Circle Theorems

Angle AOB at the centre and angle ACB at the circumference both stand on the arc AB.

1. Angle AOB and angle ACB both stand on the arc AB.
2. The angle at the centre is twice the angle at the circumference.
3. Angle AOB=2\times47
4. Answer: 94^\circ

Both angles are at the circumference and both stand on the same chord AB.

1. Angle ACB and angle ADB are both at the circumference.
2. They both stand on the chord AB, on the same side of it, so they are in the same segment.
3. No angle is at the centre, there is no tangent, and the two angles add up to 82^\circ, not 180^\circ.
4. Answer: angles in the same segment are equal

Angle TAC is between the tangent and the chord AC: look for the angle that the chord AC makes at the circumference on the other side of the chord.

1. Angle TAC is between the tangent AT and the chord AC.
2. Alternate segment theorem: it equals the angle in the other segment, at B, made by the lines from A and C.
3. That is angle ABC.
4. Angle ACB equals angle SAB (they go with the chord AB); angle BAC is what is left on the straight line SAT.
5. Answer: angle ABC

Tangents to a circle from the same point are equal in length, so triangle PAB is isosceles.

1. PA=PB (tangents from a point to a circle are equal in length).
2. So triangle PAB is isosceles and angle PAB= angle PBA.
3. Angle PAB=(180-46)\div2
4. Answer: 67^\circ

First find angle ABC using the angle at the centre, then use the fact that ABCD is a cyclic quadrilateral.

1. Angle AOC and angle ABC stand on the same arc AC.
2. The angle at the centre is twice the angle at the circumference, so angle ABC=136\div 2=68^\circ
3. ABCD is a cyclic quadrilateral, so opposite angles add up to 180^\circ.
4. Angle ADC=180-68=112^\circ

The angle in a semicircle is 90^\circ, so angle ACB is a right angle.

1. Angle ACB=90^\circ (angle in a semicircle).
2. Angles in triangle ABC: 2x+(3x+15)+90=180
3. 5x+105=180
4. 5x=75
5. Answer: x=15

The two angles are not equal: angle AOB is twice angle ACB.

1. Angle AOB is at the centre and angle ACB is at the circumference, on the same arc AB.
2. So angle AOB=2\times angle ACB
3. 5x-4=2(2x+11)
4. 5x-4=4x+22
5. Answer: x=26
6. Check: angle AOB=126^\circ and angle ACB=63^\circ

The perpendicular from the centre to a chord bisects the chord, so triangle OAM is right-angled and AM is half of AB.

1. OA is a radius, so OA=14.5 cm.
2. Triangle OAM has a right angle at M.
3. Pythagoras: AM^2=14.5^2-10^2=210.25-100=110.25
4. AM=\sqrt{110.25}=10.5 cm
5. The perpendicular from the centre to a chord bisects the chord, so AB=2\times AM
6. AB=21 cm

Use the alternate segment theorem to find angle ABC, then use the angles in triangle ABC.

1. Angle ABC= angle SAC=71^\circ (alternate segment theorem).
2. Angles in triangle ABC add up to 180^\circ.
3. Angle BAC=180-71-52
4. Answer: 57^\circ

Angle BAC stands on the same chord as angle BDC: use angles in the same segment, then look at triangle ABX.

1. Angle BAC= angle BDC=48^\circ (angles in the same segment).
2. In triangle ABX: angle AXB=180-35-48
3. Answer: 97^\circ

Opposite angles of a cyclic quadrilateral add up to 180^\circ; multiply out the bracket carefully.

1. Opposite angles of a cyclic quadrilateral add up to 180^\circ.
2. 4(x-5)+(x+50)=180
3. 4x-20+x+50=180
4. 5x+30=180
5. 5x=150
6. Answer: x=30

A tangent meets the radius at 90^\circ, and OT=OA because both are radii.

1. Angle OTP=90^\circ (a tangent is perpendicular to the radius).
2. Angle TOP=180-90-34=56^\circ
3. OT=OA (radii), so triangle OTA is isosceles.
4. Angle OTA=(180-56)\div2=62^\circ
5. Angle ATP=90-62
6. Answer: 28^\circ

Join OA and OB: each tangent meets its radius at 90^\circ, so you can find angle AOB from the quadrilateral OAPB.

1. Angle OAP= angle OBP=90^\circ (a tangent is perpendicular to the radius).
2. Angles in quadrilateral OAPB: angle AOB=360-90-90-54=126^\circ
3. The angle at the centre is twice the angle at the circumference.
4. Angle ACB=126\div2
5. Answer: 63^\circ

Triangle TAB is isosceles, and angle TAB is between a tangent and a chord.

1. TA=TB (tangents from a point are equal), so angle TAB= angle TBA.
2. Angle TAB= angle ACB=x+35 (alternate segment theorem).
3. Angles in triangle TAB: (4x-10)+2(x+35)=180
4. 6x+60=180
5. 6x=120
6. Answer: x=20

C is on the minor arc, so angle ACB is half of the reflex angle AOB.

1. Angle ACB stands on the major arc AB, so it goes with the reflex angle AOB.
2. Reflex angle AOB=2\times107=214^\circ (angle at the centre is twice the angle at the circumference).
3. Angle AOB=360-214=146^\circ
4. OA=OB (radii), so triangle OAB is isosceles.
5. Angle OAB=(180-146)\div2
6. Answer: 17^\circ

Opposite angles of a cyclic quadrilateral add up to 180^\circ, so share 180^\circ in the ratio 2:7.

1. Angle ABC+ angle ADC=180^\circ (opposite angles of a cyclic quadrilateral).
2. 180\div9=20, so angle ABC=40^\circ and angle ADC=140^\circ
3. Angle AOC=2\times40=80^\circ (angle at the centre is twice the angle at the circumference).
4. OA=OC (radii), so angle OCA=(180-80)\div2=50^\circ
5. AD=DC, so angle DCA=(180-140)\div2=20^\circ
6. Angle OCD=50+20
7. Answer: 70^\circ

One equation comes from the angle in a semicircle and another from angles in the same segment; then solve them simultaneously.

x=

1. Angle ABC=90^\circ (angle in a semicircle).
2. Triangle ABC: (x+y)+(2x-28)+90=180, so 3x+y=118
3. Angle BDC= angle BAC (angles in the same segment).
4. 3y+2=x+y, so 2y=x-2
5. Double the first equation: 6x+2y=236
6. Substitute 2y=x-2: 6x+x-2=236
7. 7x=238
8. Answer: x=34

y=

1. From 2y=x-2 with x=34:
2. 2y=32
3. Answer: y=16
4. Check: angle BAC=50^\circ, angle BDC=3\times16+2=50^\circ and angle ACB=40^\circ

Angle CDT is between the tangent and the chord DC: use the alternate segment theorem to find an angle at B, then look for a pair of opposite angles in the cyclic quadrilateral.

Part (a)

1. Angle DBC= angle CDT=2x
2. (Alternate segment theorem.)
3. So angle ABC=(x+10)+2x=3x+10
4. Angles ABC and ADC are opposite angles of the cyclic quadrilateral ABCD, so they add up to 180^\circ.
5. (3x+10)+(5x-30)=180
6. 8x-20=180
7. 8x=200, so x=25

Part (b)

1. Angle ABD=25+10=35^\circ
2. AB=AD, so triangle ABD is isosceles and angle ADB= angle ABD=35^\circ
3. (Base angles of an isosceles triangle are equal.)
4. Angle DAB=180-35-35=110^\circ
5. (Angles in a triangle add up to 180^\circ.)
6. Angle BCD=180-110=70^\circ
7. (Opposite angles of a cyclic quadrilateral add up to 180^\circ.)

Find angle AOB first, then split the isosceles triangle AOB into two right-angled triangles. Answer part (a) before attempting part (b).

Part (a)

1. Angle AOB=2\times38=76^\circ (angle at the centre is twice the angle at the circumference).
2. OA=OB (radii), so the perpendicular from O to AB bisects AB at M and bisects angle AOB.
3. In triangle OAM: AM=4.5 cm and angle AOM=38^\circ
4. \sin38^\circ=\dfrac{4.5}{OA}
5. OA=\dfrac{4.5}{\sin38^\circ}=7.309\ldots
6. Answer: 7.31 cm

Part (b)

1. Area =\frac12 ab\sin C with a=b=OA=7.309\ldots cm and C=76^\circ
2. Area =\frac12\times7.309\ldots^2\times\sin76^\circ
3. =25.92\ldots
4. Answer: 25.9 cm^2

Use the isosceles triangles OAB and OCB to write angle ABC in terms of x, then look for a pair of opposite angles. Answer part (a) before attempting part (b).

Part (a)

1. OA=OB (radii), so angle OBA= angle OAB=2x
2. OC=OB (radii), so angle OBC= angle OCB=x+25
3. So angle ABC=2x+(x+25)=3x+25
4. ABCD is a cyclic quadrilateral, so opposite angles add up to 180^\circ.
5. (3x+25)+(2x+20)=180
6. 5x+45=180, so 5x=135
7. Answer: x=27

Part (b)

1. Angle BOC=180-2(27+25)=76^\circ (angles in the isosceles triangle OBC).
2. Angle BAC is at the circumference on the same arc BC.
3. Angle BAC=76\div2
4. Answer: 38^\circ