I'm Up and Down, and Round and Round

Definition of a Circle

A **circle** is the set of all points on a plane that are equidistant from a given point. Alternatively, it is the **locus of points** equidistant from a fixed point called the **centre**.

**Key Components:**

**Centre**: The fixed point from which all points on the circle are equidistant

**Radius**: The constant distance from the centre to any point on the circle. If radius = r, then all points lie at distance r from the centre

**Chord**: A line segment connecting any two points on the circle

**Diameter**: A special chord that passes through the centre. A diameter is the longest chord and equals 2r

**Example**: A bicycle wheel with centre O and all points on the rim at equal distance r from O forms a circle.

**Exam-Important Points**:

The definition emphasizes equidistance as the fundamental property

Locus definition helps in understanding how to locate points satisfying geometric conditions

Every circle is uniquely defined by its centre and radius

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Symmetries of a Circle

**Rotational Symmetry**: A circle possesses **complete rotational symmetry**. Rotating the circle by any angle about its centre results in the same circle. No matter how much you rotate a wheel, it looks identical.

**Reflection Symmetry**: Every **diameter of a circle is a line of reflection symmetry**. If you fold a circular paper along any diameter, the two halves overlap perfectly.

**Key Properties**:

A circle has infinitely many lines of symmetry (all diameters)

A circle has rotational symmetry for all angles

These symmetries make circles appear in nature: ripples in water, cross-sections of stems, moon and sun

**Exam-Important Points**:

Reflection symmetry passes through the centre

Rotational symmetry exists at all angles (0° to 360°)

These symmetries are used to derive many circle theorems

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Circles Through Two Points

**Question**: How many circles can pass through two given points A and B?

**Answer**: **Infinitely many circles** pass through any two points.

**Explanation**:

If a circle passes through both A and B, its centre O must satisfy: OA = OB

All points equidistant from A and B lie on the **perpendicular bisector of AB**

Therefore, every point on the perpendicular bisector of AB can serve as a centre for a circle through A and B

Since the perpendicular bisector is an infinite line, there are infinitely many such circles

**Properties of These Circles**:

**Smallest circle**: Has radius = AB/2 (the midpoint of AB is the centre)

**Largest circle**: No largest circle exists; radii can be arbitrarily large

As you move away from AB along its perpendicular bisector, the radius increases

The circle appears less curved as you move farther from AB

**Example**: Two points A and B are 6 cm apart. The smallest circle through both has radius 3 cm with centre at the midpoint of AB. A circle with centre 4 cm away from AB on its perpendicular bisector would have a larger radius.

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Theorem 1: Unique Circle Through Three Non-Collinear Points

**Statement**: There is a **unique circle passing through three non-collinear points**.

**Conditions**:

The three points A, B, C must be **non-collinear** (not on the same line)

If points are collinear, no circle can pass through all three

**Why This Is True**:

If a circle passes through A, B, C with centre O:

Since OA = OB, point O lies on the perpendicular bisector of AB

Since OA = OC, point O lies on the perpendicular bisector of AC

For non-collinear points, these perpendicular bisectors meet at exactly one point O

Using O as centre and OA as radius gives the unique circle through A, B, C

**Circumcircle and Circumcentre**:

The **circumcentre** is the centre O of the circle through the triangle's vertices

The **circumcircle** is this circle

The **triangle is inscribed** in the circle

**Position of Circumcentre**:

**Acute-angled triangle**: Circumcentre lies **inside** the triangle

**Obtuse-angled triangle**: Circumcentre lies **outside** the triangle

**Right-angled triangle**: Circumcentre lies at the **midpoint of the hypotenuse**

**Example**: For a right triangle with hypotenuse 10 cm, the circumradius = 5 cm, and the circumcentre is at the midpoint of the hypotenuse.

**Exam-Important Points**:

Three collinear points cannot have a circle through them

The perpendicular bisectors of any two sides determine the circumcentre uniquely

The circumcentre's position depends on the triangle's angle type

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Theorem 2: Equal Chords Subtend Equal Angles at Centre

**Statement**: **Equal chords of a circle subtend equal angles at the centre** of the circle.

**Given**: Two chords AB and DE with AB = DE

**To Prove**: ∠ACB = ∠DCE (where C is the centre)

**Proof**:

CA = CB = radius (r)

CD = CE = radius (r)

Therefore: CA = CD and CB = CE

Given: AB = DE

By **SSS congruence**, ∆CAB ≅ ∆CDE

Corresponding angles in congruent triangles are equal

Hence: ∠ACB = ∠DCE

**Example**: Two chords of 6 cm each in a circle of radius 5 cm both subtend the same angle at the centre (approximately 73.7°).

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Theorem 3: Chords Subtending Equal Angles Are Equal

**Statement**: **Chords that subtend equal angles at the centre are equal**.

**Given**: ∠ACB = ∠DCE (where C is the centre)

**To Prove**: AB = DE

**Proof**:

AC = BC = radius

DC = EC = radius

Therefore: AC = DC and BC = EC

Given: ∠ACB = ∠DCE

By **SAS congruence**, ∆ACB ≅ ∆DCE

Corresponding sides in congruent triangles are equal

Hence: AB = DE

**Key Insight**: Theorems 2 and 3 together establish a **two-way relationship**: equal chords ⟺ equal central angles.

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Theorem 4: Perpendicularity of Line from Centre to Chord Midpoint

**Statement**: **The line joining the centre of a circle to the midpoint of a chord is perpendicular to the chord**.

**Given**: Circle with centre C; AB is a chord; M is the midpoint of AB

**To Prove**: CM ⊥ AB

**Proof**:

Triangle CAB is isosceles (CA = CB = radius)

Since CA = CB, we have ∠CAB = ∠CBA

M is the midpoint of AB, so AM = BM

By **SAS congruence**, ∆CMA ≅ ∆CMB

Therefore: ∠CMA = ∠CMB

Since ∠CMA + ∠CMB = 180° (angles on a line)

Both angles must equal 90°

Hence: CM ⊥ AB

**Diagram Application**: Draw a chord, mark its midpoint, join it to the centre; the resulting line is perpendicular to the chord.

**Example**: In a circle of radius 10 cm, if a chord is 12 cm away from the centre perpendicularly, that perpendicular line bisects the chord into two 8 cm segments (using Pythagoras theorem: √(10² - 6²) = 8).

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Theorem 5: Perpendicular from Centre Bisects the Chord

**Statement**: **The perpendicular from the centre of a circle to a chord bisects the chord**.

This is the **converse** of Theorem 4.

**Given**: CM ⊥ AB (where C is the centre)

**To Prove**: AM = BM (M bisects AB)

**Proof**:

∠CMA = ∠CMB = 90°

CA = CB (radii)

CM = CM (common side)

By **RHS congruence**, ∆CMA ≅ ∆CMB

Corresponding sides are equal

Therefore: AM = BM

**Combined Result**: The line from the centre perpendicular to a chord both **bisects the chord AND is perpendicular to it**.

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Theorem 6: Equal Chords Are Equidistant from Centre

**Statement**: **Chords of a circle having the same length are all at the same distance from the centre**.

**Given**: Chords AB and FG with AB = FG; E and H are midpoints

**To Prove**: CE = CH (distances from centre C are equal)

**Proof (Method 1 - SSS Congruence)**:

CA = CF (both radii)

CB = CG (both radii)

AB = FG (given)

By **SSS congruence**, ∆CAB ≅ ∆CFG

Altitudes from C to the chords in congruent triangles are equal

Therefore: CE = CH

**Proof (Method 2 - RHS Congruence)**:

From Theorem 5: E and H bisect chords AB and FG

AE = FH (half of equal chords)

∠CEA = ∠CHF = 90° (perpendiculars)

CA = CF (radii)

By **RHS congruence**, ∆CEA ≅ ∆CHF

Therefore: CE = CH

**Distance Definition**: Distance from centre to a chord = perpendicular distance from centre to the chord = length of perpendicular from centre to chord.

**Example**: Two chords each 8 cm long in the same circle are both equidistant from the centre, regardless of their positions.

**Exam-Important Points**:

"Same length chords" and "equidistant from centre" are equivalent properties

Distance is always measured perpendicular to the chord

This works for all chords of equal length in the same circle

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Theorem 7: Equidistant Chords Are Equal

**Statement**: **Chords of a circle that are equidistant from the centre have equal length**.

This is the **converse** of Theorem 6.

**Given**: CE = CH (equal distances from centre C)

**To Prove**: AB = FG

**Proof**:

E and H are midpoints of AB and FG respectively (from Theorem 5)

∠CEA = ∠CHF = 90° (perpendiculars)

CE = CH (given)

CA = CF (radii)

By **RHS congruence**, ∆CEA ≅ ∆CHF

Corresponding sides are equal: AE = FH

Since E and H are midpoints: AB = 2·AE and FG = 2·FH

Therefore: AB = FG

**Key Insight**: Theorems 6 and 7 together establish: **equal chords ⟺ equidistant from centre**.

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Theorem 8: Relationship Between Chord Length and Distance from Centre

**Statement**: **Among two chords of a circle, the longer chord is closer to the centre** (i.e., at a smaller distance).

Equivalently: If AB > DE, then distance from C to AB < distance from C to DE.

**Explanation Using Pythagoras Theorem**:

Let AB and DE be chords of circle with centre C and radius r.

Let perpendicular distance from C to AB = d₁

Let perpendicular distance from C to DE = d₂

M and N are midpoints of AB and DE respectively

By Pythagoras theorem:

r² = d₁² + (AB/2)²

r² = d₂² + (DE/2)²

From these:

d₁² = r² - (AB/2)²

d₂² = r² - (DE/2)²

If AB > DE, then (AB/2)² > (DE/2)², which means:

d₁² < d₂²

Therefore: d₁ < d₂

**Conclusion**: Longer chords are closer to the centre; shorter chords are farther from the centre.

**Example**: In a circle of radius 5 cm:

A chord of 8 cm is at distance √(25 - 16) = 3 cm from centre

A chord of 6 cm is at distance √(25 - 9) = 4 cm from centre

The 8 cm chord (longer) is closer to the centre

**Special Case**: The **diameter** (longest possible chord) passes through the centre, making its distance = 0 (closest possible).

**Exam-Important Points**:

This relationship holds for any two unequal chords

Can be proven using Pythagoras theorem directly

Helps determine relative positions of chords

Remember: longer chord = smaller distance from centre

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Key Formulas and Relationships

**Circle Definition Relationship**:

For any point P on circle with centre C and radius r: CP = r

**Chord and Centre Relationships**:

1. Equal chords ⟺ equal central angles (Theorems 2 & 3)

2. Equal chords ⟺ equidistant from centre (Theorems 6 & 7)

3. Longer chord ⟹ closer to centre (Theorem 8)

**Perpendicularity Property**:

Perpendicular from centre to chord bisects the chord (Theorem 4 & 5)

If M is midpoint of chord AB with centre C: CM ⊥ AB

**Pythagoras Application**:

For chord AB with centre C, radius r, perpendicular distance d from C to midpoint M:

r² = d² + (AB/2)²

AB = 2√(r² - d²)

**Three-Point Circle**:

Non-collinear points A, B, C determine unique circumcircle

Centre = intersection of perpendicular bisectors of any two sides

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Common Mistakes to Avoid

1. **Assuming collinear points can lie on a circle**: They cannot. Three collinear points have no circumcircle.

2. **Confusing "distance from centre" with "radius"**: Distance to chord ≠ radius. Distance is perpendicular from centre to chord.

3. **Forgetting equal chords must be in the same circle**: Theorem comparisons require chords to be in the same circle.

4. **Misidentifying circumcentre position**: Must check triangle type:

Acute → inside

Obtuse → outside

Right → on hypotenuse midpoint

5. **Assuming smallest circle through two points has large radius**: The smallest circle has radius = half the distance between points.

6. **Not using perpendicular bisectors correctly**: To find circumcentre, use perpendicular bisectors of **any two sides**, not just one.

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Exam Practice Applications

**Construction Problems**:

Draw circumcircle of given triangle

Locate circumcentre by perpendicular bisectors

Verify circumradius by measurement

**Proof Problems**:

Show perpendicularity using congruence

Prove equal chords have equal angles

Demonstrate equidistance property

**Calculation Problems**:

Find chord length from distance and radius: AB = 2√(r² - d²)

Find distance from chord length: d = √(r² - (AB/2)²)

Verify relationships using Pythagoras theorem

**Logical Reasoning**:

Determine number of circles through given points

Establish positions of circumcentre

Compare distances of different chords