Year 12 Trigonometry: Trigonometric relations involving circles

Circle-based trigonometry questions look geometric on the surface, but they usually become manageable once the circle is split into one or two triangles. Radii create equal lengths, tangents create right angles with radii, and central angles often control chord lengths. The trigonometry then appears inside a diagram that has more structure than a standard right triangle. The skill is not just knowing a formula. It is seeing where the hidden triangle is and deciding whether a trig ratio, a chord argument, or a symmetry argument is the fastest path.

In a circle question, look for equal radii, right angles from tangents, and central angles that let you split the diagram into congruent triangles.

Main circle patterns

A radius drawn to a tangent is perpendicular to the tangent.
Two radii and a chord often form an isosceles triangle.
Dropping a perpendicular from the centre to a chord bisects the chord.
A central angle can be used to calculate chord lengths with basic trigonometry.

Worked example 1

Problem: A circle has radius

10

cm and a central angle of

60 degrees

. Find the chord length.

Join the centre to the endpoints of the chord to form an isosceles triangle.
Bisect the central angle to create two right triangles with angle $30 degrees$ .
Half the chord is opposite the $30 degrees$ angle, with hypotenuse $10$ .
Use sine: $half chord = 10 \sin (30 degrees) = 5$ .
Therefore the full chord length is $10$ cm.

Answer: the chord length is

10

cm.

Worked example 2

Problem: A chord is

12

cm long in a circle of radius

10

cm. Find the distance from the centre to the chord.

The perpendicular from the centre to the chord bisects the chord, so each half is $6$ cm.
This forms a right triangle with hypotenuse $10$ and one leg $6$ .
Use Pythagoras or trig: $d = sqrt (10^{2} - 6^{2}) = sqrt (64) = 8$ .

Answer: the centre is

8

cm from the chord.

Turning circle geometry into trigonometry

These problems often reward clean construction. A single extra line, such as a radius to a tangent point or a perpendicular from the centre to a chord, can change a messy picture into a standard triangle problem. Once that happens, ordinary sine, cosine, or tangent ratios usually do the rest. This is why many high-scoring solutions start with a diagram annotation rather than immediate algebra.

Worked example 3

Problem: From an external point

P

, a tangent touches a circle with centre

O

T

. If

O P = 13

cm and the radius

O T = 5

cm, find the tangent length

P T

The radius to the tangent point is perpendicular to the tangent, so triangle $OPT$ is right-angled at $T$ .
Use Pythagoras: $P T = sqrt (13^{2} - 5^{2})$ .
Simplify: $P T = sqrt (169 - 25) = sqrt (144) = 12$ .

Answer: the tangent length is

12

cm.

Common traps

Forgetting that the line from the centre to a tangent is perpendicular.
Using the whole chord instead of half the chord after dropping a perpendicular from the centre.
Not labelling equal radii, which hides the isosceles triangle structure.
Jumping into formulas without first improving the diagram.

Revision focus

To revise circle relations well, redraw the same problem with different helpful construction lines and ask which one reveals the triangle fastest. This trains the geometric eye, not just the algebra. Eventually you begin to spot the hidden right triangle or isosceles triangle almost immediately.

It is also useful to describe the geometry in words before calculating. Saying "radius to tangent means ninety degrees" or "perpendicular from centre bisects the chord" often makes the numerical method obvious and prevents wasted time on less direct approaches.

Practice links

Trigonometric relations involving circles