The angle at the centre and the angle in a semicircle – Circle and Related Theorems (Senior 3 Mathematics) | Ganzaa

Circle and Related Theorems

Lessons

Senior 3 · Mathematics

Circle and Related Theorems

5 lessons

Lesson 3 of 5

The angle at the centre and the angle in a semicircle

25 min

Two viewers, one arc

Stand anywhere on the edge of a round courtyard and look at a gate $A B$ in its wall: the gate fills a certain angle of your view. Walk along the far side of the edge and, surprisingly, that angle does not change. Stand at the centre instead and the gate fills exactly twice as wide an angle. This lesson proves the second fact, the key that the rest of the unit is built on; the first one follows from it in the next lesson.

Definition

An arc $A B$ (or its chord) subtends an angle at a point $P$ when $P A$ and $P B$ are the arms of the angle $\angle APB$ .

The angle at the centre is $\angle AOB$ .
An angle at the circumference is $\angle ACB$ , where $C$ is a point of the circle not on the arc $A B$ .

Example: in the figure below, the minor arc $A B$ subtends $\angle AOB = 120^\circ$ at the centre and $\angle ACB = 60^\circ$ at $C$ .

Important

Theorem 1: the angle at the centre. The angle that an arc subtends at the centre is twice the angle it subtends at any point on the remaining part of the circumference:

\angle AOB = 2\,\angle ACB .

Proof

Join $C O$ and extend it to meet the circle at $D$ . Call $\angle OCA = a$ and $\angle OCB = b$ , so $\angle ACB = a + b$ . (In the drawing, $a = 35^\circ$ and $b = 25^\circ$ .)

Statement	Reason
$O A = O C$	(radii)
$\angle OAC = \angle OCA = a$	(base angles of an isosceles triangle)
$\angle AOD = a + a = 2a$	(exterior angle of triangle $O C A$ )
$\angle BOD = 2b$	(the same steps in triangle $O C B$ )
$\angle AOB = 2a + 2b = 2(a + b) = 2\,\angle ACB$	(adding)

The same proof works when $C$ is on the minor arc: then $\angle AOB$ is the reflex angle (more than $180^\circ$ ) facing $C$ .

Worked example 1: centre to circumference and back

(a) $\angle AOB = 130^\circ$ and $C$ is on the major arc. Find $\angle ACB$ . (b) $\angle PRQ = 47^\circ$ at the circumference. Find $\angle POQ$ .

Solution

Statement	Reason
(a) $\angle ACB = \frac{130^\circ}{2} = 65^\circ$	(angle at the centre is twice the angle at the circumference)
(b) $\angle POQ = 2 \times 47^\circ = 94^\circ$	(angle at the centre is twice the angle at the circumference)

Worked example 2: when the point is on the minor arc

$A$ , $B$ and $C$ lie on a circle with centre $O$ , with $C$ on the minor arc $A B$ . $\angle AOB = 100^\circ$ . Find $\angle ACB$ .

Solution

Statement	Reason
$\text{reflex } \angle AOB = 360^\circ - 100^\circ = 260^\circ$	(angles at a point)
$\angle ACB = \frac{260^\circ}{2} = 130^\circ$	(angle at the centre is twice the angle at the circumference: the arc facing $C$ is the major arc)

The angle in a semicircle

Let $A B$ be a diameter. The angle it subtends at the centre is a straight angle, $180^\circ$ . By Theorem 1, the angle at any point $C$ of the circle is half of that.

Important

Theorem 2: the angle in a semicircle. The angle subtended by a diameter at any point of the circle is a right angle: if $A B$ is a diameter, then $\angle ACB = 90^\circ$ .

Converse. If $\angle ACB = 90^\circ$ , then $C$ lies on the circle with diameter $A B$ . This is the median theorem of the unit "Right-Angled Triangles" seen again: the midpoint of the hypotenuse is the same distance from all three vertices, so it is the centre of a circle through them.

Worked example 3: a right angle you were not given

$A B$ is a diameter of a circle and $C$ is a point on the circle with $\angle CAB = 28^\circ$ . Find $\angle CBA$ . If $A B = 10$ cm and $A C = 8$ cm, find $B C$ .

Solution

Statement	Reason
$\angle ACB = 90^\circ$	(angle in a semicircle)
$\angle CBA = 180^\circ - 90^\circ - 28^\circ = 62^\circ$	(angle sum of a triangle)
$BC = \sqrt{10^2 - 8^2} = 6$ cm	(Pythagoras, as $\angle C = 90^\circ$ )

Worked example 4: two theorems in a row

$A$ , $B$ and $C$ lie on a circle with centre $O$ , and $\angle OAB = 25^\circ$ . Find $\angle ACB$ .

Solution

Statement	Reason
$\angle OBA = 25^\circ$	(base angles of an isosceles triangle, $O A = O B$ radii)
$\angle AOB = 180^\circ - 25^\circ - 25^\circ = 130^\circ$	(angle sum of a triangle)
$\angle ACB = \frac{130^\circ}{2} = 65^\circ$	(angle at the centre is twice the angle at the circumference)

Worked example 5: checking with coordinates

The circle $x^{2} + y^{2} = 25$ has centre $(0, 0)$ and radius 5, so $A (- 5, 0)$ and $B (5, 0)$ are the ends of a diameter. Show that $C (3, 4)$ is on the circle and that $\angle ACB = 90^\circ$ .

Solution $3^{2} + 4^{2} = 25$ , so $C$ is on the circle. Gradient of $A C$ : $\frac{4 - 0}{3 - (-5)} = \frac{1}{2}$ . Gradient of $B C$ : $\frac{4 - 0}{3 - 5} = -2$ . Their product is $\frac{1}{2} \times (-2) = -1$ , so $AC \perp BC$ (perpendicular lines), as Theorem 2 predicts.

Warning

Twice, not half, at the centre. The angle at the centre is the bigger one. A quick look at the figure should confirm it.

Both angles must stand on the same arc. If $C$ is on the minor arc, compare $\angle ACB$ with the reflex angle at $O$ , as in Worked example 2.

Tip

When a circle question has a diameter in it, look for a right angle at the circumference straight away. It is the most often missed step in exams.

Corbettmaths: Angles at centre and circumference proof. Watch how the line through the centre splits the figure into two isosceles triangles, exactly as in the proof above.

✦

Practice

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

$\angle AOB = 144^\circ$ and $C$ is on the major arc. Find $\angle ACB$ .
$\angle ACB = 38^\circ$ at the circumference. Find $\angle AOB$ .
$C$ is on the minor arc $A B$ and $\angle ACB = 115^\circ$ . Find the reflex angle $A O B$ and the angle $A O B$ .
$A B$ is a diameter and $\angle CAB = 41^\circ$ . Find $\angle ABC$ .
$A B$ is a diameter of length 13 cm, and $A C = 5$ cm. Find $B C$ .
$A$ , $B$ and $C$ are on the circle, $A$ is on the major arc $B C$ , and $\angle OBC = 34^\circ$ . Find $\angle BAC$ .
A triangle with sides 5 cm, 12 cm and 13 cm has its three vertices on a circle. Find the radius of the circle, and explain.
Prove the angle in a semicircle without Theorem 1: join $O C$ , call $\angle OAC = x$ and $\angle OBC = y$ , and use the two isosceles triangles.

Show answersHide answers

$72^\circ$ (angle at the centre is twice the angle at the circumference).
$76^\circ$ (the same theorem).
Reflex $\angle AOB = 2 \times 115^\circ = 230^\circ$ (angle at the centre); $\angle AOB = 360^\circ - 230^\circ = 130^\circ$ (angles at a point).
$\angle ACB = 90^\circ$ (angle in a semicircle), so $\angle ABC = 180^\circ - 90^\circ - 41^\circ = 49^\circ$ (angle sum of a triangle).
$\angle ACB = 90^\circ$ (angle in a semicircle), so $BC = \sqrt{13^2 - 5^2} = 12$ cm (Pythagoras).
$\angle OCB = 34^\circ$ (base angles, $O B = O C$ radii); $\angle BOC = 112^\circ$ (angle sum of a triangle); $\angle BAC = 56^\circ$ (angle at the centre is twice the angle at the circumference).
$5^{2} + 12^{2} = 169 = 13^{2}$ , so the triangle has a right angle (converse of Pythagoras). By the converse of Theorem 2, the 13 cm side is a diameter, so the radius is 6.5 cm.
$\angle OCA = x$ and $\angle OCB = y$ (base angles, $O A = O C = O B$ radii). In triangle $A B C$ : $x + y + (x + y) = 180^\circ$ (angle sum of a triangle), so $x + y = 90^\circ$ , and $\angle ACB = x + y = 90^\circ$ .

Sources and credits

Written by Ganzaa.
Video: Angles at Centre and Circumference Proof, Corbettmaths, embedded from YouTube.

The diagrams, proofs, worked examples and practice items were written by Ganzaa. Every angle in the diagrams is drawn to its stated size.

1Two viewers, one arc

2Proof

3The angle in a semicircle

Two viewers, one arc

Proof

The angle in a semicircle