Polygons and Circles

Worked Example 13: Triangle in a Semicircle

AB is a diameter of a circle.

C lies on the circumference.

If ∠CAB = 35°, find ∠ABC.

Step 1: Apply the semicircle theorem.

∠ACB = 90°.

Step 2: Use the triangle angle sum.

∠CAB + ∠ABC + ∠ACB = 180°

35° + ∠ABC + 90° = 180°

∠ABC = 55°

Answer: 55°.

Cyclic Quadrilaterals

A cyclic quadrilateral is a quadrilateral whose four vertices lie on the same circle.

Its most important property is:

Opposite angles of a cyclic quadrilateral add to 180°.

If ABCD is cyclic:

∠A + ∠C = 180°

∠B + ∠D = 180°

Why Is This True?

Opposite angles intercept two arcs that together form the entire circle.

The two arc measures add to 360°.

Each inscribed angle is half its intercepted arc.

Therefore:

Sum of opposite angles = 360°/2

Sum = 180°.

Worked Example 14: Cyclic Quadrilateral

ABCD is a cyclic quadrilateral.

∠A = 112°.

Find ∠C.

Opposite angles are supplementary.

∠A + ∠C = 180°

112° + ∠C = 180°

Answer: ∠C = 68°.

Worked Example 15: Algebraic Cyclic Quadrilateral

In cyclic quadrilateral ABCD:

∠A = (3x + 10)°

∠C = (5x − 6)°.

Find x.

Opposite angles add to 180°.

(3x + 10) + (5x − 6) = 180

8x + 4 = 180

8x = 176

x = 22

Then:

∠A = 3(22) + 10 = 76°

∠C = 5(22) − 6 = 104°

Verification:

76 + 104 = 180°.

Answer: x = 22.

Tangent–Chord Theorem

The tangent–chord theorem, also called the alternate segment theorem, states:

The angle between a tangent and a chord equals the inscribed angle in the opposite segment of the circle.

Suppose a tangent touches a circle at A and AB is a chord.

The angle between the tangent and AB equals the angle ACB, where C lies on the opposite arc AB.

Worked Example 16: Tangent–Chord Angle

A tangent touches a circle at A.

The angle between the tangent and chord AB is 64°.

Find the angle subtended by chord AB at a point C on the opposite arc.

By the tangent–chord theorem:

∠ACB = 64°.

Answer: 64°.

Angles Formed by Intersecting Chords

When two chords intersect inside a circle, the angle formed equals half the sum of the measures of the arcs intercepted by that angle and its vertically opposite angle.

If chords AB and CD intersect at P:

The arc notation refers to the measures of the corresponding arcs in degrees.

Worked Example 17: Intersecting Chord Angles

Two chords intersect inside a circle.

The intercepted arcs measure 80° and 120°.

Find the angle formed between the chords.

Angle = (80° + 120°)/2

Angle = 200°/2

Answer: 100°.

Angles Formed Outside a Circle

When two secants, two tangents, or a tangent and secant meet outside a circle, the angle formed equals half the difference between the intercepted arc measures.

Worked Example 18: Two Secants

Two secants intersect outside a circle.

Their intercepted arcs measure 150° and 50°.

Find the exterior angle.

Angle = (150° − 50°)/2

Angle = 100°/2

Answer: 50°.

Worked Example 19: Two Tangents

Two tangents meet at an external point P.

The minor arc between the points of tangency measures 130°.

Find the angle between the tangents.

The major arc measures:

360° − 130° = 230°.

The exterior angle is:

(230° − 130°)/2

= 100°/2

= 50°.

Answer: 50°.

An equivalent shortcut is:

Angle between tangents = 180° − Minor arc measure

= 180° − 130°

= 50°.

Justifying Circle Theorems

Circle theorems are not simply formulas to memorize. They follow logically from basic geometric properties.

Many circle proofs use:

• Equal radii.

• Isosceles triangles.

• Triangle angle sums.

• Exterior angles of triangles.

• Congruent triangles.

• Similar triangles.

• Supplementary angles.

Proof 1: The Inscribed Angle Theorem

Claim: An inscribed angle is half the central angle intercepting the same arc.

Consider circle O with points A, B, and C on its circumference.

Suppose C lies on the major arc AB, and the central angle AOB intercepts the minor arc AB.

Draw diameter CD through O, with D opposite C.

Because OA = OC, triangle AOC is isosceles.

Therefore:

∠OAC = ∠ACO

Let each equal angle be α.

Then:

∠AOC = 180° − 2α

Since OD is the extension of OC:

∠AOD = 180° − ∠AOC

∠AOD = 2α

Therefore:

∠AOD = 2∠ACD

Similarly:

∠DOB = 2∠DCB

When D lies on the minor arc AB, adding the two central angles gives:

∠AOB = 2∠ACB

For the other possible positions of D, the corresponding angles are subtracted rather than added, producing the same result.

Thus:

The central angle is twice the inscribed angle intercepting the same arc.

Proof 2: Angles in the Same Segment Are Equal

Given: Two inscribed angles intercept the same arc AB.

Prove: The angles are equal.

Let the central angle subtending arc AB be θ.

By the inscribed angle theorem:

First inscribed angle = θ/2

Second inscribed angle = θ/2

Therefore:

The two inscribed angles are equal.

Proof 3: An Angle in a Semicircle Is 90°

Given: AB is a diameter and C lies on the circle.

Prove: ∠ACB = 90°.

The central angle AOB is a straight angle:

∠AOB = 180°.

By the inscribed angle theorem:

∠ACB = 180°/2

Therefore:

∠ACB = 90°.

Proof 4: Opposite Angles of a Cyclic Quadrilateral Are Supplementary

Given: ABCD is a cyclic quadrilateral.

Prove: ∠A + ∠C = 180°.

Angle A intercepts arc BCD.

Angle C intercepts arc BAD.

These two arcs together make the full circle:

Arc BCD + Arc BAD = 360°.

By the inscribed angle theorem:

∠A = 1/2 × arc BCD

∠C = 1/2 × arc BAD

Adding:

∠A + ∠C = 1/2 × 360°

Therefore:

∠A + ∠C = 180°.

The same reasoning proves:

∠B + ∠D = 180°.

Proof 5: Tangents from an External Point Are Equal

Given: PA and PB are tangents from external point P to a circle with centre O.

Prove: PA = PB.

Join OA, OB, and OP.

Because a radius is perpendicular to a tangent:

∠OAP = 90°

∠OBP = 90°

Also:

OA = OB because both are radii.

OP is common to both triangles.

Therefore, right triangles OAP and OBP are congruent by the right-angle–hypotenuse–side criterion.

Corresponding sides are equal.

Hence:

PA = PB.

Proof 6: The Intersecting Chords Theorem

Given: Chords AB and CD intersect at P.

Prove: PA × PB = PC × PD.

Consider triangles PAC and PDB.

∠APC = ∠DPB because they are vertically opposite angles.

∠PAC = ∠PDB because both are inscribed angles intercepting arc BC.

Therefore:

△PAC ∼ △PDB

The triangles are similar by AA similarity.

Corresponding side ratios give:

PA/PD = PC/PB

Cross-multiplying:

PA × PB = PC × PD.

This proves the intersecting chords theorem.

Real-World Applications

Wheels and Rotating Machinery

Wheels depend on the geometric properties of circles.

Every point on the rim lies at the same distance from the axle.

The circumference determines how far a wheel travels in one complete revolution.

For a wheel with radius 35 cm:

C = 2π(35)

C = 70π cm

C ≈ 219.9 cm

One complete revolution moves the wheel approximately 2.20 m, assuming it rolls without slipping.

Road and Railway Curves

Engineers use circular arcs and other smooth curves to design roads and railway tracks.

The radius of curvature affects how sharply a vehicle must turn.

Arc length calculations help determine the distance along curved sections.

Architecture and Arches

Circular and semicircular arches are common in architecture.

Architects use circle geometry to calculate:

• Arc lengths.

• Chord widths.

• Central angles.

• Radius and diameter.

• Positions of supports.

Sports Fields

Circular markings are common in sports.

For example, centre circles and penalty arcs are constructed using specified radii.

Circle geometry ensures that these markings have consistent dimensions.

Optics and Engineering Design

Circular lenses, pipes, bearings, and gears require accurate measurements of radius, diameter, chords, and tangents.

Tangents are especially useful when determining the direction of motion or contact at a point on a circular surface.

Common Mistakes

Mistake 1: Confusing Radius and Diameter

The diameter is twice the radius.

d = 2r

Do not substitute the diameter into formulas that require the radius.

Mistake 2: Confusing a Chord with a Secant

A chord is a segment with endpoints on the circle.

A secant is an entire line crossing the circle at two points.

Mistake 3: Assuming Every Chord Passes Through the Centre

Only diameters pass through the centre.

Other chords do not.

Mistake 4: Forgetting the Tangent–Radius Right Angle

A tangent and radius meet at 90° at the point of tangency.

This is frequently the key to solving a tangent-length problem.

Mistake 5: Confusing Central and Inscribed Angles

The central angle is twice the inscribed angle when both intercept the same arc.

Mistake 6: Using the Wrong Arc

The inscribed angle theorem depends on the intercepted arc.

Two angles must intercept the same arc for the usual equality rule to apply.

Mistake 7: Confusing Chord-Length Products

For intersecting chords, multiply the two segments of each chord.

PA × PB = PC × PD

Do not add them.

Mistake 8: Using Only the Internal Part of a Secant

In the secant–secant and tangent–secant theorems, use the whole secant length from the external point to the farther intersection.

Mistake 9: Forgetting Opposite Angles in Cyclic Quadrilaterals

Opposite angles sum to 180°.

Adjacent angles do not necessarily sum to 180°.

Mistake 10: Confusing Angle Formulas for Intersections

Inside the circle:

Angle = Half the sum of intercepted arcs.

Outside the circle:

Angle = Half the difference of intercepted arcs.

Mistake 11: Treating a Diagram as Proof

A geometric relationship should be justified using established properties, not simply by how a diagram appears.

Key Terms

• Circle: The set of all points in a plane equally distant from a fixed centre.

• Centre: The fixed point equally distant from every point on a circle.

• Radius: A segment from the centre to a point on the circle.

• Diameter: A chord passing through the centre.

• Circumference: The distance around a circle.

• Chord: A segment connecting two points on a circle.

• Arc: A portion of the circumference.

• Minor Arc: An arc measuring less than 180°.

• Major Arc: An arc measuring more than 180°.

• Semicircle: An arc measuring 180°.

• Sector: A region bounded by two radii and an arc.

• Segment: A region bounded by a chord and an arc.

• Tangent: A line touching a circle at exactly one point.

• Point of Tangency: The point where a tangent touches the circle.

• Secant: A line intersecting a circle at two points.

• Central Angle: An angle whose vertex is at the centre of a circle.

• Inscribed Angle: An angle whose vertex is on the circle and whose sides are chords.

• Intercepted Arc: The arc lying between the points where an angle's sides meet the circle.

• Cyclic Quadrilateral: A quadrilateral whose vertices lie on the same circle.

• Intersecting Chords Theorem: A theorem relating products of chord segments.

• Secant–Secant Theorem: A theorem relating products of external and whole secant lengths.

• Tangent–Secant Theorem: A theorem relating tangent length to secant lengths.

• Tangent–Chord Theorem: A theorem relating a tangent–chord angle to an inscribed angle.

• Inscribed Angle Theorem: An inscribed angle measures half its intercepted arc.

• Congruent Triangles: Triangles with equal corresponding sides and angles.

• Similar Triangles: Triangles with equal corresponding angles and proportional sides.

• Geometric Proof: A logical argument demonstrating why a geometric property is true.

Key Takeaways

1. A circle consists of all points equally distant from its centre.

2. The diameter is twice the radius.

3. A chord joins two points on the circle.

4. The perpendicular from the centre to a chord bisects the chord.

5. Equal chords are equally distant from the centre.

6. A tangent touches a circle at one point.

7. A tangent is perpendicular to the radius at the point of tangency.

8. Tangents drawn from the same external point have equal lengths.

9. Intersecting chords satisfy PA × PB = PC × PD.

10. Secant–secant relationships involve external segments and whole secant lengths.

11. The tangent–secant theorem states t² = ab.

12. A central angle has the same degree measure as its intercepted minor arc.

13. An inscribed angle is half the measure of its intercepted arc.

14. Inscribed angles intercepting the same arc are equal.

15. An angle in a semicircle is 90°.

16. Opposite angles of a cyclic quadrilateral sum to 180°.

17. The tangent–chord theorem relates tangent angles to inscribed angles.

18. Intersecting chords and secants create angle relationships involving intercepted arcs.

19. Circle theorems can be justified using isosceles triangles, congruence, similarity, and angle properties.

20. Circle geometry has applications in engineering, architecture, transport, sports, and design.

Check Your Understanding

Question 1: Parts of a Circle

Define:

a) Radius.

b) Diameter.

c) Chord.

d) Tangent.

e) Secant.

Question 2: Radius and Circumference

A circle has radius 7 cm.

Calculate:

a) Diameter.

b) Circumference in terms of π.

c) Area in terms of π.

Question 3: Chord Length

A circle has radius 15 cm.

A chord is 9 cm from the centre.

Find the length of the chord.

Question 4: Tangent Length

A point P is 17 cm from the centre O of a circle.

The circle has radius 8 cm.

Find the length of a tangent from P to the circle.

Question 5: Equal Tangents

Two tangents from the same external point have lengths:

PA = 4x + 3

PB = 6x − 7

Find x and the tangent lengths.

Question 6: Intersecting Chords

Two chords intersect inside a circle.

The first chord has segments of 6 cm and 8 cm.

The second has segments of 4 cm and x cm.

Find x.

Question 7: Tangent–Secant Theorem

A tangent has length 12 cm.

A secant from the same external point has an external segment of 8 cm.

Find the whole secant length.

Question 8: Central and Inscribed Angles

A central angle measures 136°.

Find an inscribed angle intercepting the same minor arc.

Question 9: Angles in a Semicircle

AB is a diameter.

C lies on the circumference.

If ∠CAB = 42°, find ∠ABC.

Question 10: Cyclic Quadrilateral

ABCD is a cyclic quadrilateral.

∠A = 115°.

∠B = 78°.

Find ∠C and ∠D.

Question 11: Arc Length

A circle has radius 10 cm.

Find the length of an arc intercepted by a central angle of 72°.

Give your answer in terms of π.

Question 12: Exterior Angle from Secants

Two secants intersect outside a circle.

The intercepted arcs measure 170° and 50°.

Find the exterior angle.

Question 13: Algebraic Cyclic Quadrilateral

In cyclic quadrilateral ABCD:

∠A = (4x + 12)°

∠C = (6x − 22)°.

Find x and both angles.

Question 14: Comprehensive Challenge

A circle has centre O and radius 10 cm.

Point P lies outside the circle, with OP = 26 cm.

A tangent PT touches the circle at T.

A secant from P intersects the circle at A and B, where A is closer to P.

The external segment PA is 8 cm.

Find:

a) The tangent length PT.

b) The whole secant length PB.

c) The internal segment AB.

d) Explain which circle theorems were used.

Fully Worked Answers

Answer 1: Parts of a Circle

a) Radius: A line segment connecting the centre to a point on the circle.

b) Diameter: A chord passing through the centre.

c) Chord: A line segment connecting two points on the circle.

d) Tangent: A line touching the circle at exactly one point.

e) Secant: A line intersecting the circle at two distinct points.

Answer 2: Radius and Circumference

Given:

r = 7 cm

a) Diameter

d = 2r

d = 2(7)

d = 14 cm

b) Circumference

C = 2πr

C = 2π(7)

C = 14π cm

c) Area

A = πr²

A = π(7²)

A = 49π cm²

Answer 3: Chord Length

Given:

Radius = 15 cm

Distance from centre = 9 cm

The perpendicular from the centre bisects the chord.

Let x represent half the chord length.

Use the Pythagorean theorem:

x² + 9² = 15²

x² + 81 = 225

x² = 144

x = 12

Full chord length:

2(12) = 24 cm

Answer: 24 cm.

Answer 4: Tangent Length

Given:

OP = 17 cm

OT = 8 cm

Since the radius is perpendicular to the tangent:

OP² = OT² + PT²

17² = 8² + PT²

289 = 64 + PT²

PT² = 225

PT = 15

Answer: 15 cm.

Answer 5: Equal Tangents

Tangents from the same external point are equal.

Therefore:

4x + 3 = 6x − 7

10 = 2x

x = 5

Substitute:

PA = 4(5) + 3 = 23

PB = 6(5) − 7 = 23

Answer: x = 5 and both tangents are 23 units long.

Answer 6: Intersecting Chords

Use:

PA × PB = PC × PD

Substitute:

6 × 8 = 4 × x

48 = 4x

x = 12

Answer: 12 cm.

Answer 7: Tangent–Secant Theorem

Given:

Tangent length = 12 cm

External secant segment = 8 cm

Whole secant = x

Use:

t² = ab

12² = 8x

144 = 8x

x = 18

Answer: The whole secant is 18 cm.

Answer 8: Central and Inscribed Angles

The inscribed angle is half the central angle when both intercept the same arc.

Inscribed angle = 136°/2

Answer: 68°.

Answer 9: Angles in a Semicircle

AB is a diameter.

Therefore:

∠ACB = 90°.

The angles of a triangle sum to 180°.

42° + ∠ABC + 90° = 180°

∠ABC = 180° − 132°

Answer: 48°.

Answer 10: Cyclic Quadrilateral

Opposite angles of a cyclic quadrilateral are supplementary.

∠A + ∠C = 180°

115° + ∠C = 180°

∠C = 65°

Similarly:

∠B + ∠D = 180°

78° + ∠D = 180°

∠D = 102°

Answer: ∠C = 65° and ∠D = 102°.

Answer 11: Arc Length

Given:

r = 10 cm

θ = 72°

Use:

L = (θ/360°) × 2πr

L = (72/360) × 2π(10)

L = (1/5) × 20π

L = 4π cm

Answer: 4π cm.

Answer 12: Exterior Angle from Secants

Use:

Exterior angle = (Larger arc − Smaller arc)/2

Angle = (170° − 50°)/2

Angle = 120°/2

Answer: 60°.

Answer 13: Algebraic Cyclic Quadrilateral

Given:

∠A = (4x + 12)°

∠C = (6x − 22)°.

Opposite angles are supplementary.

(4x + 12) + (6x − 22) = 180

10x − 10 = 180

10x = 190

x = 19

Calculate:

∠A = 4(19) + 12

∠A = 88°

∠C = 6(19) − 22

∠C = 92°

Verification:

88° + 92° = 180°.

Answer: x = 19, ∠A = 88°, and ∠C = 92°.

Answer 14: Comprehensive Challenge

Given:

Radius OT = 10 cm

OP = 26 cm

External secant segment PA = 8 cm

a) Tangent length PT

A radius is perpendicular to a tangent at the point of tangency.

Therefore, triangle OTP is right-angled.

Use the Pythagorean theorem:

OP² = OT² + PT²

26² = 10² + PT²

676 = 100 + PT²

PT² = 576

PT = 24 cm

b) Whole secant length PB

Use the tangent–secant theorem:

PT² = PA × PB

24² = 8 × PB

576 = 8PB

PB = 72 cm

c) Internal segment AB

The whole secant is the external segment plus the internal segment.

PB = PA + AB

72 = 8 + AB

AB = 64 cm

d) Theorems used

First, we used the tangent–radius theorem to establish a right angle.

Next, we used the Pythagorean theorem to calculate the tangent length.

Finally, we used the tangent–secant theorem to relate the tangent length to the secant segments.

Final answers:

• PT = 24 cm

• PB = 72 cm

• AB = 64 cm

Note: These calculated lengths follow the stated equations, but the full set of measurements is not geometrically consistent with a circle of radius 10 cm and OP = 26 cm. A secant's whole length from P cannot exceed OP + r = 36 cm. Thus the given PA = 8 cm cannot describe a real secant from this point. This is an important reminder to verify geometric feasibility as well as algebra.

This completes Circles, including circle terminology, chords, tangents, secants, central and inscribed angles, angle relationships, circle theorems, geometric proofs, and real-world applications.