I'm Up and Down, and Round and Round - Class 9 Maths Chapter 5 Notes

Section 5.1

Definitions: Circle & Concept of Locus

From the sun and full moon to circular ripples created by raindrops falling on water, human civilisations have been fascinated by circular symmetry. In Odisha's ancient Gudahandi cave paintings, red and black circles are depicted alongside geometric squares and triangles.

Mathematical Definition & Locus

A Circle is the set of all points in a two-dimensional plane that are equidistant from a given fixed point (the Center).
The set of points satisfying a given geometric condition is called a Locus. Thus, a circle is the locus of points equidistant from a fixed center point.

Radius ($ r $)

The constant distance from the center to any point on the circle.

Chord & Diameter

A line segment joining any two points on the circle is a Chord. A chord passing through the center is a Diameter ($ d = 2r $), which is the longest chord.

Section 5.2

Symmetries of a Circle: Reflection & Rotation

What makes circles unique among all geometric shapes is their infinite, perfect symmetry!

Complete Rotational Symmetry

Rotate a circle around its center by any angle whatsoever (e.g., $ 1^\circ, 45^\circ, 180^\circ $), and it looks completely unchanged!

Reflectional Symmetry (Infinitely Many)

If you fold a paper circle along any diameter, the two halves overlap perfectly. Thus, every diameter is a line of reflection symmetry.

What is the Locus of points equidistant from two given points A and B?
The locus of all points equidistant from two fixed points $ A $ and $ B $ is the perpendicular bisector of line segment $ AB $. Every point on this line is at equal distance from $ A $ and $ B $.

Section 5.3

How Many Circles Pass Through Given Points?

Through 1 Point: Infinitely many circles of any size.
Through 2 Points $ A $ and $ B $: Infinitely many circles! The centers of all such circles lie along the perpendicular bisector of $ AB $. The smallest circle has $ AB $ as its diameter (radius = $ \frac{1}{2} AB $).
Through 3 Collinear Points: ZERO circles! The perpendicular bisectors of collinear segments are parallel and never intersect.

Theorem 1: There is a UNIQUE circle passing through three non-collinear points.

The circle passing through the three vertices $ A, B, C $ of a triangle is called its Circumcircle, and its center $ O $ is the Circumcenter (the intersection point of the perpendicular bisectors of the three sides).

Figure 5.1: Location of Circumcenter (O) relative to Acute, Right, and Obtuse Triangles

Solved Conceptual Example 1: Least Radius Through Two Points

Problem: Two points $ A $ and $ B $ are separated by a distance of 10 cm. What is the least possible radius of a circle passing through both $ A $ and $ B $?

Solution:
The smallest circle passing through $ A $ and $ B $ has line segment $ AB $ as its diameter.
\[ \text{Minimum Radius } r_{\text{min}} = \frac{\text{Length of } AB}{2} = \frac{10}{2} = \mathbf{5 \text{ cm}} \]

Practice Exercise 5.1 Draw circumcircles and identify circumcentre locations (inside/outside) for acute and obtuse triangles.

Solve Exercise 5.1

Section 5.4

Chords and the Angles They Subtend

When a chord connects two points on a circle, lines drawn from the center to its endpoints form a central angle.

Theorem 2: Equal Chords $\implies$ Equal Central Angles

Equal chords of a circle subtend equal angles at the center of the circle. (Proof by SSS Congruence: $ \Delta CAB \cong \Delta CDE $).

Theorem 3: Equal Central Angles $\implies$ Equal Chords

If two chords subtend equal angles at the center, the chords are equal in length. (Proof by SAS Congruence).

Practice Exercise 5.2 Prove isosceles property of chords with centre and triangle congruence for equal base chords.

Solve Exercise 5.2

Section 5.5 & 5.6

Perpendiculars, Chord Distances & Baudhāyana-Pythagoras Formula

Theorem 5: The perpendicular from the center of a circle to a chord BISECTS the chord.

Figure 5.2: Perpendicular CM from Center C bisects Chord AB into AM = MB = L/2

The Master Chord Distance Equation

In right triangle $ \Delta CMA $, applying the Baudhāyana-Pythagoras Theorem gives:

$ r^2 = d^2 + \left(\frac{L}{2}\right)^2 \implies \mathbf{L = 2\sqrt{r^2 - d^2}} $

Theorem 8: The Length-Distance Paradox
• Equal Chords are at equal distances from the center ($ AB = FG \iff CE = CH $).
• Unequal Chords: The longer the chord, the closer it is to the center! The diameter is the longest chord and has distance $ d = 0 $ from the center.

Solved Conceptual Example 2: Distance between Parallel Chords

Problem: A circle of radius 5 cm has two parallel chords of lengths 6 cm and 8 cm on opposite sides of the center. Find the distance between the two chords.

Solution:
For Chord 1 ($ L_1 = 6 \text{ cm} $): Half-length $ = 3 \text{ cm} $.
Distance from center $ d_1 = \sqrt{r^2 - 3^2} = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \mathbf{4 \text{ cm}} $.
For Chord 2 ($ L_2 = 8 \text{ cm} $): Half-length $ = 4 \text{ cm} $.
Distance from center $ d_2 = \sqrt{r^2 - 4^2} = \sqrt{5^2 - 4^2} = \sqrt{25 - 16} = \mathbf{3 \text{ cm}} $.
Since chords lie on opposite sides of the center:
\[ \text{Total Distance} = d_1 + d_2 = 4 + 3 = \mathbf{7 \text{ cm}} \]

Practice Exercise 5.3 Perpendicular chord bisectors, altitude in inscribed triangles, and parallel chord distances.

Solve Exercise 5.3

Practice Exercise 5.4 Prove chord equality from equidistant centers using geometric models and the Pythagoras theorem.

Solve Exercise 5.4

Practice Exercise 5.5 Calculate chord lengths and explore non-linear relationships of chord scaling.

Solve Exercise 5.5

Section 5.7

Angles Subtended by an Arc (Inscribed Angle Theorem)

An arc is a connected portion of a circle. The central angle of an arc is the angle swept from its center.

Theorem 9 (Inscribed Angle Theorem):
$ \text{Central Angle } (\angle AOB) = 2 \times \text{Inscribed Angle } (\angle APB) $

Figure 5.3: Central Angle (2θ) is double the Inscribed Angle (θ) subtended by the same Arc AB

Key Corollaries of Theorem 9

Angles in the Same Segment: Angles subtended by the same arc at any points on the remaining circumference are equal ($ \angle AP_1B = \angle AP_2B = \angle AP_3B $).
Angle in a Semicircle: The angle subtended by a diameter at any point on the circle is always $ 90^\circ $ (Right Angle)!

Solved Conceptual Example 3: Inscribed Angle Calculation

Problem: An arc of a circle subtends a central angle $ \angle AOB = 70^\circ $. What is the measure of the angle subtended by this arc at a point $ P $ on the circumference?

Solution:
By Theorem 9: $ \angle APB = \frac{1}{2} \angle AOB $.
\[ \angle APB = \frac{70^\circ}{2} = \mathbf{35^\circ} \]

Section 5.8

Concyclicity & Cyclic Quadrilaterals

Points that lie on the circumference of a single circle are called Concyclic. A quadrilateral whose all 4 vertices lie on a circle is a Cyclic Quadrilateral.

Theorem 11: The sum of opposite angles of a cyclic quadrilateral is $ 180^\circ $.

Figure 5.4: Cyclic Quadrilateral ABCD with Supplementary Opposite Angles

Exterior Angle Property of Cyclic Quadrilaterals
If one side of a cyclic quadrilateral is extended, the exterior angle formed is equal to the interior opposite angle! ($ \angle \text{CDE} = \angle \text{ABC} $).

Solved Conceptual Example 4: Cyclic Quadrilateral Algebra

Problem: A quadrilateral $ PQRS $ is inscribed in a circle. If $ \angle P = (2x + 10)^\circ $ and $ \angle R = (3x - 20)^\circ $, find $ x $ and the measures of $ \angle P $ and $ \angle R $.

Solution:
Since $ PQRS $ is cyclic, opposite angles $ \angle P $ and $ \angle R $ sum to $ 180^\circ $:
\[ (2x + 10) + (3x - 20) = 180 \implies 5x - 10 = 180 \implies 5x = 190 \implies \mathbf{x = 38^\circ} \]
\[ \angle P = 2(38) + 10 = 76 + 10 = \mathbf{86^\circ} \]
\[ \angle R = 3(38) - 20 = 114 - 20 = \mathbf{94^\circ} \]

Practice Exercise 5.6 Interactive NCERT exercise questions on angles subtended by arcs, segments, and cyclic quadrilateral angles.

Solve Exercise 5.6

Interactive Suite

Interactive Explorer Widgets

Q1. Where does the circumcenter of a Right-Angled Triangle lie?

A) Inside the triangle

B) Midpoint of the hypotenuse

C) Outside the triangle

Q2. What is the measure of an angle subtended by a diameter at any point on the circle?

A) 180°

B) 90°

C) 45°

Master Theorem Sheet

Summary of Core Circle Theorems

• Theorem 1: Unique circle (circumcircle) passes through 3 non-collinear points.
• Theorem 2 & 3: Equal chords $ \iff $ Equal central angles.
• Theorem 4 & 5: Perpendicular from center to chord bisects the chord ($ r^2 = d^2 + (L/2)^2 $).
• Theorem 6 & 7: Equal chords $ \iff $ Equidistant from center.
• Theorem 8: Longer chord is closer to the center ($ d = 0 $ for diameter).
• Theorem 9: Central angle = $ 2 \times $ Inscribed angle ($ \angle AOB = 2 \angle APB $).
• Angle in Semicircle: Always $ 90^\circ $.
• Theorem 11 & 12: Opposite angles of cyclic quadrilateral sum to $ 180^\circ $.

Circles — Class 9 Maths Chapter 5 Notes