Worked examples of circle theorems become easier when you identify the chord or arc before naming a theorem. An angle answer needs both a number and a reason whose conditions actually hold. A line that looks like a diameter may be only a chord; a line touching a drawing may not have been given as a tangent.
All configurations below are original. Each sketch is accompanied by complete written givens, so its appearance is never the evidence for an angle. Angles are in degrees. The middle letter names the vertex: angle ABC is at B, between BA and BC. All named circle points are distinct unless explicitly stated otherwise.
Match a theorem to its conditions
| Fact to use | Required condition |
|---|---|
| An angle subtended by a diameter is 90°. | The vertex is on the circle; its arms end at the diameter’s endpoints. |
| A central angle is twice a circumference angle. | Both correspond to the same intercepted arc; identify a reflex central angle if needed. |
| Angles in the same segment are equal. | They stand on the same chord, with vertices on the same side of it. |
| Opposite angles of a cyclic quadrilateral total 180°. | All four vertices lie on one circle, in the stated cyclic order. |
| A radius is perpendicular to a tangent. | The radius ends at the point of contact. |
| A tangent–chord angle equals an alternate-segment angle. | The chord starts at the point of contact; select the appropriate tangent ray and opposite segment. |
These standard conditions can be checked against the Australian Mathematical Sciences Institute’s circle geometry reference. In your own working, name the actual letters as well as the theorem. “Same segment, chord AB” is more informative than “circle rule”.
Worked case 1: a diameter starts an angle chain
Givens: O is the centre of a circle. A, O and B lie on a straight line in that order, and A and B are on the circle, so AB is a diameter. C is on the upper semicircle from A to B. Chords AC and BC and radius OC are drawn. Angle CAB = 34°. Find angles ACB, ABC, BOC and OCB, taking BOC to be the non-reflex angle.
| Angle | Value and reason |
|---|---|
| ACB | 90°; AB is a diameter and C lies on its circle. |
| ABC | 180° − 90° − 34° = 56°; angles of triangle ABC. |
| BOC | 2 × 34° = 68°; centre and circumference angles intercept the minor arc BC. |
| OCB | (180° − 68°)/2 = 56°; OB = OC, so triangle BOC is isosceles. |
There are two checks. Triangle ABC has angles 34°, 56° and 90°, totalling 180°. Triangle BOC has 68°, 56° and 56°, also totalling 180°. At B, rays BA and BO coincide because O lies between A and B; this explains why ABC and OBC are the same angle. Equality here is not an extra circle theorem.
Worked case 2: opposite angles, then an exterior angle
Givens: A, B, C and D lie on one circle in that cyclic order, forming a convex quadrilateral. Angle BAD = 72° and angle ABC = 104°. Extend side BC through C to E, so B, C and E are collinear with C between B and E. Find angles BCD, CDA and DCE.
BAD and BCD are opposite angles of cyclic quadrilateral ABCD. Therefore BCD = 180° − 72° = 108°. ABC and CDA form the other opposite pair, so CDA = 180° − 104° = 76°. The interior-angle check is 72° + 104° + 108° + 76° = 360°.
At C, rays CB and CE point in opposite directions. Angles BCD and DCE form a straight angle, giving DCE = 180° − 108° = 72°. This also agrees with the cyclic exterior-angle result: the exterior angle equals the opposite interior angle BAD.
A common wrong step is 180° − 104° for BCD. Those two angles are adjacent, not opposite. Draw a small arrow from the requested vertex to the vertex across the quadrilateral before subtracting. The order of the four circle points matters because it tells you which vertices are opposite.
Worked case 3: choose the correct tangent ray
Givens: O is the centre, and T, U and V lie on the circle. T is its lowest point. TP is a horizontal tangent ray pointing right from T. U is above and right of T, while V lies on the major arc TU. Angle PTU = 58° and angle TUV = 47°. Chords TU, UV and VT and radius OT are drawn. Find TVU, VTU and OTU.
Chord TU starts at the tangent’s point of contact. The angle between TP and TU equals the angle subtended by chord TU at V in the alternate segment. Thus TVU = 58°. In triangle TUV, VTU = 180° − 58° − 47° = 75°.
OT is perpendicular to TP because a radius meets the tangent at a right angle. Ray TU lies between TP and TO, so OTU = 90° − 58° = 32°. The position statement is important: without knowing which rays bound the angle, blindly subtracting from 90° may select the wrong angle.
The tangent ray pointing left would make 180° − 58° = 122° with TU. It does not give a second value for TVU. Changing tangent rays changes the tangent–chord angle and the relevant segment. The word “alternate” is a condition about the geometry, not permission to choose either nearby angle.
Three independent practice tasks with solutions
Practice 1: a fresh diameter
Givens: O is the centre; P, O and Q are collinear in that order; P and Q lie on the circle. R lies on the upper semicircle, and angle PQR = 47°. Find PRQ, QPR and the non-reflex angle POR.
Solution: PRQ = 90° because PQ is a diameter. Triangle PQR gives QPR = 180° − 90° − 47° = 43°. POR = 2 × 47° = 94° because POR and PQR intercept the same minor arc PR. The triangle check is 47° + 90° + 43° = 180°.
Practice 2: an exterior angle at another vertex
Givens: W, X, Y and Z form a convex cyclic quadrilateral in that order. Angle WXY = 112° and XYZ = 67°. Extend WZ through Z to E, with Z between W and E. Find YZW, XWZ and YZE.
Solution: YZW = 180° − 112° = 68° and XWZ = 180° − 67° = 113°, using opposite angles of a cyclic quadrilateral. Because ZW and ZE are opposite rays, YZE = 180° − 68° = 112°. The exterior angle agrees with opposite interior angle WXY.
Practice 3: diagnose an incorrect same-segment claim
Givens: A and B are endpoints of a chord. C lies on the major arc AB and D lies on the minor arc AB, so C and D are on opposite sides of chord AB. All four points lie on one circle, in cyclic order A, D, B, C. Angle ACB = 41°. A learner claims ADB is also 41°. Find its correct value.
Solution: In cyclic quadrilateral ADBC, angles ADB and ACB are opposite. Therefore ADB = 180° − 41° = 139°. The equal-angle theorem fails because the vertices are in opposite segments. The two angles intercept the two different arcs joining the same endpoints.
Use circle relationships to justify lengths
An angle theorem does not by itself give a length. First identify a radius, chord midpoint or tangent that creates a justified right triangle. The circle and right-angle conditions below are givens; they are not inferred from the sketches.
Worked chord case: the perpendicular from the centre bisects the chord
A circle has centre O and radius 13 cm. A chord AB is 5 cm from O, measured along perpendicular OM, with M on AB. Find AB. Since OA = OB and OM is common, the right triangles OMA and OMB have equal hypotenuse and one equal side. They are congruent, so AM = MB.
Pythagoras in triangle OMA gives AM² = OA² − OM² = 13² − 5² = 144. Thus AM = 12 cm and AB = 24 cm. The negative square root is not a length. Check with coordinates O(0,0), M(0,5), A(−12,5), B(12,5): both endpoints have squared distance 144 + 25 = 169 from O. Reporting 12 cm would give only half the chord.
Worked tangent case: name the point of contact
From a point P outside a circle of centre O and radius 7 cm, tangents touch the circle at T and U. OP = 25 cm. Find PT and PU. OT is perpendicular to PT at T, so triangle OTP is right-angled at T and OP is its hypotenuse. Hence PT² = 25² − 7² = 576 and PT = 24 cm.
Triangle OUP is also right-angled, with the same hypotenuse OP and radius OU = OT. Therefore the triangles are congruent and PU = PT = 24 cm. Alternatively apply Pythagoras separately to OUP. Do not subtract 7 directly from 25: OP and OT are different sides of a right triangle, not adjacent parts of one straight segment.
Independent length tasks
- A chord is 16 cm long in a circle of radius 10 cm. Find its perpendicular distance from the centre, and justify using half of 16.
- A tangent PT has length 12 cm, and the radius OT at its contact point is 5 cm. Find OP. State which side is the hypotenuse.
Explained answers
- The perpendicular from the centre bisects the chord, so the half chord is 8 cm. Distance d satisfies d² + 8² = 10², giving d = 6 cm. A second parallel chord may lie on the opposite side of the centre, but its perpendicular distance is still 6 cm.
- OT ⟂ PT because PT is tangent at T. OP is the hypotenuse, so OP = √(5² + 12²) = 13 cm. Adding squared lengths is required here; subtracting would treat the wrong side as hypotenuse.
Review every answer by attaching its relationship: “centre perpendicular bisects chord”, “radius perpendicular to tangent at contact”, “equal radii”, or the right-triangle calculation. A line perpendicular somewhere other than the circle’s contact point does not establish the required tangent.
Frequently asked questions
May I measure an angle on the sketch?
Use given information and theorems for the answer. A sketch supports identification; measurement is at most a rough check.
Does every chord give a right angle?
No. The right-angle result requires a diameter, which passes through the centre.
When are two radii useful?
They create an isosceles triangle with equal base angles. State the centre and the two radius endpoints.
Can the central angle be reflex?
Yes. Match the arc intercepted by the circumference angle; doubling it can give a central angle greater than 180°.
Must opposite cyclic angles both be acute?
No. They add to 180°, so if one is acute the other is obtuse. Both may be right angles.
Is every perpendicular line a tangent?
A line perpendicular to a radius at its circle endpoint is tangent there. A perpendicular elsewhere is insufficient.
What if the givens contradict each other?
Check the labels and conditions. If the stated relations remain inconsistent, explain the contradiction rather than forcing a diagram or answer.
Make your next answer auditable
Write each angle on a separate line with its reason and relevant chord, arc or triangle. For additional resource navigation, visit the existing Mathematics topical revision collection. Member access may be required for its contents; no uninspected geometry file is promised.
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