# Circle Geometry: Tangents, Chords and Secants

> Tangent and radius, tangents from an external point, the alternate segment theorem, and intersecting chords, secants and tangents.

From HSC Maths Advanced on WhizzNotes, free study notes for NSW students: https://whizznotes.com/mathematics-advanced/tangents-and-secants/. Updated 3 November 2026.
Year 11 and Year 12 boxes go deeper than the Year 10 text around them.

### Story time: The bike chain
Look at a bike chain where it leaves the front gear. It comes off in a perfectly straight line, touching the gear at just one point. That straight part is a **tangent** to the circle of the gear.

Engineers designing belts, pulleys and even running tracks need the tangent rules: a tangent always meets the radius at a right angle, and two tangents from the same point are always the same length. Those two facts work out how long a belt must be.

**In short:** A tangent touches a circle at exactly one point and is perpendicular to the radius there. Two tangents from the same outside point are equal in length, and the alternate segment theorem says the angle between a tangent and a chord equals the angle in the alternate segment.

## Starting point: Years 9 and 10

### Tangent words

- A **tangent** touches a circle at exactly one point, the **point of contact**.
- A **secant** is a line that cuts the circle at two points.

### Tangent theorems

*Diagram: Two tangents from external point P touch the circle at A and B. Each tangent meets the radius at a right angle*
*PA = PB, and each tangent meets its radius at 90°.*

### Key idea: Tangent facts
1. A tangent is **perpendicular to the radius** at the point of contact.
2. The two tangents from an external point are **equal in length**.
3. **Alternate segment theorem:** the angle between a tangent and a chord equals the angle in the alternate segment.

### Worked example: Tangent length
A circle has radius 5 cm. Point $P$ is 13 cm from the centre $O$. The tangent from $P$ meets the radius at $90^\circ$, so its length is $\sqrt{13^2 - 5^2} = 12$ cm.

### Worked example: Alternate segment
A tangent at $A$ makes a $64^\circ$ angle with chord $AB$. Then any angle $ACB$ at the circumference on the other side of $AB$ is also $64^\circ$.

### Intersecting chords, secants and tangents

### Key idea: Three product rules
- **Two chords** $AB$ and $CD$ crossing at $P$ inside the circle: $AP \times PB = CP \times PD$.
- **Two secants** from an outside point $P$, cutting at $A, B$ and $C, D$: $PA \times PB = PC \times PD$ (measure both lengths from $P$).
- **Tangent and secant** from $P$, with tangent $PT$: $PT^2 = PA \times PB$.

### Worked example: Crossing chords
Two chords cross. One is split into 4 cm and 6 cm. The other has one part 3 cm. The other part $x$ satisfies $3x = 24$, so $x = 8$ cm.

### Worked example: Tangent and secant
From $P$, a secant cuts the circle 4 cm and 9 cm from $P$. The tangent from $P$ has $PT^2 = 4 \times 9 = 36$, so $PT = 6$ cm.

### Watch out: Measure from the outside point
For secants, use the whole length $PB$ from $P$ to the far point, not just the part inside the circle.

**Mind map**

```text
Tangents and secants
  Tangent
    Touches once
    Perpendicular to radius
    Two tangents equal
  Alternate segment
    Tangent-chord angle
    Equals angle opposite
  Products
    Chords: AP × PB = CP × PD
    Secants: PA × PB = PC × PD
    Tangent: PT² = PA × PB
```

### Year 11: Tangents to circles on the number plane
The tangent to $x^2 + y^2 = 25$ at $(3, 4)$ is perpendicular to the radius. The radius has gradient $\dfrac{4}{3}$, so the tangent has gradient $-\dfrac{3}{4}$:

$$y - 4 = -\frac{3}{4}(x - 3) \;\Rightarrow\; 3x + 4y = 25$$

A line is a tangent to a circle if substituting it into the circle's equation gives a quadratic with discriminant 0, meaning exactly one meeting point.

### Year 12: Tangents to any curve
For circles, the tangent comes from the radius. For any other curve, the gradient of the tangent at $x = a$ is the derivative $f'(a)$. Check that both methods agree for the top half of the circle, $y = \sqrt{25 - x^2}$:

$$\frac{dy}{dx} = \frac{-x}{\sqrt{25 - x^2}}$$

At $x = 3$ this is $-\dfrac{3}{4}$, the same gradient as above.

## Quick summary

- A tangent touches the circle once and is perpendicular to the radius there.
- Two tangents from one outside point are equal.
- Alternate segment theorem: tangent-chord angle equals the angle in the alternate segment.
- Chords: $AP \times PB = CP \times PD$. Secants: $PA \times PB = PC \times PD$. Tangent: $PT^2 = PA \times PB$.

## Quiz

Try these questions from the chapter quiz.

1. A tangent meets the radius at the point of contact at:
   - A. $45^\circ$
   - B. $90^\circ$
   - C. $180^\circ$
   - D. any angle
   - Answer: B. $90^\circ$. Tangent and radius are perpendicular.

2. A line that cuts a circle at two points is a:
   - A. tangent
   - B. secant
   - C. radius
   - D. diameter only
   - Answer: B. secant. A secant passes through the circle.

3. Radius 6, and $P$ is 10 from the centre. The tangent length from $P$ is:
   - A. 4
   - B. 8
   - C. 16
   - D. $\sqrt{136}$
   - Answer: B. 8. $\sqrt{100 - 36} = 8$.

The whole chapter quiz is on the page, for signed-in students: https://whizznotes.com/mathematics-advanced/tangents-and-secants/#quiz