Circles & Cyclic Quads
Interactive curriculum lessons, worked examples, and geometric problem-solving techniques designed to help Kenyan students master CBC, KPSEA, KCSE, and IGCSE mathematics.
First Principles
Objective: Master the angle properties of circles, chords, tangents, and cyclic quadrilaterals from first geometric principles.
Concrete Scenario: Picture the famous globe roundabout at Nyayo Stadium in Nairobi. Four radial access roads meet the perimeter at points \(A\), \(B\), \(C\), and \(D\). Traffic engineers design sightlines and zebra crossings along chords of the circle. Why do drivers entering at opposite roads have complementary visual sweeping angles adding up to \(180^\circ\)? The geometry of circles dictates these exact relationships.
Geometric Insight: Every point on a circle is equidistant from its center \(O\) by radius \(r\). Because the radii from \(O\) to any two boundary points form an isosceles triangle, we can prove foundational angle theorems:
- Angle at the Centre: The angle subtended by an arc at the centre is twice the angle subtended by the same arc at any point on the circumference: \[ \angle AOB = 2\angle ACB \]
- Angles in the Same Segment: Inscribed angles subtended by the same arc are equal: \[ \angle ACB = \angle ADB \]
- Angle in a Semicircle: A diameter subtends a right angle (\(90^\circ\)) on the circumference.
- Cyclic Quadrilateral: When all four vertices lie on a circle, the opposite angles sum to \(180^\circ\) (\(\angle A + \angle C = 180^\circ\)).
Key Formulas
Diameter: The longest chord passing through circle centre \(O\), exactly twice the radius \(r\).
Chord Length: Length of a chord subtended by central angle \(\theta\).
Inscribed Angle Theorem: An angle subtended at the circumference is half the angle subtended at the centre by the same arc.
Cyclic Quadrilateral Property: Opposite interior angles of an inscribed quadrilateral are supplementary.
Intersecting Chords / Secants Theorem: When two lines intersect at \(P\) (internally or externally) cutting the circle at \(A, B\) and \(C, D\).
Tangent-Secant Theorem: For tangent segment \(PT\) and secant line \(PAB\) drawn from external point \(P\).
Ptolemy's Theorem: In a cyclic quadrilateral, the sum of products of opposite sides equals the product of diagonals.
Worked Examples
- Reason: Form the isosceles triangle \(\triangle OAB\). The two radii \(OA = OB = 15\text{ cm}\).
- Since \(\angle AOB = 60^\circ\), the base angles are \(\frac{180^\circ - 60^\circ}{2} = 60^\circ\), making \(\triangle OAB\) equilateral.
- Alternatively, use the chord formula: \[ L_{\text{chord}} = 2r\sin\left(\frac{\theta}{2}\right) = 2(15)\sin\left(30^\circ\right) = 30 \times 0.5 = 15\text{ cm} \]
Answer: \(15\text{ cm}\)
- Reason: The perpendicular line from the centre to a chord bisects the chord into two equal halves.
- Half-chord length \(= \frac{10}{2} = 5\text{ cm}\).
- Form a right-angled triangle with base \(5\text{ cm}\), height \(6\text{ cm}\), and hypotenuse \(r\).
- Apply Pythagoras' Theorem: \[ r^2 = 5^2 + 6^2 = 25 + 36 = 61 \]
- Solve for \(r\): \[ r = \sqrt{61} \approx 7.8102... \approx 7.81\text{ cm} \]
Answer: \(7.81\text{ cm}\)
- Reason: Since vertices \(A, B, C, D\) all lie on the circle, apply Ptolemy's Theorem: \[ AB \cdot CD + BC \cdot AD = AC \cdot BD \]
- Substitute the known side lengths: \[ (5 \times 8) + (7 \times 6) = AC \times 10 \]
- Compute the products: \[ 40 + 42 = 10 \cdot AC \implies 82 = 10 \cdot AC \]
- Solve for diagonal \(AC\): \[ AC = \frac{82}{10} = 8.2\text{ cm} \]
Answer: \(8.2\text{ cm}\)
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Practice