Chord, Secant and Tangent Properties
When two chords cross inside a circle, the products of their parts are equal. When two secants are drawn from a point outside a circle, the products of each whole secant and its outside part are equal, and the square of a tangent from the point equals that product. In NSW Year 10 Advanced maths these results find unknown lengths, often through a quadratic (NESA outcome MA5‑CIR‑P‑01).
You will find missing lengths for intersecting chords, two secants, and a tangent with a secant, and form and solve quadratics such as \(x(x+3)=40\). You will find the radius of a pipe from the width and depth of the water, and use a tangent to find how far the light of a NSW lighthouse can be seen. Harder questions prove the intersecting chords result with similar triangles and the tangent-secant result. It is part of Year 10 Advanced Circle Geometry.
Every question with a fully worked solution.
- Chord, Secant and Tangent Properties - Video - Power Theorems - Chords, Secants & Tangents - Circle Theorems Watch
Theory
When two chords cross, or two lines from one point cut a circle, the products of the intercepts are equal: \(AP \times PB = CP \times PD\) for intersecting chords, and \(PA \times PB = PC \times PD\) or \(PT^2 = PA \times PB\) for secants and tangents from an external point. This Year 10 Advanced maths page shows how to use them, including quadratic equations and practical problems.
A secant is a line that cuts a circle at two points. The parts of a chord or secant between the points of intersection are called intercepts.
For two chords crossing at \(P\) inside the circle, each chord is split into two parts, and the products of the parts are equal.
For two secants from a point \(P\) outside the circle, use the outside part (from \(P\) to the near point) and the whole secant (from \(P\) to the far point). A tangent is the limiting case where the near and far points join, so its product is \(PT \times PT = PT^2\).
All three results come from similar triangles, which is how they are proved.
It comes up if you are heading for Advanced or Extension maths in Years 11 and 12.
Intersecting chords \(AB\) and \(CD\) meeting at \(P\):
Two secants \(PAB\) and \(PCD\) from an external point \(P\):
A tangent \(PT\) and a secant \(PAB\) from an external point \(P\):
How to use the product theorems
- Decide where the lines meet: inside the circle (chords) or outside (secants or a tangent).
- Write each length you need from the meeting point \(P\). For a secant, add the outside part and the chord to get the whole secant.
- Form the equation and solve. If it is quadratic, rearrange to \(= 0\), factorise (or use the formula), and reject a negative or zero length.
- Answer the question asked: often a chord length, which is a difference or sum of the parts.
Problem types
- Parts in a ratio: if \(CP : PD = 2 : 5\), write \(CP = 2k\) and \(PD = 5k\), solve for \(k\) (positive), then find the lengths.
- Expressions on both sides: when both products contain \(x^2\), expanding makes the \(x^2\) terms cancel and leaves a linear equation. Check by substituting back.
- A line through the centre: the secant through \(O\) has parts \(PA\) and \(PA + 2r\). You can check a tangent length with Pythagoras' theorem, since \(OT \perp PT\).
- Two circles sharing a chord: a point on the common chord produced has the same \(PA \times PB\) for both circles, so its tangents to the two circles are equal.
- Combined with angles: some questions mix a tangent with the alternate segment theorem, straight-line angles and the angle at the centre. Find the angles one triangle at a time.
- Proofs: join two pairs of points to make two triangles. Show they are similar (AA) using vertically opposite or common angles plus angles on the same arc (or the alternate segment theorem for a tangent). Then write matching sides in proportion and cross-multiply.
Intersecting chords: the products of the parts are equal.
| \(x(x + 5)\) | \(=\) | \(4 \times 9\) |
| \(x^2 + 5x - 36\) | \(=\) | \(0\) |
| \((x + 9)(x - 4)\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(4 \ \text{or}\ -9\) |
A length cannot be negative, so \(x = \mathbf{4}\) and \(AB = 4 + 9 = \mathbf{13}\).
Check: \(4 \times 9 = 36\) and \(4 \times 9 = 36\).
Secants from an external point: (outside part) \(\times\) (whole secant) is the same for both. Here \(PB = 6 + 10 = 16\).
| \(PC \times PD\) | \(=\) | \(PA \times PB\) |
| \(8 \times PD\) | \(=\) | \(6 \times 16\) |
| \(PD\) | \(=\) | \(12\) |
| \(CD\) | \(=\) | \(12 - 8 = \mathbf{4}\) |
Watch out: \(PA \times AB = 60\) is not the right product. Always multiply the outside part by the whole secant.
The span is a chord. The vertical line through the top of the arch is a diameter, and it bisects the span into \(12\) m and \(12\) m. The diameter is cut into \(4\) and \(2r - 4\).
Intersecting chords (the span and the diameter):
| \(12 \times 12\) | \(=\) | \(4(2r - 4)\) |
| \(144\) | \(=\) | \(8r - 16\) |
| \(8r\) | \(=\) | \(160\) |
| \(r\) | \(=\) | \(\mathbf{20}\text{ m}\) |
Check with Pythagoras' theorem: the centre is \(20 - 4 = 16\) m below the span, and \(16^2 + 12^2 = 400 = 20^2\).
Tangent–secant theorem: the square of the tangent equals the outside part times the whole secant.
| \(PT^2\) | \(=\) | \(PA \times PB\) |
| \(14^2\) | \(=\) | \(7 \times PB\) |
| \(PB\) | \(=\) | \(\dfrac{196}{7} = 28\) |
| \(AB\) | \(=\) | \(28 - 7 = \mathbf{21}\text{ cm}\) |
Common pitfalls
Frequently asked questions
What is the intersecting chords theorem?
If two chords of a circle cross at a point \(P\), the product of the two parts of one chord equals the product of the two parts of the other: \(AP \times PB = CP \times PD\).
How do you use the secant-secant theorem?
From the external point, multiply the outside part of each secant by its whole length. The two products are equal, so you can solve for an unknown length.
What is the tangent-secant theorem?
For a tangent \(PT\) and a secant \(PAB\) from the same external point, \(PT^2 = PA \times PB\), where \(PB\) is the whole secant.
How do you find the radius of a pipe from the water level?
The water surface is a chord and the vertical diameter bisects it. The diameter is split into the water depth and the rest. Half the surface width squared equals the depth times the rest of the diameter, which gives the diameter.
Why are the products of intercepts equal?
Joining the ends of the chords or secants makes two similar triangles. Their matching sides are in proportion, and cross-multiplying the proportion gives the equal products.