Circle and Related Theorems

Lessons

Senior 3 · Mathematics

Circle and Related Theorems

5 lessons

Lesson 5 of 5

Tangents and the alternate segment theorem

25 min

Mud from a back wheel

Ride a bicycle or a moto through a puddle and look at the spray: drops leave the wheel in straight lines that only just graze the rim. Each drop flies off along a tangent, the line that touches the circle at the point where the drop let go. Tangents are where circles meet straight lines, in wheels, belts round pulleys, and roads that curve into straights. Three theorems describe them.

Definition

A tangent to a circle is a straight line that meets the circle at exactly one point, the point of contact. A point outside the circle is an external point; from it, the length of a tangent is the distance to the point of contact.

Example: in the first figure below, the line through $A$ is a tangent with point of contact $A$ ; in the second, $T A$ and $T B$ are the two tangent lengths from the external point $T$ .

The tangent and the radius

Important

Theorem 6: tangent and radius. A tangent is perpendicular to the radius drawn to the point of contact: $OA \perp$ tangent.

Reason. Every point $Q$ of the tangent other than $A$ lies outside the circle, so $O Q > O A$ . So $A$ is the point of the tangent closest to $O$ , and the shortest segment from a point to a line is the perpendicular one.

Converse. A line through a point of the circle, perpendicular to the radius there, is a tangent.

Worked example 1: a tangent length

$T$ is 13 cm from the centre of a circle of radius 5 cm. A tangent from $T$ touches the circle at $A$ . Find $T A$ .

Statement	Reason
$\angle OAT = 90^\circ$	(tangent is perpendicular to the radius)
$TA = \sqrt{13^2 - 5^2} = \sqrt{144} = 12$ cm	(Pythagoras in triangle $O A T$ )

Two tangents from one point

Important

Theorem 7: tangents from an external point. The two tangents from an external point $T$ are equal in length: $T A = T B$ . Also, $O T$ bisects the angle $A T B$ and the angle $A O B$ .

Proof. In triangles $O A T$ and $O B T$ : $\angle OAT = \angle OBT = 90^\circ$ (tangent is perpendicular to the radius); $O A = O B$ (radii); $O T$ is common. So the triangles are congruent (RHS), which gives $T A = T B$ , $\angle OTA = \angle OTB$ and $\angle AOT = \angle BOT$ .

A useful by-product. In the quadrilateral $O A T B$ the angles at $A$ and $B$ are right angles, so $\angle AOB + \angle ATB = 180^\circ$ (angle sum of a quadrilateral).

Worked example 2: angles round two tangents

In the figure above, $\angle ATB = 50^\circ$ . Find $\angle AOB$ , $\angle OTA$ and $\angle TAB$ .

Statement	Reason
$\angle AOB = 360^\circ - 90^\circ - 90^\circ - 50^\circ = 130^\circ$	(tangents perpendicular to radii; angle sum of a quadrilateral)
$\angle OTA = \frac{50^\circ}{2} = 25^\circ$	( $O T$ bisects the angle between the tangents)
$T A = T B$	(tangents from an external point)
$\angle TAB = \frac{180^\circ - 50^\circ}{2} = 65^\circ$	(base angles of isosceles triangle $T A B$ )

Worked example 3: a circle inside a triangle

A circle touches all three sides of triangle $A B C$ , where $A B = 7$ cm, $B C = 8$ cm and $C A = 9$ cm. Find the tangent lengths from each vertex.

Solution Let the tangent lengths from $A$ , $B$ and $C$ be $x$ , $y$ and $z$ . Each vertex has two equal tangents (tangents from an external point), so each side is the sum of two of them:

x + y = 7, \qquad y + z = 8, \qquad x + z = 9 .

Adding: $2 (x + y + z) = 24$ , so $x + y + z = 12$ . Then $x = 12 - 8 = 4$ , $y = 12 - 9 = 3$ , $z = 12 - 7 = 5$ (all in cm).

The angle between a tangent and a chord

A chord $A B$ drawn from the point of contact makes an angle with the tangent. The chord cuts the circle into two segments; the one on the other side of $A B$ from that angle is the alternate segment.

Important

Theorem 8: the alternate segment theorem. The angle between a tangent and a chord through the point of contact equals any angle subtended by the chord in the alternate segment. In the figure, both angles are $58^\circ$ .

Proof

Draw the diameter $A D$ and join $B D$ . Call the angle between the tangent and the chord $v$ , and $\angle DAB = u$ .

Statement	Reason
$u + v = 90^\circ$	(tangent is perpendicular to the radius $O A$ )
$\angle ABD = 90^\circ$	(angle in a semicircle)
$\angle ADB = 180^\circ - 90^\circ - u = 90^\circ - u = v$	(angle sum of triangle $A B D$ )
$\angle ACB = \angle ADB = v$ for any $C$ in the alternate segment	(angles in the same segment)

Worked example 4: all three angles of a triangle

The tangent at $A$ makes $64^\circ$ with $A B$ and $50^\circ$ with $A C$ . Find the angles of triangle $A B C$ .

Statement	Reason
$\angle ACB = 64^\circ$	(alternate segment theorem, with chord $A B$ )
$\angle ABC = 50^\circ$	(alternate segment theorem, with chord $A C$ )
$\angle BAC = 180^\circ - 64^\circ - 50^\circ = 66^\circ$	(angles on a straight line, the tangent)
Check: $64^\circ + 50^\circ + 66^\circ = 180^\circ$	(angle sum of a triangle)

Warning

Match the angle with the right segment. The tangent–chord angle on the right of $A B$ equals the angle in the segment on the left of $A B$ . A common error is to equate it with an angle on its own side.

Only the radius to the point of contact is perpendicular to the tangent. Other radii are not.

Tip

Where there is a tangent, draw the radius to the point of contact and mark the right angle at once. Where there are two tangents from a point, mark the two equal lengths. Most tangent problems then become an isosceles triangle or Pythagoras.

Corbettmaths: Alternate segment theorem proof. Watch how the right angles from the tangent and from the semicircle combine, the same chain of reasons as the proof above.

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Practice

Give a reason for each step. $O$ is always the centre. The last two are harder.

$T$ is 17 cm from the centre of a circle of radius 8 cm. Find the length of a tangent from $T$ .
A tangent from $T$ has length 24 cm and the radius is 7 cm. How far is $T$ from the centre?
Tangents from $T$ touch the circle at $A$ and $B$ , and $\angle ATB = 64^\circ$ . Find $\angle AOB$ and $\angle TAB$ .
Tangents at $A$ and $B$ meet at $T$ , and $\angle AOB = 140^\circ$ . Find $\angle ATB$ .
The tangent at $A$ makes an angle of $71^\circ$ with the chord $A B$ . $C$ is in the alternate segment. Find $\angle ACB$ .
A circle touches the three sides of triangle $A B C$ , where $A B = 10$ cm, $B C = 12$ cm and $C A = 14$ cm. Find the tangent lengths from $A$ , $B$ and $C$ .
The tangent at $A$ makes $62^\circ$ with the chord $A B$ , and $C$ is in the alternate segment with $A B = A C$ . Find $\angle BAC$ .
Prove that the tangents at the two ends of a diameter are parallel.

Show answersHide answers

$\sqrt{17^2 - 8^2} = 15$ cm (tangent perpendicular to radius; Pythagoras).
$\sqrt{24^2 + 7^2} = \sqrt{625} = 25$ cm.
$\angle AOB = 180^\circ - 64^\circ = 116^\circ$ (angles at $A$ and $B$ are right angles; angle sum of a quadrilateral). $\angle TAB = \frac{180^\circ - 64^\circ}{2} = 58^\circ$ ( $T A = T B$ , base angles of an isosceles triangle).
$180^\circ - 140^\circ = 40^\circ$ .
$71^\circ$ (alternate segment theorem).
With $x$ , $y$ , $z$ from $A$ , $B$ , $C$ : $x + y = 10$ , $y + z = 12$ , $x + z = 14$ , so $x + y + z = 18$ , giving $x = 6$ cm, $y = 4$ cm, $z = 8$ cm.
$\angle ACB = 62^\circ$ (alternate segment theorem); $\angle ABC = 62^\circ$ (base angles, $A B = A C$ ); $\angle BAC = 180^\circ - 124^\circ = 56^\circ$ (angle sum of a triangle).
Let $A B$ be a diameter, with tangents at $A$ and $B$ . Each tangent makes $90^\circ$ with $A B$ (tangent perpendicular to the radius). The two co-interior angles on the same side of $A B$ add up to $90^\circ + 90^\circ = 180^\circ$ , so the tangents are parallel.

1Mud from a back wheel

2The tangent and the radius

3Two tangents from one point

4The angle between a tangent and a chord

5Proof

Mud from a back wheel

The tangent and the radius

Two tangents from one point

The angle between a tangent and a chord

Proof