# Tangent Of A Circle

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.

## What is the tangent of a circle?

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

### Key parts of a circle needed for these theorems

• 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.

### Proving that if two tangents meet, they are the same length

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.

## How to use the tangent theorems

In order to use the tangent of a circle:

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

### How to use the tangent theorems

Tangent 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.

## Tangent of a circle examples

### Example 1: standard diagram

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 .

1. 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

2Use 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° ,

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

3Use 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 :

$BAD=90-38\\ BAD=52^{\circ}$

### Example 2: angles in the same segment

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 .

Here we have:

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

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 :

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

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

$BCF=90-42\\ BCF=48^{\circ}$

### Example 3: angles at the centre

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 .

Here we have:

• The angle BAC = 21°
• DE is a tangent
• The angle BCE = θ

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 :

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

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

$BCE=90-69\\ BCE=21^{\circ}$

### Example 4: tangent of a circle

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.

Here we have:

• The angle DCE = 80°
• AF is a tangent
• The angle OBD = θ

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!

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

$DBF=90-62\\ DBF=28^{\circ}$

### Example 5: alternate segment theorem

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 .

Here we have:

• The angle ABC = 56°
• DE is a tangent
• The angle OAC = θ

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° .

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

$OAC=90-56\\ OAC=34^{\circ}$

### Example 6: complex diagram

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 .

Here we have:

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

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° .

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° ,

$FPG=180-52\\ FPG=128^{\circ}$

### Common misconceptions

• 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.

### Practice tangent of a circle questions

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)

### Tangent of a circle GCSE questions

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)

(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)

x = OAC = 360 - (60+30+240) = 30^{\circ}

(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)

(a)
BCO = 4^{\circ}

(1)

Tangent meets the radius at 90^{\circ}

(1)

COB = 180 - (4+4) = 172^{\circ}

(1)

Triangle COB is isosceles as OC = OB

(1)

x= 360 - 172 = 188^{\circ}

(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)

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)

BAC = 3\times15-10=35^{\circ}

(1)

## Learning checklist

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

## Still stuck?

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#### GCSE Maths Papers - November 2022 Topics

Practice paper packs based on the November advanced information for Edexcel 2022 Foundation and Higher exams.

Designed to help your GCSE students revise some of the topics that are likely to come up in November exams.