Using Circle Theorems

Seeing which part of the circle an angle uses

Circle theorems connect angles and lengths in a circle. Their usefulness comes from recognising the relevant points and lines, even when a diagram contains several overlapping triangles.

A chord joins two points on the circumference. An arc is a curved part of the circumference between two points. A chord divides the circle into two segments, each bounded by the chord and one of the arcs.

Suppose $A$ and $B$ lie on a circle and you join them to another point $P$ on the circumference. The angle $\angle APB$ is said to be subtended by chord $AB$, or by the arc $AB$ that does not contain . In three-letter angle notation, the middle letter is the vertex: is the angle at .

This gives you a reliable way to read a diagram: start at the vertex of an angle and follow its two arms to their endpoints on the circle. Those endpoints tell you which chord and arc the angle uses.

The angle at the centre is twice the angle at the circumference

Let $O$ be the centre, with $A$, $B$ and $P$ on the circumference. If the central angle $\angle AOB$ and the circumference angle subtend the , then

$$\angle AOB=2\angle APB.$$

You double the circumference angle to find the central angle, or halve the central angle to find the circumference angle. For example, a central angle of $112^\circ$ gives a circumference angle of $56^\circ$ on the same arc.

There are two angles between the radii $OA$ and $OB$: a smaller angle and a reflex angle. Choose the one that follows the same arc as the circumference angle. A circumference angle standing on the major arc must be paired with the reflex angle at the centre.

A useful special case occurs when $AB$ is a diameter, a chord passing through the centre. The central angle is then $180^\circ$, so the angle at the circumference is $90^\circ$. This is the theorem .

Angles in the same segment are equal

If $P$ and $Q$ lie on the circumference on the same side of chord $AB$, then

$$\angle APB=\angle AQB.$$

Both angles stand on the same chord and subtend the same arc. Each is half the same central angle, which explains why they are equal.

The angles need not be next to one another, and their arms may cross inside the circle. The centre need not be drawn. To recognise the relationship, trace one angle back to endpoints $A$ and $B$, then look for another angle formed by lines from those same endpoints, with its vertex in the same segment.

Opposite angles in a cyclic quadrilateral

A cyclic quadrilateral has all four vertices on one circle. If its vertices are $A$, $B$, $C$ and $D$ in order around the circumference, its opposite angles satisfy

$$\angle DAB+\angle BCD=180^\circ$$

and

$$\angle ABC+\angle CDA=180^\circ.$$

For example, an interior angle opposite $134^\circ$ is $180^\circ-134^\circ=46^\circ$.

Why does this work? The two opposite angles subtend the two arcs between the other pair of vertices. Those arcs make a complete circle, so their central angles total $360^\circ$. The circumference angles are half those central angles, giving a total of $180^\circ$.

The alternate segment theorem connects a tangent to an angle inside the circle

A tangent touches a circle at exactly one point. Suppose it touches at $T$, and $TA$ is a chord. The angle between the tangent and chord $TA$ equals the angle subtended by in the .

To locate the matching angle, choose a point $B$ on the arc in that opposite segment and join it to both ends of the chord. The matching angle is $\angle TBA$: its arms end at $T$ and $A$. It is not simply any angle near the tangent.

The four arrangements below show which points and lines each theorem connects.

Combining theorems to find an unknown

A circle problem may require you to transfer an angle using one theorem before calculating with another. Ordinary angle facts still apply: angles in a triangle sum to $180^\circ$, angles on a straight line sum to $180^\circ$, and angles around a point sum to $360^\circ$.

For example, consider a circle with centre $O$, a tangent at $T$, and a triangle $TAB$ whose three vertices lie on the circumference. The angle between the tangent and chord is , and the central angle on the same as is .

The alternate segment theorem gives $\angle TBA=(2x+5)^\circ$. The angle at the centre is twice this angle, so

$$5x-2=2(2x+5).$$

Expanding gives $5x-2=4x+10$, hence $x=12$. The tangent–chord angle is therefore $29^\circ$ and the central angle is , consistent with the doubling relationship.

The important connection is that both the transferred circumference angle and the central angle use the endpoints $T$ and $A$.

Radii, chords and tangents provide further deductions

A radius joins the centre to the circumference. All radii of one circle have equal length, so two radii and a chord form an isosceles triangle. Its two base angles are equal, which can help you find angles before applying another circle theorem.

A radius to the point where a tangent touches the circle is perpendicular to the tangent. Thus $OT$ and the tangent at $T$ meet at $90^\circ$. This right angle can be split into smaller angles by a chord through $T$.

Two tangents drawn from the same external point have equal lengths. If tangents from $P$ touch at $T$ and $U$, then $PT=PU$. Joining the centre to $T$ and gives two right-angled triangles with equal radii and a shared hypotenuse .

The perpendicular from the centre to a chord bisects the chord. If $OM$ meets chord $AB$ at $90^\circ$, then $AM=MB$. This splits the isosceles triangle formed by radii , and chord into two matching right-angled triangles. You can therefore use half the chord length in either triangle when applying Pythagoras’ theorem or trigonometry. Equivalently, the of a chord passes through the centre.

Angle relationships

• Centre and circumference: central angle $=2\times$ circumference angle, for the same arc. Check whether the central angle is reflex.
• Semicircle: the angle opposite a diameter is $90^\circ$.