Teach Me Maths

Circle Theorems

Question 12 marks

O is the centre. \angle ACB=47^\circ. Find the non-reflex angle AOB in degrees.

Hint

Compare the angles standing on the same arc.

Worked solution

O is the centre. \angle ACB=47^\circ. Find the non-reflex angle AOB in degrees.

  1. The angle at the centre is twice the angle at the circumference standing on the same arc.
  2. \angle AOB=2\times47^\circ=94^\circ.

Question 22 marks

O is the centre. The non-reflex angle AOB is 134^\circ. Find \angle ACB in degrees.

Hint

Identify the angle at the circumference on arc AB.

Worked solution

O is the centre. The non-reflex angle AOB is 134^\circ. Find \angle ACB in degrees.

  1. The angle at the circumference is half the angle at the centre standing on the same arc.
  2. \angle ACB=134^\circ\div2=67^\circ.

Question 31 mark

AC is a diameter. B lies on the circle. Find \angle ABC in degrees.

Hint

Consider the angle standing on a diameter.

Worked solution

AC is a diameter. B lies on the circle. Find \angle ABC in degrees.

  1. An angle in a semicircle is a right angle.
  2. So \angle ABC=90^\circ.

Question 42 marks

ABCD is a cyclic quadrilateral. \angle ABC=68^\circ. Find \angle ADC in degrees.

Hint

Use the sum of the opposite angles of a cyclic quadrilateral.

Worked solution

ABCD is a cyclic quadrilateral. \angle ABC=68^\circ. Find \angle ADC in degrees.

  1. Opposite angles of a cyclic quadrilateral add to 180^\circ.
  2. \angle ADC=180^\circ-68^\circ=112^\circ.

Question 51 mark

C and D are on the same side of chord AB. Which reason proves \angle ACB=\angle ADB?

Hint

Identify the chord on which both angles stand.

Worked solution

C and D are on the same side of chord AB. Which reason proves \angle ACB=\angle ADB?

  1. Opposite angles of a cyclic quadrilateral add to 180^\circ.
  2. Angles in the same segment are equal.
  3. The angle at the centre is half the angle at the circumference.
  4. The angle in a semicircle is a right angle.
  1. Both angles stand on the chord AB, and C and D are on the same side of AB, so they are in the same segment.
  2. Angles in the same segment are equal, so \angle ACB=\angle ADB.
  3. The cyclic quadrilateral fact says opposite angles add to 180^\circ, so it can't show two angles are equal. The third statement is the wrong way round: the angle at the centre is twice the angle at the circumference.

Question 61 mark

PT is tangent to the circle at P. O is the centre. Which statement must be true?

Hint

Use the relationship between a tangent and the radius at contact.

Worked solution

PT is tangent to the circle at P. O is the centre. Which statement must be true?

  1. OP=PT
  2. \angle OPT=45^\circ
  3. \angle OPT=90^\circ
  4. OT=OP
  1. A tangent is perpendicular to the radius at the point of contact.
  2. So \angle OPT=90^\circ.

Question 72 marks

O is the centre and \angle AOB=116^\circ. Find \angle OAB in degrees.

Hint

The two sides from the centre to the circle have equal lengths.

Worked solution

O is the centre and \angle AOB=116^\circ. Find \angle OAB in degrees.

  1. OA=OB because they are radii, so triangle OAB is isosceles.
  2. The two base angles are equal: (180^\circ-116^\circ)\div2=32^\circ.

Question 83 marks

AT is tangent at A and O is the centre. \angle AOB=124^\circ. Find \angle TAB in degrees.

Hint

First find a base angle in the triangle made from two radii.

Worked solution

AT is tangent at A and O is the centre. \angle AOB=124^\circ. Find \angle TAB in degrees.

  1. Triangle OAB is isosceles, so \angle OAB=(180^\circ-124^\circ)\div2=28^\circ.
  2. OA is perpendicular to tangent AT.
  3. \angle TAB=90^\circ-28^\circ=62^\circ.

Question 93 marks

PA and PB are tangents. O is the centre and \angle APB=54^\circ. Find the non-reflex angle AOB in degrees.

Hint

Look at the four angles of quadrilateral OAPB.

Worked solution

PA and PB are tangents. O is the centre and \angle APB=54^\circ. Find the non-reflex angle AOB in degrees.

  1. The radius and tangent meet at a right angle at both A and B.
  2. The angles of quadrilateral OAPB total 360^\circ.
  3. \angle AOB=360^\circ-90^\circ-90^\circ-54^\circ=126^\circ.

Question 102 marks

AC is a diameter and \angle BAC=31^\circ. Find \angle BCA in degrees.

Hint

Use the angle in a semicircle before the triangle angle sum.

Worked solution

AC is a diameter and \angle BAC=31^\circ. Find \angle BCA in degrees.

  1. \angle ABC=90^\circ because AC is a diameter.
  2. \angle BCA=180^\circ-90^\circ-31^\circ=59^\circ.

Question 113 marks

ABCD is cyclic. \angle ABC=(3x+16)^\circ and \angle ADC=(2x+9)^\circ. Find x.

Hint

Form an equation using the two opposite angles.

Worked solution

ABCD is cyclic. \angle ABC=(3x+16)^\circ and \angle ADC=(2x+9)^\circ. Find x.

  1. Opposite angles add to 180^\circ, so 3x+16+2x+9=180.
  2. 5x+25=180, giving 5x=155 and x=31.

Question 123 marks

O is the centre. The non-reflex angle AOB is 124^\circ. C lies on the minor arc AB. Find \angle ACB in degrees.

Hint

The angle at C stands on the arc that does not contain C.

Worked solution

O is the centre. The non-reflex angle AOB is 124^\circ. C lies on the minor arc AB. Find \angle ACB in degrees.

  1. C is on the minor arc AB, so \angle ACB stands on the major arc AB (the arc that does not contain C).
  2. The angle at the centre on the major arc is the reflex angle AOB=360^\circ-124^\circ=236^\circ.
  3. The red major arc does not contain C. It is paired with the reflex centre angle, so the circumference angle uses 236^\circ, not 124^\circ.
    Use the arc that excludes CThe unchanged circle configuration highlights the major arc AB and its reflex central angle 236 degrees. The angle at C stands on that major arc, not the minor arc containing C.Use the arc that excludes CABCO236°θReflex AOB = 360° − 124° = 236°236° = 2θ
    The major arc AB and reflex central angle are highlighted; θ is the angle ACB standing on that arc.
  4. The angle at the centre is twice the angle at the circumference, so \angle ACB=236^\circ\div2=118^\circ.

Question 133 marks

A, B, C and D lie on the circle. \angle ADB=50^\circ and \angle BAC=45^\circ. Find \angle ABC in degrees.

Hint

Transfer an angle on chord AB into triangle ABC.

Worked solution

A, B, C and D lie on the circle. \angle ADB=50^\circ and \angle BAC=45^\circ. Find \angle ABC in degrees.

  1. \angle ACB=50^\circ because angles in the same segment are equal.
  2. In triangle ABC, \angle ABC=180^\circ-50^\circ-45^\circ=85^\circ.

Question 143 marks

AT is tangent at A. \angle BAC=50^\circ and \angle ABC=60^\circ. Find \angle TAB in degrees.

Hint

Find the angle opposite chord AB in triangle ABC.

Worked solution

AT is tangent at A. \angle BAC=50^\circ and \angle ABC=60^\circ. Find \angle TAB in degrees.

  1. Angles in triangle ABC add to 180^\circ: \angle ACB=180^\circ-50^\circ-60^\circ=70^\circ.
  2. Alternate segment theorem: the angle between a tangent and a chord is equal to the angle in the alternate segment. Here the angle between tangent AT and chord AB equals the angle at C standing on AB.
  3. So \angle TAB=\angle ACB=70^\circ.

Question 153 marks

O is the centre. OA=13 cm and chord AB=24 cm. OM is perpendicular to AB. Find OM in cm.

Hint

A perpendicular from the centre bisects a chord.

Worked solution

O is the centre. OA=13 cm and chord AB=24 cm. OM is perpendicular to AB. Find OM in cm.

  1. AM=24\div2=12 cm because the perpendicular from the centre bisects the chord.
  2. Triangle OMA is right angled, so OM^2=13^2-12^2=25.
  3. OM=5 cm.

Question 16Challenge3 marks

A, B and C are points on a circle, centre O. AC is a diameter. \angle OAB=x^\circ.

Find \angle OBC in terms of x.

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

Use the isosceles triangle OAB and the angle in a semicircle.

Worked solution

A, B and C are points on a circle, centre O. AC is a diameter. \angle OAB=x^\circ.

Find \angle OBC in terms of x.

  1. OA=OB because they are radii, so triangle OAB is isosceles and \angle OBA=\angle OAB=x^\circ.
  2. AC is a diameter, so \angle ABC=90^\circ (the angle in a semicircle is a right angle).
  3. The radius BO splits the derived right angle at B into x^\circ and the required angle \theta.
    Split the semicircle angleOA and OB are equal radii. The angle OBA equals the given angle OAB, x. ABC is right angled because AC is a diameter. The remaining angle OBC is theta.Split the semicircle angleABCOxxθ∠ABC = 90°θ = 90° − x
    Ticks mark radii OA and OB. The right angle ABC follows from the diameter; the blue x is derived by isosceles-triangle angles.
  4. \angle OBC=\angle ABC-\angle OBA, so \angle OBC=(90-x)^\circ.

Question 17Challenge4 marks

BT is tangent at B. ABCD is cyclic. \angle BCD=81^\circ and \angle CAD=38^\circ. Find \angle TBC in degrees.

Hint

Use the cyclic quadrilateral to find the whole angle BAD.

Worked solution

BT is tangent at B. ABCD is cyclic. \angle BCD=81^\circ and \angle CAD=38^\circ. Find \angle TBC in degrees.

  1. Opposite angles of a cyclic quadrilateral add to 180^\circ, so \angle BAD=180^\circ-81^\circ=99^\circ.
  2. \angle BAD is split by AC into \angle BAC and \angle CAD, so \angle BAC=99^\circ-38^\circ=61^\circ.
  3. The derived 61^\circ angle at A stands on chord BC. The angle marked \theta between that chord and the tangent is equal to it.
    Carry the angle to the tangentAngle BAC is derived as 99 minus 38 equals 61 degrees. The given CAD is 38 degrees. The tangent-chord angle TBC is theta and is paired with BAC.Carry the angle to the tangentABCDT61°38°θ∠BAC = 99° − 38° = 61°∠TBC = ∠BAC = 61°
    Grey 38° is given; blue 61° is derived. The red tangent angle uses the alternate segment theorem.
  4. Alternate segment theorem: the angle between tangent BT and chord BC equals the angle in the alternate segment, \angle BAC. So \angle TBC=61^\circ.

Question 18Challenge4 marks

PA and PB are tangents to the circle, centre O. C is a point on the major arc AB and D is a point on the minor arc AB. \angle APB=48^\circ.

Find \angle ADB in degrees.

Hint

Find the angle at the centre first. Then use C, and the cyclic quadrilateral ACBD.

Worked solution

PA and PB are tangents to the circle, centre O. C is a point on the major arc AB and D is a point on the minor arc AB. \angle APB=48^\circ.

Find \angle ADB in degrees.

  1. A tangent meets the radius at 90^\circ, so \angle OAP=\angle OBP=90^\circ.
  2. The angles of quadrilateral OAPB add to 360^\circ: \angle AOB=360^\circ-90^\circ-90^\circ-48^\circ=132^\circ.
  3. Joining the radii gives quadrilateral OAPB. Its two tangent-contact angles are right angles, leaving 132^\circ at O.
    1 Find the centre angleQuadrilateral OAPB has right angles at tangent contacts A and B and given angle APB of 48 degrees, so its centre angle AOB is 132 degrees.1. Find the centre angleOABP48°132°360° − 90° − 90° − 48° = 132°
    The first stage isolates the tangent quadrilateral and the derived non-reflex centre angle.
  4. The angle at the centre is twice the angle at the circumference, so \angle ACB=132^\circ\div2=66^\circ.
  5. ACBD is a cyclic quadrilateral, so its opposite angles add to 180^\circ: \angle ADB=180^\circ-66^\circ=114^\circ.
  6. The angle 66^\circ at C and the required angle at D are opposite angles in cyclic quadrilateral ACBD.
    2 Use opposite cyclic anglesOn the same circle, A C B D is a cyclic quadrilateral. The derived angle ACB is 66 degrees and its opposite angle ADB is theta, so theta is 114 degrees.2. Use opposite cyclic anglesABCD66°θθ = 180° − 66° = 114°
    The second stage isolates ACBD in the same configuration; C is on the major arc and D on the minor arc.

Question 19Challenge4 marks

AT is tangent at A. ABCD is cyclic. \angle ABC=(2x+5)^\circ, \angle ADC=(3x+25)^\circ and \angle BAC=75^\circ. Find \angle TAB in degrees.

Hint

Find x using opposite angles, then use triangle ABC.

Worked solution

AT is tangent at A. ABCD is cyclic. \angle ABC=(2x+5)^\circ, \angle ADC=(3x+25)^\circ and \angle BAC=75^\circ. Find \angle TAB in degrees.

  1. Opposite angles of a cyclic quadrilateral add to 180^\circ: (2x+5)+(3x+25)=180, so 5x+30=180 and x=30.
  2. \angle ABC=2\times30+5=65^\circ. Angles in triangle ABC add to 180^\circ: \angle ACB=180^\circ-65^\circ-75^\circ=40^\circ.
  3. After finding x=30, the triangle gives \angle ACB=40^\circ. This is the angle paired with tangent AT and chord AB.
    Find the angle on chord ABIn triangle ABC the given 75 degree angle at A and derived 65 degree angle at B leave 40 degrees at C. The tangent angle TAB equals the angle ACB.Find the angle on chord ABABCDT75°65°40°θx = 30 gives ∠ABC = 65°∠TAB = ∠ACB = 40°
    Grey 75° is given; blue 65° and 40° are derived. The red angle θ is the required tangent angle.
  4. Alternate segment theorem: the angle between tangent AT and chord AB equals the angle in the alternate segment, \angle ACB. So \angle TAB=40^\circ.

Question 20Challenge4 marks

O is the centre and AC is a diameter. DT is tangent at D. \angle OBC=40^\circ and \angle CAD=35^\circ.

(a)

Find \angle BAC in degrees.

2 marks

(b)

Find \angle TDB in degrees.

2 marks

Hint

Start with the isosceles triangle formed by two radii.

Worked solution

O is the centre and AC is a diameter. DT is tangent at D. \angle OBC=40^\circ and \angle CAD=35^\circ.

Part (a)

Find \angle BAC in degrees.

  1. OB=OC, so \angle BOC=180^\circ-2\times40^\circ=100^\circ.
  2. The angle at the circumference on arc BC is half the angle at the centre, so \angle BAC=50^\circ.
  3. The equal radii give 100^\circ at O, then the angle on the same arc BC is 50^\circ at A.
    Centre angle, then circumference angleOB and OC are equal radii. The given 40 degree base angle gives centre angle BOC of 100 degrees, and angle BAC is 50 degrees.Centre angle, then circumference angleABCDTO40°100°50°∠BOC = 180° − 2 × 40° = 100°∠BAC = 100° ÷ 2 = 50°
    Part (a): the derived centre and circumference angles are annotated on the original configuration.

Part (b)

Find \angle TDB in degrees.

  1. \angle BAD=\angle BAC+\angle CAD=50^\circ+35^\circ=85^\circ.
  2. The angles at A combine to 85^\circ. This complete angle BAD is the one paired with tangent DT and chord DB.
    Combine the two angles at AThe derived 50 degree BAC and given 35 degree CAD combine to angle BAD of 85 degrees. The tangent angle TDB is paired with BAD.Combine the two angles at AABCDTO50°35°θ∠BAD = 50° + 35° = 85°∠TDB = ∠BAD = 85°
    Part (b): blue 50° comes from part (a), grey 35° is given, and the red θ marks the target tangent angle.
  3. By the alternate segment theorem, \angle TDB=\angle BAD=85^\circ.

More on this topic: Circle Theorems worksheet with full solutions

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