14.1 Circle Geometry

14.1 Circle Geometry

Quick Answer: On the MAT spatial-perception strand you apply circle theorems — especially cyclic quadrilaterals and the angle relationships among tangents, chords, and arcs — with no calculator. The live NBTP Mathematics page currently lists this work under the broader heading properties of shapes (2D and 3D). The April 2015 CETAP booklet still names §3.3 Circle Geometry explicitly. Independent OpenExamPrep practice here trains those booklet theorems; it is not an official NBTP paper.

What the papers actually name

The current Test Content and Mathematics pages on nbt.ac.za describe spatial perception as transformations, properties of shapes (2D and 3D), perimeter/area/volume modelling, and analytic geometry. They do not list “cyclic quadrilateral” or “tangent-chord theorem” by name.

Dr Carol Bohlmann’s April 2015 booklet The National Benchmark Tests: Preparing your learners for the Mathematics (MAT) test still does. Under 3. Spatial perception, after 3.1 geometric objects and 3.2 analytic geometry, 3.3 Circle Geometry names:

• cyclic quadrilaterals
• relationships between tangents, chords, and angles in a circle

The same booklet’s achievement-level table expects Basic writers to recognise the relevant axioms and theorems, Intermediate writers to apply them to typical problems, and Proficient writers to solve more complex circle-geometry problems. That is the spatial content this section teaches. Treat the current web wording as the broader envelope; treat the 2015 §3.3 list as the theorems you must still fire without scaffolding.

MAT items do not cue you with “hence, using the theorem…”. A diagram appears and four numerical or algebraic options follow. You choose the theorem. Calculators stay packed away; numbers are chosen so exact arithmetic — often 30°-60°-90°, isosceles triangles, or 3-4-5 radii — is enough. A formula sheet will not whisper “opposite angles of a cyclic quadrilateral.”

The working kit

Worked applications (original figures)

1. Cyclic quadrilateral

ABCD is cyclic. Angle ABC = 78° and angle BCD = 101°. Opposite angles of a cyclic quadrilateral are supplementary:

• angle ADC = 180° − 78° = 102°
• angle DAB = 180° − 101° = 79°

A popular trap is to treat ABCD as a parallelogram and set opposite angles equal (copying 78° or 101°). Cyclic and parallelogram overlap only in special cases such as a rectangle. Do not import side-parallel habits onto a circle diagram.

If side AD of this cyclic quadrilateral is produced to E, the exterior angle CDE equals the opposite interior angle ABC, so angle CDE = 78°. MAT often parks the needed angle outside the quadrilateral so that writers who only look at the four interior corners miss it.

2. Centre and circumference — choose the arc

O is the centre. Points P and Q lie on the circle. Angle POQ = 124°. Point R lies on the major arc PQ. Angle PRQ stands on minor arc PQ, the same arc as the central angle, so angle PRQ = 124° / 2 = 62°.

If S instead lies on the minor arc, angle PSQ stands on the major arc and equals half of (360° − 124°) = half of 236° = 118°. The paper will not stamp “major” or “minor” on the diagram. You decide from which side of chord PQ the circumference point sits. Drawing radii OP and OQ and marking the reflex central angle when the point is on the minor arc is the whole method.

Same-segment check: if T is another point on the major arc PQ, then angle PTQ = angle PRQ = 62°, because both stand on the same minor arc PQ. A trap option 124° copies the centre; a trap 248° doubles the centre instead of halving it.

3. Diameter

AB is a diameter, C is on the circumference, and angle CAB = 28°. The angle in a semicircle is 90°, so angle ACB = 90°, and angle ABC = 180° − 90° − 28° = 62°.

The trap 28° copies the given angle as if AC = BC (never stated). The trap 56° doubles 28° as if it were a centre-circumference pair. The trap 90° names the right angle but not the angle asked.

4. Two tangents and the central angle

Tangents from T touch the circle at A and B. O is the centre. Angle ATB = 48°. Radii OA and OB are perpendicular to the tangents, so angle OAT = angle OBT = 90°. Quadrilateral OATB has angle sum 360°, hence angle AOB = 360° − 90° − 90° − 48° = 132°.

Triangles OAT and OBT are congruent: right angles at A and B, equal tangents TA = TB, equal radii OA = OB, shared OT. The two angles at O are equal, each 66°, and 66° + 66° = 132° is a useful check. A trap 48° copies the given angle; a trap 96° doubles it as if T were on the circle.

5. Alternate segment

A tangent at A meets chord AB. The angle between the tangent and AB is 37°. C is a point on the circumference on the side of AB away from that tangent angle (the alternate segment). Then angle ACB = 37°.

If D is in the same segment as the tangent angle, angle ADB is not 37°. Sketch the tangent as a line, draw chord AB, and shade the segment on the other side of the chord before you copy 37° onto a circumference angle.

6. Perpendicular from the centre to a chord (3-4-5)

The radius is 5 and chord AB has length 8. M is the foot of the perpendicular from O to AB. The perpendicular from the centre to a chord bisects the chord, so AM = 4. Then triangle OMA is right-angled at M:

OM = √(5² − 4²) = √(25 − 16) = √9 = 3.

That is a radius-half-chord-distance triple. MAT also likes 5-12-13 and 6-8-10. You are not being asked for π or a decimal square root; you are being asked to see a right triangle hiding in the circle. Equal chords are equidistant from the centre — the converse of the same picture.

Exam-day method

1. Mark every given equal length, right angle, and shared arc on the diagram, even though the answer is multiple choice.
2. Name the arc an angle stands on before you double or halve.
3. For cyclic quads, hunt the pair of opposite angles, then check whether an exterior angle is the one the stem actually asks.
4. For tangents, draw the two radii to the contact points first; the right angles appear.
5. If the arithmetic looks ugly, you probably used the wrong arc or treated a circumference angle as a central one.

Independent OpenExamPrep checks below use original values. They are not copies of NBTP exemplar items.