MathematicsExam code: 0580

Circle Theorems

Geometry

PractiseFlashcards

The angle at the centre is twice the angle at the circumference

  • Take a pair of points on the circumference and join each of them to the centre, and also to some third point on the circumference
  • Whatever angle appears at the centre is double the one appearing at that third point
  • The shape often looks like an arrowhead, and the two angles must be subtended by the same arc
The same theorem in a less obvious arrangement: P and Q are joined to the centre and to a third point close to P, so the shape no longer looks like an arrowhead, but the angle at the centre is still twice the angle at the circumference
Source: Angles at centre & circumference by Save My Exams
  • Where the centre angle is reflex, it is that reflex value which doubles the circumference angle, not the smaller one beside it
A circle where the third point sits on the short arc, so the angle at the centre facing the same arc is the reflex one, marked 2x, and it is that reflex angle rather than the smaller one beside it which is double the angle x at the circumference
Source: Angles at centre & circumference by Save My Exams
  • An angle of 33° at the circumference gives 66° at the centre
A circle with P and Q on the circumference and the arc between them picked out in red: both are joined to the centre, giving an angle of 2x there, and both are joined to a third point on the circumference, where the angle is x
Source: Angles at centre & circumference by Save My Exams

Angles in the same segment are equal

  • Two points on the circumference joined to two different third points give two equal angles
  • Both angles must start from the same pair of points, and both third points must lie on the same side of the chord joining them
  • Drawing in that chord makes the segment visible and settles whether the two angles really match
P and Q on a circle, each joined to two different points on the circumference above them, giving two equal angles marked y; the same two points are also joined to a second pair below, giving two more equal angles marked x
Source: Circles & segments by Save My Exams

Opposite angles in a cyclic quadrilateral add to 180°

  • A cyclic quadrilateral has all four of its corners on the circumference
  • Its two pairs of opposite angles each total 180°
  • An angle of 104° therefore sits opposite one of 76°
  • Check all four corners are on the circle before using it, since a quadrilateral with only three on the circumference does not qualify
A quadrilateral ABCD drawn inside a circle with centre O, every one of its four corners sitting on the circumference, which is what makes it cyclic
Source: Cyclic quadrilaterals by Save My Exams

Worked example

The centre angle and a cyclic quadrilateral

A circle has centre O, and A, B and C are points on it. Angle AOC at the centre is 130°, and B is on the major arc. (a) Find angle ABC. (b) D is a fourth point on the circle, on the minor arc AC, so ABCD is a cyclic quadrilateral. Angle BAD is 85°. Find angle BCD.

A circle with centre O marked. Points A, B and C lie on the circumference, with B on the major arc AC. The radii OA and OC are drawn and the angle between them at O, on the side away from B, is marked 130 degrees. The chords AB and BC are drawn, and the angle at B is not labelled. The figure is marked NOT TO SCALE.

Solution:

  • (a) Angle AOC at the centre and angle ABC at the circumference both stand on the same arc AC, so angle AOC is twice angle ABC
  • Angle ABC = 130 ÷ 2 = 65°
  • The reason is that the angle at the centre is twice any angle at the circumference standing on the same arc
  • (b) BAD and BCD are opposite angles of a cyclic quadrilateral, so they add to 180°
  • Angle BCD = 180 − 85 = 95°
  • The reason is "opposite angles of a cyclic quadrilateral add to 180°"
  • Check that all four corners really sit on the circumference before using that second theorem: a quadrilateral with one vertex at the centre, such as OABC, is not cyclic

The alternate segment theorem

  • The angle between a tangent and a chord equals the angle in the alternate segment, that is the angle at the circumference on the other side of that chord
  • Look for a triangle with all three corners on the circle and one corner touching a tangent
  • Find the side that forms the tangent angle, and the matching angle is the one facing it from across the triangle
  • An angle of 58° between tangent and chord matches one of 58° in the alternate segment
  • The name is the reason, so write "alternate segment theorem" in full
A triangle with all three corners on a circle, one corner resting on a tangent: the angle between the tangent and one side of the triangle is marked in red, and so is the equal angle facing it from the opposite corner of the triangle, on the far side of that chord
Source: Alternate segment theorem by Save My Exams

Exam tip

Write each reason as the whole theorem

A circle question asks for a reason in 8 of the 21 papers, and only the complete theorem earns that mark: "opposite angles of a cyclic quadrilateral add to 180°", "the angle in a semicircle is 90°", "alternate segment theorem". Schemes have refused "alternate segment" on its own, "AST" and "alternate angles" for the alternate segment theorem, and "right-angled triangle in a semicircle". "Supplementary" or "add to" in place of "sum to" is fine.

Worked example

Using the alternate segment theorem

Triangle ABC has all three corners on a circle. The straight line PAQ is the tangent at A, with P nearer to B and Q nearer to C. Angle QAC is 59° and angle ACB is 49°. Find angle BAC, giving a reason for each step.

A circle with triangle ABC drawn inside it, its three corners on the circle, A at the top. A straight line PAQ touches the circle at A, with P to the left on the side of B and Q to the right on the side of C. The angle between AQ and the chord AC is marked 59 degrees and the angle ACB at C is marked 49 degrees; angle BAC is not marked. The figure is marked NOT TO SCALE.

Solution:

  • Angle QAC is the angle between the tangent AQ and the chord AC
  • By the alternate segment theorem it equals the angle at the circumference on the other side of AC, which is angle ABC
  • So angle ABC = 59°, and the reason is "alternate segment theorem"
  • The three angles of triangle ABC add to 180°
  • Angle BAC = 180 − 59 − 49 = 72°
  • The second reason is that the angles in a triangle add to 180°
  • Check: the same theorem with the chord AB gives angle PAB = angle ACB = 49°, and the angles on the straight line PAQ add up: 49 + 72 + 59 = 180

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