One to one maths interventions built for KS4 success

Weekly online one to one GCSE maths revision lessons now available

In order to access this I need to be confident with:

Parts of a circle Angles in polygons Angles on a straight line Angles around a point Angles in parallel lines Triangles Congruence and similarityThis topic is relevant for:

Here we will learn about the circle theorems involving *tangents of a circle*, including their application, proof, and using them to solve more difficult problems.

There are also *circle theorem *worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.

A **tangent** of a circle is a straight line that touches the circumference of the circle at only one point.

**The angle between a tangent and radius is 90 degrees**.**Tangents which meet at the same point are equal in length**.

In **diagram 1** above, the tangent meets the circle at point A , which is perpendicular to ( 90 degrees to) the radius of the circle at that point (the point of tangency is at point A ).

In **diagram 2**, two tangents meet the circle at two different points ( B and D ) and they intersect at point A . If the points B and D are linked by a chord, AB and AD are the same length so ABD is an isosceles triangle. If the points B and D join to the centre of the circle C , they form a kite ABCD . This means that we have the two circle theorems:

*Isosceles triangle*

*Kite*

- The
**radius**of a circle is the distance from the**centre**to the**circumference**of the circle. The radius is half of the**diameter**. - The
**centre**of the circle is a point that locates the middle of a circle. - The
**circumference**of the circle is the distance around the edge of the circle.

To be able to prove this theorem, you do not need to know any other circle theorem. You just need to be confident with **angles in a triangle**. You also need to understand **congruence.**

Step | Diagram | Description |

1 | Firstly we take an arbitrary point labelled A (a random point in space) outside of the circle. | |

2 | The point A can be connected to the circle by two tangents. One line touches the circle at B, the other tangent touches the circle at C. We need to prove that the length AB=AC. We do this by using triangles. | |

3 | If we join OA together and then connect OB and OC, we construct two triangles. If we can prove that these two triangles are congruent, then AC will be equal to AB. | |

4 | The angles OBA and OCA are 90 degrees each as tangents meet a circle at 90 degrees. This means we have two right angle triangles. We can also see that as OB and OC are radii of the circle, they must be the same length. | |

5 | Triangles AOB and AOC are both right angles. They share the same side length AO and another side length of each triangle is the same as it is the radius of the circle (OB=OC). This means that the two triangles are congruent and so AC = AB. This means that tangents that meet at the same point are equal in length. |

In order to use the tangent of a circle:

**Locate the key parts of the circle for the theorem.****Use other angle facts to determine the remaining angle(s) made with the tangent.****Use the tangent theorem to state the other missing angle.**

Get your free Tangent of a circle worksheet of 20+ questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREEGet your free Tangent of a circle worksheet of 20+ questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREETangent of a circle is one of 7 circle theorems you will need to know. You may find it helpful to start with our main circle theorems page and then look in detail at the rest.

Points A , B , and C are on the circumference of a circle with centre O . DE is a tangent at point A . Calculate the size of angle BAD .

**Locate the key parts of the circle for the theorem**.

Here we have:

- The angle BCA = 52°
- AC is a diameter
- DE is a tangent
- The angle BAD = θ

2**Use other angle facts to determine the remaining angle(s) made with the tangent**.

As AC is a diameter and **the angle in a semicircle is 90° **, angle ABC = 90° . As **angles in a triangle total 180° **,

\[CAB=180-(90+52)\\
CAB=38^{\circ}\]

3**Use the tangent theorem to state the other missing angle**.

As **the angle between the tangent and the radius is 90° **, we can now calculate angle BAD :

\[BAD=90-38\\
BAD=52^{\circ}\]

A, B, C and D are points on the circumference of a circle with centre O . AC and BD intersect at the point G . EF is a tangent at point C and is parallel to BD . Calculate the size of angle BCF .

**Locate the key parts of the circle for the theorem**.

Here we have:

- The angle BDC = 48°
- AC is a diameter
- EF is a tangent
- The angle BCF = θ

**Use other angle facts to determine the remaining angle(s) made with the tangent**.

**Angles in the same segment are equal** and so angle BDC = angle BAC = 48° . As AC is a diameter and **angles in a semicircle are 90° **, angle ABC = 90° .

**Angles in a triangle total 180° ** so we can now calculate angle ACB :

\[ACB=180-(90+48)\\
ACB=42^{\circ}\]

**Use the tangent theorem to state the other missing angle**.

As **the angle between the tangent and the radius is 90° **, we can now calculate angle BCF :

\[BCF=90-42\\
BCF=48^{\circ}\]

A circle with centre O has three points on the circumference, A, B , and C . The tangent DE passes through the point C . Calculate the size of angle BCE .

**Locate the key parts of the circle for the theorem**.

Here we have:

- The angle BAC = 21°
- OC is a radius
- DE is a tangent
- The angle BCE = θ

**Use other angle facts to determine the remaining angle(s) made with the tangent**.

**The angle at the centre is twice the angle at the circumference** and so the angle BOC = 21\times 2 = 42° . Triangle OBC is isosceles because OB and OC are radii.

As OBC is an isosceles triangle, we can calculate the size of angle OCB :

\[OCB=(180-42)\div2\\
OCB=69^{\circ}\]

**Use the tangent theorem to state the other missing angle**.

As **the angle between the tangent and the radius is 90° **, we can now calculate angle BCE :

\[BCE=90-69\\
BCE=21^{\circ}\]

B, C and D are points on the circumference of a circle with centre O . AE and AF are tangents to the circle. Calculate the size of angle DBF.

**Locate the key parts of the circle for the theorem**.

Here we have:

- The angle DCE = 80°
- OB is a radius
- AF is a tangent
- The angle OBD = θ

**Use other angle facts to determine the remaining angle(s) made with the tangent**.

We need to find a way of calculating angle OBD as this angle added to θ is equal to 90° . COBD is a quadrilateral and so if we can calculate all of the angles within this quadrilateral, we can then find angle DBF .

ABOC is a kite because **tangents that meet at the same point are equal** and the two other sides OB and OC are radii and so they are the same length. The line OA bisects the angle BOC and so we can state that the angle COA = 72° as it is the same size as angle AOB . As both tangents meet the radius at 90 degrees, angles ACO and ABO are also equal to 90° .

**The angle at the centre is twice the angle at the circumference **and so as the angle at the centre would equal 144°(72+72=144), the angle BDC is half this angle and so angle BDC = 72° .

We can also calculate the size of the angle OCE because **the angle between the tangent and the radius is 90° **. This means angle OCD = 90 - 80 = 10° .

As angles around a point total 360° , we can say that the reflex angle BOC is equal to:

\[BOC=360-144\\
BOC=216^{\circ}\]

As **angles in a quadrilateral total 360° **, we can calculate the size of angle OBD as COBD is a quadrilateral.

\[OBD=360-(10+216+72)\\
OBD=62^{\circ}\]

Phew!

**Use the tangent theorem to state the other missing angle**.

As **the angle between the tangent and the radius is 90° **, we can now calculate angle DBF :

\[DBF=90-62\\
DBF=28^{\circ}\]

The triangle ABC is inscribed in a circle with centre O . DE is a tangent at the point A . Calculate the size of the angle OAC .

**Locate the key parts of the circle for the theorem**.

Here we have:

- The angle ABC = 56°
- OA is a radius
- DE is a tangent
- The angle OAC = θ

**Use other angle facts to determine the remaining angle(s) made with the tangent**.

The Alternate Segment Theorem states that **the angle made with the tangent and a chord is equal to the angle in the alternate segment.** This means that the angle CAE is equal to the angle ABC , so CAE = 56° .

**Use the tangent theorem to state the other missing angle**.

As **the angle between the tangent and the radius is 90° **, we can now calculate angle OAC :

\[OAC=90-56\\
OAC=34^{\circ}\]

ABCD is an arrowhead inscribed inside a circle with centre C . The two tangents EF and GH meet at the external point P . Calculate the size of angle FPG .

**Locate the key parts of the circle for the theorem**.

Here we have:

- The angle BAD = 64°
- BC and CD are radii
- EF and GH are tangents
- The angle FPG = θ

**Use other angle facts to determine the remaining angle(s) made with the tangent**.

The **angle at the centre is twice the angle at the circumference **so the angle BCD is twice the size of angle BAD . BCD = 128° .

**Use the tangent theorem to state the other missing angle**.

As **the angle between the tangent and the radius is 90° **, we can now calculate angle BPD which is on a straight line with FPG :

\[BPD=360-(90+90+128)\\
BPD=52^{\circ}\]

As **angles on a straight line total 180° **,

\[FPG=180-52\\
FPG=128^{\circ}\]

**The angle between the tangent and the radius**

Either through miscalculation or an assumption, the angle between the tangent and the radius is not 90° because “it doesn’t look like it”, it must be proven.

**The alternate segment theorem**

There are cases when the Alternate Segment Theorem is used to describe the angle at the tangent or the angle in the alternate segment at the circumference but neither are true. Take Example 5 above. The angle OAC is assumed to be equal to 56° whereas angle CAE is equal to 56° .

**The angle is double or half the opposing angle**

The kite that is formed when two tangents meet has two angles of 90° and 90° because they meet the radius at 90° . The other two angles are assumed to be double or half whereas they should total 180° (it is a unique case for the cyclic quadrilateral).

Below is Example 4. The angle COB is correctly equal to 144° as 72×2=144° . The angle CAB is incorrectly equal to 144÷2 = 72° . This is incorrect because angles in a quadrilateral should total 360° , whereas shape ABOC has a total of 90+90+144+72=396° . The only case when this is correct is when ABOC is a square.

1. The right triangle ABC is inscribed in the circle with centre O . The line DE is a tangent to the circle with the point of contact at A . Angle BAD = 78^{\circ} . Calculate the size of the angle ABC .

78^{\circ}

12^{\circ}

90^{\circ}

68^{\circ}

- BAC = 90 – 78 = 12^{\circ} (the tangent meets the radius at 90^{\circ} )
- ABC = 90^{\circ} (angles in a semicircle)
- ABC = 180 – (90+12) = 78^{\circ} (angles in a triangle)

2. A, B, C, and D are points on the circumference of a circle with centre O . AC and BD are endpoints of two perpendicular lines with AC passing through the center of the circle. EF is a tangent to the circle at point C . Calculate the size of the angle BCF .

56^{\circ}

22^{\circ}

44^{\circ}

68^{\circ}

- DBC = 68^{\circ} ( BD is parallel to EF as BD and AC are perpendicular)
- BGC = 90^{\circ} (chord of a triangle)
- BCG = 180 – (90+68) = 22^{\circ}
- 90 – 22 = 68^{\circ} (the tangent meets the radius at 90^{\circ} )

3. The triangle ABC is inscribed into a circle with centre O . The line DE is a tangent at the point C . The angle BCE = θ. Calculate the size of angle BCE .

45^{\circ}

61^{\circ}

29^{\circ}

90^{\circ}

OCA = 29^{\circ} (triangle OAC is isosceles as OA = OC )

ABC = 90^{\circ} (angles in a semicircle)

OCB = 90 – 29 = 61^{\circ}

BCE = 90 – 61 = 29^{\circ} (the tangent meets the radius at 90^{\circ} )

4. A circle with centre O has three points on the circumference, B, C and D . The lines AE and AF are tangents to the circle at the points B and C . The tangents meet at the point A . Angle BAC = 40^{\circ} . Calculate the size of the angle ODC .

60^{\circ}

80^{\circ}

40^{\circ}

55^{\circ}

- OAC = 40\div2 = 20^{\circ} ( OA bisects the angle BAC )
- ACO = 90^{\circ} (angles in a semicircle)
- AOC = 180 – (90+20) = 70^{\circ} (angles in a triangle)
- OCD is an isosceles triangle as OC = OD
- ODC = (180-70)\div2=55^{\circ} (angles in an isosceles triangle)

5. A circle with centre O has three points on the circumference, A, B , and C . The line DE is a tangent to the circle at the point A . Angle BAD = 78^{\circ} . Calculate the size of the angle ABC .

86^{\circ}

4^{\circ}

90^{\circ}

88^{\circ}

- EAC = 4^{\circ} (alternate segment theorem)
- OAC = 90^{\circ} (angles in a semicircle)
- OAC = 90 – 4 = 86^{\circ}

6. A, B, and D are points on the circle with centre C . The tangents EF and GH intersect at the point P at an angle of 112^{\circ} . Calculate the size of angle BAD .

112^{\circ}

56^{\circ}

68^{\circ}

34^{\circ}

- BPD = 180 – 112 = 68^{\circ} (angles on a straight line)
- CDP = CBP = 90^{\circ} (the tangent meets the radius at 90^{\circ} )
- BCD = 180 – (90+90+68) = 112^{\circ} (angles in a quadrilateral)
- BAD = 112\div2 = 56^{\circ} (angle at the centre is twice the angle at the circumference)

1. (a) The circle below has centre O . The triangle ABC is inscribed into the triangle. The tangent DE goes through the point A . BC = AC . Calculate the size angle x .

(b) Hence find the value of y .

c) What type of triangle is ABC ?

**(7 marks)**

Show answer

(a)

BCA = 60^{\circ}

**(1)**

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

**(1)**

Reflex angle AOB = 360 - 120 = 240^{\circ}

**(1)**

Angles in a quadrilateral total 360^{\circ}

**(1)**

**(1)**

(b)

y=CAE=90 - 30 = 60^{\circ}

**(1)**

(c)

Equilateral

**(1)**

2. (a) ABC is a triangle inscribed in a circle with centre O . The tangent DE passes through the point C . Calculate the size of angle x . Explain your answer.

(b) Hence or otherwise, calculate the value of y . Give a reason for your answer.

**(7 marks)**

Show answer

(a)

BCO = 4^{\circ}

**(1)**

Tangent meets the radius at 90^{\circ}

**(1)**

**(1)**

Triangle COB is isosceles as OC = OB

**(1)**

**(1)**

(b)

y=172\div2=86^{\circ}

**(1)**

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

**(1)**

3. The circle below has centre C . A and B are points on the circumference of the circle with the tangent DE going through the circle at B . The line AE is a straight line through the centre of the circle. The angle BAC = 3x-10^{\circ} and the angle BEC = 20^{\circ} . By calculating the value of x , calculate the size of angle BAC . State any angle facts used in your answer.

**(7 marks)**

Show answer

ABC is an isosceles triangle as AC = BC are radii

**(1)**

CBE = 90^{\circ} as the tangent meets the radius at 90^{\circ}

**(1)**

BCE = 180 - (90 + 20) = 70^{\circ} as Angles in a triangle total 180^{\circ}

**(1)**

ACB = 180 - 70 = 110^{\circ} as angles on a straight line total 180^{\circ}

**(1)**

3x-10+3x-10+110=180 or 6x+90=180

**(1)**

6x=90 so x=15

**(1)**

**(1)**

You have now learned how to:

- Apply and prove the standard circle theorems concerning angles, radii, tangents and chords, and use them to prove related results

Prepare your KS4 students for maths GCSEs success with Third Space Learning. Weekly online one to one GCSE maths revision lessons delivered by expert maths tutors.

Find out more about our GCSE maths tuition programme.