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Cambridge IGCSE Mathematics 0580 · Geometry

Circle theorems I

The words for the parts of a circle, the angle in a semicircle and the right angle between a tangent and a radius, used with the angle facts from 4.6 and with Pythagoras and trigonometry. Extended: the angle at the centre, angles in the same segment, cyclic quadrilaterals and the alternate segment theorem, and how to write each as a reason.

C4.7E4.7Core and ExtendedFree lesson

What the syllabus says

C4.7 (Core)

Calculate unknown angles and give explanations using the following geometrical properties of circles:
angle in a semicircle = 90°
angle between tangent and radius = 90°.

Notes and examples

Candidates will be expected to use the geometrical properties listed in the syllabus when giving reasons for answers.

E4.7 (Extended)

Calculate unknown angles and give explanations using the following geometrical properties of circles:
angle in a semicircle = 90°
angle between tangent and radius = 90°
angle at the centre is twice the angle at the circumference
angles in the same segment are equal
opposite angles of a cyclic quadrilateral sum to 180° (supplementary)
alternate segment theorem.

Notes and examples

Candidates are expected to use the geometrical properties listed in the syllabus when giving reasons for answers.

Quoted from the Cambridge IGCSE Mathematics 0580 syllabus for 2025 to 2027.

Circle words you need

The centre is usually labelled $O$ . Every radius (plural radii) goes from the centre to the circle, and all radii of a circle are equal.
A chord is a straight line joining two points on the circle. A diameter is a chord through the centre: it is two radii long.
An arc is part of the circumference. A chord splits the circle into a minor arc and a major arc, and splits the disc into two segments.
A tangent is a straight line that touches the circle at exactly one point, the point of contact.
A quadrilateral with all four vertices on a circle is a cyclic quadrilateral.

The most useful fact is the simplest: radii are equal. Any triangle with two radii as sides, such as $O A B$ with $A$ and $B$ on the circle, is isosceles. Look for these triangles first in every circle question: they turn one known angle into two.

Throughout this topic the syllabus expects you to use the circle properties as your reasons, in words. A number with no reason scores the angle mark only.

The angle in a semicircle

The angle in a semicircle is $90^{°}$ . If $A B$ is a diameter and $C$ is any point on the circle, then angle $A C B = 90^{°}$ .

Why it is true. Join $C$ to the centre $O$ . Now $O A = O B = O C$ , because they are all radii, so the diagram contains two isosceles triangles:

in triangle $O A C$ , the base angles are equal: call each one $a$ ;
in triangle $O B C$ , the base angles are equal: call each one $b$ .

The big triangle $A B C$ has angles $a$ , $b$ and $a + b$ (the angle at $C$ is made of both). So $a + b + (a + b) = 180$ , which gives $2 (a + b) = 180$ and $a + b = 90$ . The angle at $C$ is $a + b = 90^{°}$ , wherever $C$ is.

How to spot it: a triangle whose longest side is a diameter, with its third vertex on the circle. The right angle is always at the vertex on the circle, opposite the diameter. Once you know there is a right angle, everything from right-angled triangles is available: the angle sum, Pythagoras' theorem, and sine, cosine and tangent.

Tangent and radius

The angle between a tangent and a radius is $90^{°}$ . The radius drawn to the point of contact is perpendicular to the tangent.

Why it is true. The point of contact $P$ is the only point of the tangent on the circle; every other point of the tangent is outside the circle, so it is further from $O$ than the radius. That makes $O P$ the shortest distance from $O$ to the line, and the shortest distance from a point to a line is the perpendicular one. You can also see it by symmetry: the diagram is symmetrical about the line $O P$ , so the two angles between $O P$ and the tangent are equal, and they add up to $180^{°}$ .

How to spot it: a straight line touching the circle at one point, with the centre joined to that point. The right angle is at the point of contact. A common exam set-up joins the centre to a point $T$ on the tangent: triangle $O P T$ then has a right angle at $P$ , so $O T$ is the hypotenuse.

The right angle is only there if the line goes to the point of contact. A radius to some other point does not meet the tangent at $90^{°}$ .

Worked example 1

$A B$ is a diameter of a circle, centre $O$ , and $C$ is on the circle. Angle $C A B = 25^{°}$ . Find (a) angle $A B C$ and (b) angle $B O C$ . Give reasons.

(a) Angle $A C B = 90^{°}$ (the angle in a semicircle is $90^{°}$ ).
Angle $A B C = 180 - 90 - 25 = 65^{°}$ (angles in a triangle add up to $180^{°}$ ).
(b) $O B = O C$ (radii), so triangle $O B C$ is isosceles and angle $O C B$ = angle $O B C = 65^{°}$ (base angles of an isosceles triangle are equal).
Angle $B O C = 180 - 65 - 65 = 50^{°}$ (angles in a triangle).
Notice that $50^{°}$ is exactly twice the $25^{°}$ at $A$ . That is not a coincidence: it is the Extended theorem about the angle at the centre.

Answer: (a) $65^{°}$ (b) $50^{°}$

Worked example 2

This question is from a calculator paper. $T P$ is the tangent at $P$ to a circle, centre $O$ , radius 5 cm. $O T = 13$ cm. Find (a) the length $T P$ and (b) angle $O T P$ , correct to 1 decimal place.

(a) Angle $O P T = 90^{°}$ (the tangent is perpendicular to the radius $O P$ ), so $O T$ is the hypotenuse.
$T P^{2} = 13^{2} - 5^{2} = 169 - 25 = 144$ , so $T P = 12$ cm.
(b) At angle $O T P$ : the opposite side is $O P = 5$ and the hypotenuse is $O T = 13$ .
$\sin O T P = \frac{5}{13}$ , so angle $O T P = \sin^{- 1} (\frac{5}{13}) = {22.61...}^{°}$ , which is ${22.6}^{°}$ .

Answer: (a) $12$ cm (b) ${22.6}^{°}$

Check yourself

$A B$ is a diameter of a circle and $C$ is a point on the circle. Angle $A B C = 41^{°}$ . Find angle $B A C$ .

Answer it in the portal, where your answer is checked as you type →

Check yourself

$T P$ is a tangent to a circle, centre $O$ , at $P$ . Angle $O T P = 29^{°}$ . Find angle $P O T$ , and give the reason for the right angle you use.

Answer it in the portal, where your answer is checked as you type →

The angle at the centre Extended

The angle at the centre is twice the angle at the circumference, when both stand on the same arc. If $A$ , $B$ and $C$ are on a circle with centre $O$ , and $C$ is on the major arc, then angle $A O B = 2 \times$ angle $A C B$ .

Why it is true. Join $C$ to $O$ and extend the line. Triangles $O A C$ and $O B C$ are isosceles (radii). Call their base angles $p$ and $q$ , so angle $A C B = p + q$ . In triangle $O A C$ the angle at $O$ is $180 - 2 p$ , so the angle between $O A$ and the extended line is $2 p$ (angles on a straight line). In the same way the angle between $O B$ and the extended line is $2 q$ . So angle $A O B = 2 p + 2 q = 2 (p + q)$ , twice angle $A C B$ .

Things to watch:

The two angles must stand on the same arc. Trace the arc from $A$ to $B$ that does not pass through $C$ : the angle at $O$ that faces that arc is the one that is double.
If $C$ is on the minor arc, the angle at the centre on the same arc is reflex. For example, if angle $A D B = 115^{°}$ with $D$ on the minor arc, the reflex angle $A O B$ is $230^{°}$ , and the ordinary angle $A O B$ is $360 - 230 = 130^{°}$ .
The angle in a semicircle is a special case: a diameter makes a straight angle of $180^{°}$ at the centre, and half of it is $90^{°}$ .

Angles in the same segment Extended

Angles in the same segment are equal. If $C$ and $D$ are on the circle, on the same side of the chord $A B$ , then angle $A C B$ = angle $A D B$ .

Why it is true. Both angles stand on the same arc $A B$ , so each is half of the same angle at the centre.

How to spot it: look for a "bow tie" or "butterfly": two triangles on the same chord, with their third vertices on the same arc, often drawn so that their sides cross. The chord can be any chord, including one that is not drawn. Check carefully which chord each angle stands on: angle $C A D$ stands on chord $C D$ , and pairs with angle $C B D$ , not with an angle on $A B$ .

If the two points are on opposite sides of the chord, the angles are not equal: they are opposite angles of a cyclic quadrilateral and add up to $180^{°}$ .

Cyclic quadrilaterals Extended

Opposite angles of a cyclic quadrilateral add up to $180^{°}$ (they are supplementary).

Why it is true. Join two opposite vertices, $B$ and $D$ , to the centre. Angle $D A B$ stands on one arc $B D$ , so the angle at the centre on that arc is $2 \times$ angle $D A B$ . Angle $B C D$ stands on the other arc, so the angle at the centre on that arc is $2 \times$ angle $B C D$ . The two angles at the centre make a full turn, $360^{°}$ , so the two angles of the quadrilateral add up to $180^{°}$ .

Two consequences:

The exterior angle of a cyclic quadrilateral (made by extending a side) equals the interior angle at the opposite vertex: both are $180^{°}$ minus the same angle.
The quadrilateral must be cyclic: all four vertices on the circle. If one vertex is the centre $O$ , it is not a cyclic quadrilateral and this fact does not apply.

The alternate segment theorem Extended

The alternate segment theorem: the angle between a tangent and a chord, at the point of contact, is equal to the angle in the alternate segment.

In the diagram, $T A$ is the tangent at $A$ and $A B$ is a chord. The angle $T A B$ is between them. The alternate segment is the segment on the other side of the chord $A B$ from that angle, and any angle at the circumference standing on $A B$ in that segment, such as angle $A C B$ , equals angle $T A B$ .

Why it is true. Draw the diameter from $A$ through $O$ to a point $D$ on the circle. The tangent meets the diameter at $90^{°}$ , so angle $T A B = 90 -$ angle $B A D$ . In triangle $A B D$ the angle at $B$ is $90^{°}$ (angle in a semicircle), so angle $A D B = 90 -$ angle $B A D$ as well. And angle $A C B$ = angle $A D B$ (same segment).

How to spot it: a tangent and a chord meeting at the same point on the circle, with a triangle drawn on that chord. Each chord from the point of contact gives one pair: chord $A B$ pairs the tangent angle on its side with the angle at $C$ ; chord $A C$ pairs the tangent angle on the other side with the angle at $B$ . In an answer, "alternate segment theorem" is the reason, and it must not be confused with alternate angles in parallel lines.

Worked example 3 Extended

$A$ , $B$ , $C$ and $D$ are on a circle, centre $O$ . $T A$ is the tangent at $A$ . Angle $T A D = 64^{°}$ and angle $B D C = 30^{°}$ . Find angle $A B D$ , angle $A C D$ , angle $B A C$ and angle $A O D$ , giving a reason each time.

Angle $A B D = 64^{°}$ : alternate segment theorem (tangent $T A$ , chord $A D$ ).
Angle $A C D = 64^{°}$ : angles in the same segment are equal (both stand on chord $A D$ ).
Angle $B A C = 30^{°}$ : angles in the same segment are equal (angles $B A C$ and $B D C$ both stand on chord $B C$ ).
Angle $A O D = 2 \times 64 = 128^{°}$ : the angle at the centre is twice the angle at the circumference (both stand on arc $A D$ ).
Check with the tangent: angle $O A T = 90^{°}$ , so angle $O A D = 90 - 64 = 26^{°}$ ; triangle $O A D$ is isosceles, so angle $A O D = 180 - 2 \times 26 = 128^{°}$ . The two routes agree.

Answer: $64^{°}$ , $64^{°}$ , $30^{°}$ , $128^{°}$

Worked example 4 Extended

$P Q R S$ is a cyclic quadrilateral. Angle $S P Q = 2 y$ , angle $Q R S = 3 y - 20$ . Side $R S$ is extended to $U$ . Find $y$ , and the exterior angle $P S U$ if angle $P Q R = 95^{°}$ .

Angles $S P Q$ and $Q R S$ are opposite, so $2 y + 3 y - 20 = 180$ (opposite angles of a cyclic quadrilateral add up to $180^{°}$ ).
$5 y = 200$ , so $y = 40$ . Angle $S P Q = 80^{°}$ and angle $Q R S = 100^{°}$ .
Angle $P S R$ is opposite angle $P Q R$ , so angle $P S R = 180 - 95 = 85^{°}$ .
$R S U$ is a straight line, so angle $P S U = 180 - 85 = 95^{°}$ . It equals the interior opposite angle $P Q R$ , as it always will.

Answer: $y = 40$ ; angle $P S U = 95^{°}$

Check yourself Extended

$A$ , $B$ , $C$ and $D$ lie on a circle. Angle $A B C = 93^{°}$ . Find angle $A D C$ .

Answer it in the portal, where your answer is checked as you type →

Check yourself Extended

$T A$ is a tangent to a circle at $A$ . $A B$ is a chord, and $C$ is a point on the major arc $A B$ . Angle $T A B = 47^{°}$ . Find angle $A C B$ , and give the reason.

Answer it in the portal, where your answer is checked as you type →

Key points

Radii are equal, so a triangle with two radii as sides is isosceles.
The angle in a semicircle is $90^{°}$ : the right angle is at the point on the circle, opposite the diameter.
The angle between a tangent and a radius is $90^{°}$ , at the point of contact.
Extended: the angle at the centre is twice the angle at the circumference on the same arc.
Extended: angles in the same segment are equal.
Extended: opposite angles of a cyclic quadrilateral add up to $180^{°}$ .
Extended: the angle between a tangent and a chord equals the angle in the alternate segment.
Give each reason in words, using the property as the syllabus states it.

Where marks are lost

Putting the right angle of a semicircle at the centre, or at an end of the diameter. It is at the vertex on the circle.
Using a right angle between a tangent and a line that does not go to the point of contact.
Using Pythagoras with the wrong hypotenuse. In triangle $O P T$ with the tangent at $P$ , the hypotenuse is $O T$ .
Extended: halving when the angle at the centre is asked for, or doubling when the angle at the circumference is asked for.
Extended: using the non-reflex angle at the centre when the angle at the circumference is on the minor arc. The angle on the same arc is reflex.
Extended: setting opposite angles of a cyclic quadrilateral equal. They add up to $180^{°}$ .
Extended: writing "alternate angles" when you mean the alternate segment theorem. They are different facts, and only one of them is about circles.
Giving a reason such as "circle theorem" or "angle at the centre". The examiner wants the full property: "the angle at the centre is twice the angle at the circumference".

Related IGCSE Mathematics 0580 lessons

E4.8

All IGCSE Mathematics 0580 lessons →

An original lesson written to the Cambridge IGCSE Mathematics 0580 syllabus (2025 to 2027). The worked examples and questions are our own, not Cambridge questions. Back to every IGCSE Mathematics 0580 lesson.

Circle theorems I

Circle words you need

The angle in a semicircle

Tangent and radius

Worked example 1

Worked example 2

Check yourself

Check yourself

The angle at the centre Extended

Angles in the same segment Extended

Cyclic quadrilaterals Extended

The alternate segment theorem Extended

Worked example 3 Extended

Worked example 4 Extended

Check yourself Extended

Check yourself Extended

Key points

Where marks are lost

Related IGCSE Mathematics 0580 lessons

Circle theorems II

Angles

Symmetry

Similarity