# 圆的性质

> 爱德思国际GCSE数学A · 4MA1
> 来源: https://www.owlsprep.com/zh/study/edexcel-igcse-math-a-s4-circle-properties/

本指南覆盖爱德思IGCSE数学A（4MA1）要求的全部圆的性质，包括基础层级的弦/切线规则和高阶的圆角定理，配套面向考试的详细例题。

**先修:** [基础角与三角形性质](https://www.owlsprep.com/zh/study/edexcel-igcse-math-a-s3-basic-angle-rules/); [用于求解长度/角度方程的代数知识](https://www.owlsprep.com/zh/study/edexcel-igcse-math-a-s1-algebra-basics/)

## 学习目标

- 识别圆的标准组成部分，包括半径、弦、切线、弧、扇形和弓形
- 应用基础层级的弦与切线性质，通过正确推导计算未知长度
- （仅高阶）使用相交弦规则、圆内接四边形性质以及全部5条圆角定理求解问题
- 为每一步写出明确的几何推导，以获得考试全部分数

## 核心圆定义（全部层级）

所有圆相关问题的起点都是正确识别考试图标注的标准圆组成部分。以下所有术语都是基础和高阶层级都要求熟记的内容。

**圆的标准组成部分** — - 圆心：到圆上所有点距离相等的定点
- 半径：从圆心到圆周任意一点的线段
- 直径：过圆心的弦，长度 = 2 × 半径
- 周长：圆的边界总长度
- 弓形：由弦和弧围成的区域

**例题:** 请标注圆示意图上的W、X、Y、Z部分：W仅与圆有一个交点，X是由两条半径和一段弧围成的区域，Y连接圆周上的两个点，Z是圆周的一段。

1. W是切线：它仅与圆有一个交点
2. X是扇形：由两条半径和一段弧围成
3. Y是弦：连接圆周上的两个点
4. Z是弧：圆周的一段

**考试命令词**

圆相关问题的常见指令术语：

- **Find** — Calculate the missing value, show working and give a reason for each step *(Find the length of the tangent from point P to the circle)*

- **Give a reason** — State the exact circle theorem/property you used *(Give a reason for your answer to part (a))*

*计算器:* allowed

## 基础层级弦与切线性质

这3条性质是基础层级唯一考察的圆规则，也在高阶问题中频繁使用。你必须准确写出对应的性质作为推导依据，才能获得全部分数。

**基础层级圆性质** — 1. 从同一点引出的两条切线长度相等
2. 切线与过切点的半径垂直（夹角为90°）
3. 从圆心向弦作的垂线平分该弦（反之亦然）

**例题:** 从点P向圆心为O、半径为5 cm的圆引一条长度为12 cm的切线。点M是长度为8 cm的弦AB的中点。计算(a) OP的长度，(b) OM的长度。

1. 第(a)部分：切线与过切点的半径垂直，因此三角形OTP（T为切点）为直角三角形。
2. $$OP^2 = OT^2 + TP^2 = 5^2 + 12^2 = 25 + 144 = 169$$
3. OP = \sqrt{169} = 13 cm。推导依据：切线与过切点的半径垂直，勾股定理。
4. 第(b)部分：圆心到弦的垂线平分弦，因此AM = ½ AB = 4 cm。
5. $$OM^2 = OA^2 - AM^2 = 5^2 - 4^2 = 25 - 16 = 9$$
6. OM = 3 cm。推导依据：圆心到弦的垂线平分弦，勾股定理。

> **tip**
>
> 基础层级题目每写出一条正确的推导依据即可获得1分，因此切勿省略你所用性质的说明。

**概念自测**

1. Two tangents are drawn from point X to a circle, touching at Y and Z. If XY = 7 cm, what is XZ?

   *解析:* Correct: Tangents from the same external point are equal in length.

*计算器:* allowed

## 高阶层级相交弦性质

高阶学生需要应用两条相交弦规则，分别适用于在圆内相交的弦，以及在圆外相交的直线。

**相交弦性质** — 1. 圆内相交情况：若两条弦在圆内相交，则一条弦被交点分成的两段长度的乘积等于另一条弦被交点分成的两段长度的乘积。
2. 圆外相交情况：若一条切线和一条割线在圆外相交，则切线长度的平方等于割线的全长与其圆外段长度的乘积。

**例题:** 两条弦AB和CD在圆内交于点E。已知AE = 3 cm，EB = 8 cm，CE = 4 cm，求ED的长度。

1. 应用圆内相交弦性质：AE × EB = CE × ED
2. $$3 × 8 = 4 × ED$$
3. 24 = 4 ED → ED = 6 cm。推导依据：圆内相交弦被交点分成的两段长度的乘积相等。

*计算器:* allowed

## 高阶层级圆角定理

高阶层级共有5个核心圆角定理，全部需要熟记。作答给出推导依据时，必须准确写出对应定理的名称。

**圆角定理** — 1. 圆心角定理：同一段弧对应的圆心角等于2倍的圆周角。
2. 半圆角定理：直径所对的圆周角为90°。
3. 同弓形定理：同一段弧在同一弓形内所对的圆周角相等。
4. 圆内接四边形定理：圆内接四边形的对角之和为180°。
5. 弦切角定理：切线与弦的夹角等于交错弓形内的圆周角。

**例题:** 在圆心为O的圆中，∠AOB = 100°（A、B均在圆周上）。点C位于圆周上，且与O分属弦AB的两侧弓形。(a) 求∠ACB的度数。(b) 若AB为直径，重新求∠ACB的度数。

1. 第(a)问：应用圆心角定理：圆心角的大小是同弧对应圆周角的两倍。
2. $$Angle ACB = \frac{1}{2} × 100 = 50°$$
3. 推导依据：同一段弧所对的圆心角是其对应圆周角的两倍。
4. 第(b)问：AB为直径，因此直径所对的圆周角为90°。
5. ∠ACB = 90°。推导依据：半圆内的圆周角为直角。

> **考试提示**
>
> 弦切角定理是高阶圆定理中最高频的考点，请多练习快速在几何图中识别对应交错弓形的方法。

*计算器:* allowed

## 常见错误

- **错误做法:** Measuring angles/lengths from diagrams labelled 'not to scale'
  - 原因: Exam diagrams are never drawn to scale, so measurements will give incorrect answers, and you will lose all reasoning marks.
  - 正确做法: Only use circle theorems and given values to calculate missing values, and always quote your reasoning.
- **错误做法:** Assuming any radius and tangent form a right angle, not just at the point of contact
  - 原因: The perpendicular rule only applies where the tangent touches the circle at its single point of contact.
  - 正确做法: Only apply the 90° angle rule for tangents and radii at the exact point of contact.
- **错误做法:** (Higher) Using addition instead of multiplication for intersecting chord properties
  - 原因: The rule uses products of segments, not sums, leading to incorrect length calculations.
  - 正确做法: Always multiply the lengths of the two segments of each chord when applying intersecting chord rules.
- **错误做法:** (Higher) Applying cyclic quadrilateral rules to non-cyclic quadrilaterals
  - 原因: The opposite angles sum to 180° rule only applies if all four vertices lie on the circumference of the circle.
  - 正确做法: Confirm the quadrilateral is explicitly labelled as cyclic, or all vertices lie on the circle, before using the cyclic quadrilateral angle rule.
- **错误做法:** Quoting vague reasoning like 'circle rule' instead of the specific theorem name
  - 原因: Exam answers require explicit reasoning to award full marks, vague statements get zero reasoning marks.
  - 正确做法: Always name the exact theorem you used, e.g. 'angle in a semicircle is 90°' instead of 'circle angle rule'.

## 速查表

| Tier | Property / Theorem | Rule | Exam Reasoning Phrase |
| --- | --- | --- | --- |
| Foundation | Equal tangents | Two tangents from same external point are equal length | Tangents from a common external point are equal |
| Foundation | Tangent-radius perpendicularity | Tangent ⊥ radius at point of contact | Tangent is perpendicular to radius at point of contact |
| Foundation | Chord bisector rule | Perpendicular from centre to chord bisects chord | Perpendicular from centre of circle to chord bisects the chord |
| Higher | Internal intersecting chords | AE × EB = CE × ED for intersecting chords AB, CD at E | Product of segments of intersecting chords are equal |
| Higher | Centre angle theorem | Angle at centre = 2 × angle at circumference | Angle subtended at centre is twice angle at circumference |
| Higher | Semicircle angle theorem | Angle subtended by diameter = 90° | Angle in a semicircle is a right angle |
| Higher | Same segment theorem | Angles in same segment equal | Angles subtended by same arc in same segment are equal |
| Higher | Cyclic quadrilateral theorem | Opposite angles sum to 180° | Opposite angles of a cyclic quadrilateral sum to 180° |
| Higher | Alternate segment theorem | Angle between tangent and chord = angle in alternate segment | Alternate segment theorem |

## 下一步

Once you have mastered all circle properties covered in this guide, you are ready to apply them to multi-step geometry problems and mixed topic exam questions. Circle properties are frequently combined with triangle rules (Pythagoras, trigonometry, congruence) and angle facts to make up 5-8 mark extended response questions on both Foundation and Higher tier papers. You should also practice past paper questions to familiarise yourself with common diagram layouts and reasoning phrasing expected by Edexcel examiners. Next, move on to sector and arc mensuration (including perimeter and area calculations for sectors and segments) to complete the geometry unit for your IGCSE Maths exam.

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