Sec 3 E-Math Properties of Circles notes

Circle problems are written in precise vocabulary, and each rule names its pieces. A chord joins two points on the circumference; the longest chord is a diameter. An arc is part of the circumference. A sector is the pizza-slice region between two radii and an arc; a segment is the region cut off between a chord and an arc. A tangent touches the circle at exactly one point. When an angle’s arms pass through two circle points, the arc between them subtends the angle.

• Angle in a semicircle: the angle subtended by a diameter is 90°. If AB is a diameter and C sits anywhere on the circle, angle ACB = 90°.
• Tangent ⊥ radius: a tangent meets the radius at the point of contact at 90° — the single most-used fact in tangent questions.
• Centre beats circumference: the angle at the centre is twice the angle at the circumference standing on the same arc. Centre 76° → circumference 38°.
• Same segment: angles in the same segment (standing on the same chord, on the same side) are equal — 55° and 55°.
• Opposite segments: angles in opposite segments are supplementary — for a cyclic quadrilateral, opposite angles sum to 180°.

The circle’s symmetry makes three chord facts and two tangent facts true. Chords: the perpendicular from the centre to a chord bisects the chord (and bisects its arc); the perpendicular bisector of any chord passes through the centre; equal chords sit at equal distances from the centre. Tangents: the two tangents from an external point have equal lengths, and the straight line from the point to the centre bisects the angle between the tangents.

1. 1Chord length and position

   A chord 16 cm long in a circle of radius 10 cm: halve the chord (8 cm), then radius² = 8² + d² gives d = √(100 − 64) = 6 cm from the centre.

2. 2Reverse the question

   A chord 24 cm long at 5 cm from the centre: radius = √(12² + 5²) = √169 = 13 cm.

3. 3Equal chords

   If two chords both measure 24 cm, both sit exactly 5 cm away — equidistance follows from equality.

Drop the perpendicular from the centre O to a chord AB and it bisects AB at M. Triangle OMA is right-angled at M, with OA the radius r, AM the half-chord, and OM the distance d from the centre. The three satisfy r² = (half-chord)² + d² — so any two of {radius, chord length, distance} determine the third.

The same triangle explains the equidistance rule: equal chords have equal half-lengths, so the same r and half-chord force the same d. Equal chords cannot sit at different depths.

Circle examination questions rarely need just one property. The strategy is a relay scan: mark the centre, the tangents, any diameters and any chords; ask which single fact fires first; write that angle down with its reason; then re-scan. Reasons earn their own marks, and the chain is short — usually two to four hops.

1. 1Scan for free facts

   Diameter → right angle ready. Tangent → 90° with radius ready. Centre angle → double available.

2. 2Fire the first fact

   Write the angle with its named reason.

3. 3Re-scan with new numbers

   Triangle sums, isosceles triangles (two radii!) and the new angles unlock the rest.

4. 4Close the loop

   Check totals: opposite cyclic angles should sum to 180°; triangle angles to 180°.

Two radii make an isosceles triangle every time they appear with a chord — spotting OA = OB triangles is the single most profitable habit in this chapter.