Tangent of a Circle - Math Steps, Examples & Questions

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Tangent of a circle

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

Students will first learn about the tangent of a circle as part of geometry in high school.

What is the tangent of a circle?

A tangent of a circle is a straight line that touches the circumference of the circle at a single point. This point is called the point of tangency.

Diagram $1$ – The angle between a tangent and radius is $90$ degrees.
Diagram $2$ – Tangents which meet at the same point are equal in length.

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

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 center of the circle $C,$ they form a kite $ABCD.$ This means that you have the two circle theorems:

Isosceles triangle

Kite

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Key parts of a circle needed for these theorems

The radius of a circle is the distance from the center to the circumference of the circle. The radius is half of the diameter.
The center of the circle is a point that is located in 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.

What is the tangent of a circle?

Common Core State Standards

How does this relate to high school math?

High school: Geometry – Circles (HS.G.C.A.4)
Construct a tangent line from a point outside a given circle to the circle.

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.

Tangent of a circle examples

Example 1: standard diagram

Points $A, B,$ and $C$ are on the circumference of a circle with point $O$ as the center. $($ This can also be called circle $O.) \, DE$ is a tangent at point $A.$ Calculate the size of angle $BAD.$

Locate the key parts of the circle for the theorem.

Here you have:

Angle $BCA=52^{\circ}$
$AC$ is the diameter
$DE$ is a tangent
Angle $BAD=\theta$

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 $90^{\circ},$ angle $ABC=90^{\circ}.$ As angles in a triangle total $180^{\circ},$

\begin{aligned} \text{ Angle } CAB&=180-(90+52) \\\\ &=38^{\circ} \end{aligned}

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

As the angle between the tangent and the radius is $90^{\circ},$ you can now calculate angle $BAD\text{:}$

\begin{aligned}BAD&=90-38 \\\\ &=52^{\circ} \end{aligned}

Example 2: angles in the same segment

$A, B, C,$ and $D$ are points on the circumference of a circle with center $O. \, AC$ and $BD$ intersect at 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 you have:

Angle $BDC=48^{\circ}$
$AC$ is the diameter
$EF$ is a tangent
Angle $BCF=\theta$

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

Angles in the same segment are equal, so angle $BDC=$ angle $BAC=48^{\circ}.$ As $AC$ is a diameter and angles in a semicircle are $90^{\circ},$ angle $ABC=90^{\circ}.$

Angles in a triangle total $180^{\circ}$ so you can now calculate angle $ACB\text{:}$

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

Use the tangent theorem to state the other missing angle.

As the angle between the tangent and the radius is $90^{\circ},$ you can now calculate angle $BCF\text{:}$

\begin{aligned}BCF&=90-42 \\\\ &=48^{\circ} \end{aligned}

Example 3: angles at the center

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

Locate the key parts of the circle for the theorem.

Here you have:

Angle $BAC=21^{\circ}$
$OC$ is the radius
$DE$ is a tangent
Angle $BCE=\theta$

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

The angle at the center is twice the angle at the circumference and so angle $BOC=21\times{2}=42^{\circ}.$ Triangle $OBC$ is isosceles because $OB$ and $OC$ are radii.

As $OBC$ is an isosceles triangle, you can calculate the size of angle $OCB\text{:}$

$\begin{aligned}OCB&=(180-42)\div{2} \\\\ &=69^{\circ} \end{aligned}$

Use the tangent theorem to state the other missing angle.

As the angle between the tangent and the radius is $90^{\circ},$ you can now calculate angle $BCE\text{:}$

\begin{aligned}BCE&=90-69 \\\\ &=21^{\circ} \end{aligned}

Example 4: tangent of a circle

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

Locate the key parts of the circle for the theorem.

Here you have:

Angle $DCE=80^{\circ}$
$OB$ is the radius
$AF$ is a tangent
Angle $OBD=\theta$

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

You need to find a way of calculating angle $OBD$ as this angle added to $\theta$ is equal to $90^{\circ}. \, COBD$ is a quadrilateral, so if you can calculate all of the angles within this quadrilateral, you 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, so they are the same length.

The line $OA$ bisects the angle $BOC$ and so you can state that the angle $COA=72^{\circ}$ 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 $90^{\circ}.$

The angle at the center is twice the angle at the circumference and so as the angle at the center would equal $144^{\circ} \, (72+72=144^{\circ}),$ the angle $BDC$ is half this angle and so angle $BDC=72^{\circ}.$

You can also calculate the size of the angle $OCE$ because the angle between the tangent and the radius is $90^{\circ}.$ This means angle $OCD=90-80=10^{\circ}.$

As angles around a point total $360^{\circ},$ you can say that the reflex angle $BOC$ is equal to:

\begin{aligned}BOC&=360-144 \\\\ &=216^{\circ} \end{aligned}

As angles in a quadrilateral total $360^{\circ},$ you can calculate the size of angle $OBD$ as $COBD$ is a quadrilateral.

$\begin{aligned}OBD&=360-(10+216+72) \\\\ &=62^{\circ} \end{aligned}$

Use the tangent theorem to state the other missing angle.

As the angle between the tangent and the radius is $90^{\circ},$ you can now calculate angle $DBF\text{:}$

\begin{aligned}DBF&=90-62 \\\\ &=28^{\circ} \end{aligned}

Example 5: alternate segment theorem

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

Locate the key parts of the circle for the theorem.

Here you have:

Angle $ABC=56^{\circ}$
$OA$ is a radius
$DE$ is a tangent
The angle $OAC=\theta$

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 $CAE=56^{\circ}.$

Use the tangent theorem to state the other missing angle.

As the angle between the tangent and the radius is $90^{\circ},$ you can now calculate angle $OAC\text{:}$

\begin{aligned}OAC&=90-56 \\\\ &=34^{\circ} \end{aligned}

Example 6: complex diagram

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

Locate the key parts of the circle for the theorem.

Here you have:

Angle $BAD=64^{\circ}$
$BC$ and $CD$ are radii
$EF$ and $GH$ are tangents
Angle $FPG=\theta$

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

The angle at the center is twice the angle at the circumference so the angle $BCD$ is twice the size of angle $BAD. \, BCD=64\times{2}=128^{\circ}.$

Use the tangent theorem to state the other missing angle.

As the angle between the tangent and the radius is $90^{\circ},$ you can now calculate angle $BPD$ which is on a straight line with $FPG\text{:}$

\begin{aligned}BPD&=360-(90+90+128) \\\\ &=52^{\circ} \end{aligned}

As angles on a straight line total $180^{\circ},$

\begin{aligned}FPG&=180-52 \\\\ &=128^{\circ} \end{aligned}

Teaching tips for tangent of a circle

Provide worksheets with real-world examples where tangents are encountered, such as the point where a road meets a circular roundabout or the point where a billiard ball strikes the cushion of a table. Relating abstract concepts to tangible situations enhances understanding.

Start with simple practice questions and gradually increase complexity. This approach enhances critical thinking and problem-solving skills.

Highlight connections between tangents and other mathematical concepts, such as the Pythagorean theorem (Pythagoras’ theorem), similar triangles, and trigonometry.

Easy mistakes to make

Misrepresenting the angle between the tangent and the radius
Either through miscalculation or an assumption, the angle between the tangent and the radius is not $90^{\circ}$ because “it doesn’t look like it”, it must be proven.

Misusing 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 is true.

Take Example $5$ above. The angle $OAC$ is assumed to be equal to $56^{\circ}$ whereas angle $CAE$ is equal to $56^{\circ}.$

Thinking the angle is double or half the opposing angle
The kite that is formed when two tangents meet has two angles of $90^{\circ}$ and $90^{\circ}$ because they meet the radius at $90^{\circ}.$

The other two angles are assumed to be double or half whereas they should total $180^{\circ}$ (it is a unique case for the cyclic quadrilateral).

In the example below, the angle $COB$ is correctly equal to $144^{\circ}$ as $72\times{2}=144^{\circ}.$ The angle $CAB$ is incorrectly equal to $144\div{2}=72^{\circ}.$

This is incorrect because angles in a quadrilateral should total $360^{\circ},$ whereas shape $ABOC$ has a total of $90+90+144+72=396^{\circ}.$ The only case when this is correct is when $ABOC$ is a square.

Central angle
Chords of a circle
Subtended
Circle chord theorems

Practice tangent of a circle questions

12^{\circ}

78^{\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 are $90^{\circ})$
$ACB=180-(90+12)=78^{\circ} \, ($ angles in a triangle sum to $180^{\circ})$

68^{\circ}

44^{\circ}

22^{\circ}

56^{\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 $90^{\circ})$

61^{\circ}

29^{\circ}

45^{\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 $90^{\circ})$

40^{\circ}

80^{\circ}

60^{\circ}

55^{\circ}

$OAC=40\div{2}=20^{\circ} \, (OA$ 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 $OC=OD$
$ODC=(180-70)\div{2}=55^{\circ} \, ($ 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 are $90^{\circ})$
$OAC=90-4=86^{\circ}$

112^{\circ}

68^{\circ}

56^{\circ}

34^{\circ}

$BPD=180-112=68^{\circ} \, ($ angles on a straight line total $180^{\circ})$
$CDP=CBP=90^{\circ} \, ($ the tangent meets the radius at $90^{\circ})$
$BCD=360-(90+90+68)=112^{\circ} \, ($ angles in a quadrilateral total $360^{\circ})$
$BAD=112\div{2}=56^{\circ} \, ($ angle at the center is twice the angle at the circumference $)$

Tangent of a circle FAQs

What is a tangent to a circle?

A tangent to a circle is a straight line that touches the circle at exactly one point, without crossing it. This single point where the circle touches the line is called the point of tangency.

What is the tangent-secant theorem?

The Tangent-Secant Theorem states that if a tangent and a secant are drawn from a point outside a circle, then the square of the length of the tangent segment is equal to the product of the lengths of the whole secant segment and its external segment.

How do you find the length of a tangent segment from a point to a circle?

If you know the distance from the point to the center of the circle $(d)$ and the radius of the circle $(r),$ the length of the tangent segment can be found using the Pythagorean theorem: Length of tangent $=\sqrt{d^2-r^2}$

How do you find the point of tangency?

To find the point of tangency, you need the coordinates of the point and the equation of the circle. The point of tangency will satisfy both the equation of the circle and the equation of the tangent line.

What is the difference between the tangent of a circle and a chord of a circle?

A tangent is a straight line that touches a circle at exactly one point. A chord is a line segment with both endpoints on the circumference of the circle.

The next lessons are

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