Tangent of a circle

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GCSE Maths Geometry and Measure Circle Theorems

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

What is the tangent of a circle?

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.

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.

How to use the tangent theorems

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.

How to use the tangent theorems

Tangent of a circle worksheet

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

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Tangent of a circle worksheet

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

DOWNLOAD FREE

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 $BAD$ .

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 $BCF$ .

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 $BCF$ :

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

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 $BCE$ .

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 $BCE$ :

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

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 $DBF.$

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 $DBF$ :

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

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 $OAC$ .

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 $OAC$ :

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

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 $FPG$ .

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 $180°$ ,

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

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 $72\times2=144°$ . The angle $CAB$ is incorrectly equal to $144\div2 = 72°$ . 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

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)

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 $90^{\circ}$ )

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 $90^{\circ}$ )

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)

86^{\circ}

4^{\circ}

90^{\circ}

88^{\circ}

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

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)

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)

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

(1)

(b)
$y=CAE=90 - 30 = 60^{\circ}$

(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)

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)

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)

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

The next lessons are

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