (same segment) or on opposite sides (supplementary angles).
    (iii) Yes. If two points X and Y on the same side of AB make equal angles, then A, B, X, Y all lie on the same circle.
  3. Solution: ABCD is cyclic. Opposite angles add to 180°. Angle D = 100°, so angle B = x = 180° - 100° = 80°.
A B O 60° OA = OB = AB A B C D 100° 80° Opposite angles add to 180°
Answer sketches for Set 5.6: an equilateral radius-chord case and a cyclic quadrilateral angle pair.

8. Concyclic Points and Cyclic Quadrilaterals

Points that lie on the same circle are called concyclic.

Theorem 9: Equal angles on the same side mean concyclic points. If a line segment AB subtends equal angles at points C and D on the same side of AB, then A, B, C, and D all lie on one circle.
Theorem 10: Opposite angles of a cyclic quadrilateral sum to 180°. In a quadrilateral inscribed in a circle, ∠A + ∠C = 180° and ∠B + ∠D = 180°.
A B C D∠A + ∠C = 180° ∠B + ∠D = 180°
In every cyclic quadrilateral, each pair of opposite angles is supplementary.
Theorem 11 (Converse): Supplementary opposite angles mean cyclic. If the opposite angles of a quadrilateral add up to 180°, the quadrilateral can be inscribed in a circle.
Imp: A rectangle is always cyclic because every angle is 90°, so each opposite pair adds to 180°.

9. End-Chapter Exercises — Full Solutions

Solutions
  1. Solution: r = 13 cm, d = 5 cm. Chord = 2√(13² - 5²) = 2√(169 - 25) = 2√144 = 24 cm.
  2. Solution: Angle at centre = 70°. Angle on circumference = 70° ÷ 2 = 35°.
  3. Solution: Diameter = 26 cm, so r = 13 cm. Chord = 24 cm, half = 12 cm. Distance = √(13² - 12²) = √25 = 5 cm.
  4. Solution: r = 15 cm, d = 9 cm. Chord = 2√(15² - 9²) = 2√144 = 24 cm.
  5. Solution: The centre is equidistant from both ends of any chord. By definition, the perpendicular bisector of a segment contains every point that is equidistant from its ends. Therefore, the centre must lie on the perpendicular bisector of every chord.
  6. Solution: 90°. AB is the diameter, so the arc ACB is a semicircle. The angle subtended by a semicircle at any point on the edge is always a right angle.
  7. Solution: In a cyclic quadrilateral, opposite angles are supplementary. ∠C = 180° - 75° = 105°. ∠D = 180° - 110° = 70°.
  8. Solution: (2x + 10) + (3x - 20) = 180 → 5x - 10 = 180 → 5x = 190 → x = 38°.
    ∠P = 2(38) + 10 = 86°.
    ∠R = 3(38) - 20 = 94°.
  9. Solution: Half chord = 8 cm, distance = 6 cm. Radius = √(8² + 6²) = √100 = 10 cm.
  10. Solution: Semi-perimeter s = (5 + 5 + 12 + 12)/2 = 17. Using Brahmagupta's formula: Area = √((17-5)(17-5)(17-12)(17-12)) = √(12 × 12 × 5 × 5) = 60 square units.
  11. Solution: Draw the perpendicular bisectors of any two sides of the quadrilateral. If their meeting point lies inside the quadrilateral, the circumcentre is inside. If the meeting point lies outside, the circumcentre is outside.
  12. Solution: Let equal chords AB and CD meet at P. By the intersecting-chords rule, AP × PB = CP × PD. Since AB = CD, the matching pieces must be equal: AP = CP and BP = DP (or AP = DP and BP = CP). Thus, the segments of one chord equal the corresponding segments of the other.
  13. Solution: Half chord = 3 cm, distance = 3 cm. Radius = √(3² + 3²) = 3√2 cm. Draw a circle of radius 3√2 cm, draw a line 3 cm from the centre, and mark a 6 cm chord along it.
  14. Solution: In a parallelogram, opposite angles are equal. For it to be cyclic, opposite angles must also sum to 180°. Therefore, each angle must be 90°. A parallelogram with four right angles is a rectangle. Hence, only a rectangle can be inscribed in a circle.
  15. Solution: In a rectangle, diagonals are equal and bisect each other. When inscribed in a circle, each diagonal subtends 90° at the other two vertices, so each diagonal must be a diameter. Therefore, their intersection is the midpoint of both diameters, which is the centre of the circle.
  16. Solution: All chords of fixed length L are at the same distance d = √(r² - (L/2)²) from the centre. Their midpoints all lie on a circle concentric with the original circle, with radius d. The shape is a circle.
  17. Solution: Since chords AB and AC are equal, triangle ABC is isosceles with AB = AC. The perpendicular bisector of BC passes through the centre. In an isosceles triangle, the angle bisector of ∠A is also the perpendicular bisector of BC. Therefore, the centre lies on the angle bisector of ∠BAC.
  18. Solution: Let distances from centre be d₁ (for 10 cm chord) and d₂ (for 24 cm chord). d₁ = √(r² - 5²), d₂ = √(r² - 12²). Given d₁ - d₂ = 7. Also d₁² - d₂² = 119. Factor: (d₁ - d₂)(d₁ + d₂) = 119 → 7(d₁ + d₂) = 119 → d₁ + d₂ = 17. Solving: d₁ = 12, d₂ = 5. Then r = √(12² + 5²) = √169 = 13 cm.
  19. Solution: A regular hexagon divides the circle into 6 equal parts. Each central angle = 60°. The triangle formed by two radii and a side is equilateral. So side length = r. Distance from centre to side (apothem) = r × cos(30°) = r√3/2.
  20. Solution: Since MN is a diameter, ∠MPN = 90° (angle in a semicircle). ∠MOP is the central angle subtended by arc MP, while ∠MNP is the angle at the circumference subtended by the same arc. Therefore, ∠MOP = 2 × ∠MNP.
  21. Solution: Exterior angle CDE = 180° - ∠ADC. Since ABCD is cyclic, ∠ADC + ∠ABC = 180°. Therefore, ∠CDE = 180° - (180° - ∠ABC) = ∠ABC.
  22. Solution: Any chord other than the diameter is at some distance d > 0 from the centre. Its length = 2√(r² - d²), which is always less than 2r (the diameter). Hence, no chord is longer than the diameter.
  23. Solution: For any chord through A, let its distance from the centre be d. Length = 2√(r² - d²). To make the chord shortest, d must be largest. The maximum possible distance for a chord through A is OA itself, which happens when the chord is perpendicular to OA at point A. Therefore, the chord perpendicular to OA is the shortest.
  24. Solution: In the figure, OA = OD = OB = radius. So triangles OAD and OBD are isosceles. Let ∠OAD = ∠ODA = a and ∠OBD = ∠ODB = b. In triangle ABD, the three angles are a, b, and (a + b). Their sum is a + b + (a + b) = 180° → 2(a + b) = 180° → a + b = 90°. Thus ∠ADB = a + b = 90°.
  25. Solution: Since chords CC' and DD' are both perpendicular to diameter AB, they are parallel to each other. The perpendicular from the centre O to any chord bisects it. Because the chords are perpendicular to AB, the perpendicular from O to each chord lies along AB. Hence the midpoints M and M' lie on AB. Therefore, segment MM' lies along the diameter AB.
  26. Solution: Let the base angles of the four isosceles triangles at the centre be p, q, u, v as marked. The total angle sum of the four triangles is 4 × 180° = 720°. Subtracting the 360° around the centre leaves 2(p + q + u + v) = 360°, so p + q + u + v = 180°. Now ∠DAB = p + v and ∠BCD = q + u. Therefore ∠DAB + ∠BCD = p + q + u + v = 180°.
A B O Chord-distance triangle A B P 90° Angle in a semicircle A B C D x 180° - x Cyclic quadrilateral
a chord-distance right triangle, a semicircle angle, and a cyclic quadrilateral angle pair.