Circle Geometry – Key Theorems for CCEA IGCSE | IGCSE CCEA 数学：圆周运动考点精讲

📚 Circle Geometry – Key Theorems for CCEA IGCSE | IGCSE CCEA 数学：圆周运动考点精讲

Circle geometry is a major topic in the CCEA IGCSE Mathematics syllabus. It involves a set of angle, chord, tangent and cyclic quadrilateral theorems that allow you to calculate unknown angles, prove relationships and solve multi-step problems. Mastering these theorems and knowing when to apply them is essential for success in both the calculator and non-calculator papers.

圆的几何是 CCEA IGCSE 数学考纲中的一大重点。它涵盖一系列关于角、弦、切线和圆内接四边形的定理，能帮助你计算未知角度、证明几何关系并解决多步应用题。掌握这些定理并学会灵活运用，是通过计算器与非计算器试卷的关键。

1. Angle at the Centre Theorem | 圆心角定理

The angle subtended by an arc at the centre of a circle is twice the angle subtended by the same arc at any point on the remaining circumference. If ∠AOB is at the centre and ∠APB is at the circumference, both standing on the same arc AB, then ∠AOB = 2 × ∠APB.

同一段弧所对的圆心角等于它所对的圆周角的两倍。如果 ∠AOB 是圆心角，∠APB 是同弧 AB 上的圆周角，那么 ∠AOB = 2 × ∠APB。

This theorem is used directly to find missing angles when one of them is known. Always check whether the given angle sits at the centre or on the circumference before applying the relationship.

当已知其中一个角时，可直接使用该定理求未知角。应用前务必确认已知角是圆心角还是圆周角。

2. Angle in a Semicircle | 半圆上的圆周角

An angle inscribed in a semicircle is always a right angle (90°). If A, B and C are points on a circle and AB is a diameter, then ∠ACB = 90°.

半圆所对的圆周角永远是直角（90°）。若 A、B、C 是圆上的点，且 AB 为直径，则 ∠ACB = 90°。

This special case of the angle-at-the-centre theorem is extremely common in CCEA exam questions. Look for diameters or statements like ‘O is the midpoint of AB’ to identify right angles hidden in diagrams.

这是圆心角定理的特例，在 CCEA 试题中极其常见。寻找直径或“O 是 AB 的中点”之类的表述，就能发现图中隐藏的直角。

3. Angles in the Same Segment | 同弧上的圆周角

Angles subtended by the same arc (or chord) in the same segment of a circle are equal. If points P and Q lie on the same side of chord AB, then ∠APB = ∠AQB.

同一段弧（或同一条弦）在圆的同一弓形上所对的圆周角相等。若点 P 和 Q 在弦 AB 的同侧，则 ∠APB = ∠AQB。

This theorem is often used to transfer an angle from one part of the diagram to another. In complex figures, shading the relevant segment can help you spot equal angles quickly.

该定理常用于将角度从一个位置“转移”到另一个位置。在复杂图形中，给相关弓形涂上阴影能帮你快速锁定相等的角。

4. Cyclic Quadrilateral Theorem | 圆内接四边形定理

The opposite angles of a cyclic quadrilateral add up to 180°. If quadrilateral ABCD is cyclic, then ∠A + ∠C = 180° and ∠B + ∠D = 180°.

圆内接四边形的对角互补，即对角之和为 180°。如果四边形 ABCD 内接于圆，则 ∠A + ∠C = 180°，∠B + ∠D = 180°。

You can also use the exterior angle property: an exterior angle of a cyclic quadrilateral equals the interior opposite angle. For example, the exterior angle at A equals ∠C.

你也可以利用外角性质：圆内接四边形的外角等于内对角。例如，顶点 A 处的外角等于 ∠C。

5. Tangent-Radius Theorem | 切线与半径定理

A tangent to a circle is perpendicular to the radius drawn to the point of contact. If line AT is tangent at point A, and O is the centre, then ∠OAT = 90°.

圆的切线垂直于过切点的半径。如果直线 AT 切圆于点 A，O 为圆心，则 ∠OAT = 90°。

This 90° angle is the starting point for many right-triangle trigonometry and Pythagoras problems involving tangents. Always draw the radius to the point of tangency to create a right-angled triangle.

这个 90° 角是许多涉及切线的直角三角形三角学问题和毕达哥拉斯定理问题的起点。务必连接圆心与切点，构造直角三角形。

6. Tangents from an External Point | 圆外一点引两条切线

Two tangents drawn from the same external point to a circle are equal in length. If PA and PB are tangents from point P to the circle, then PA = PB.

从圆外同一点引圆的两条切线长度相等。若 PA 和 PB 是从点 P 引出的两条切线，则 PA = PB。

In addition, the line joining the external point to the centre bisects the angle between the two tangents and the angle between the two radii. This creates two congruent right-angled triangles (ΔOAP ≅ ΔOBP).

此外，连接外部点与圆心的直线平分两切线之间的夹角以及两半径之间的夹角，从而形成两个全等的直角三角形（ΔOAP ≅ ΔOBP）。

7. Alternate Segment Theorem | 弦切角定理

The angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment. If line PAT is a tangent at A, and chord AB divides the circle, then ∠BAT = ∠BCA (angle in the alternate segment).

切线与过切点的弦所夹的角等于该弦在另一弓形中所对的圆周角。若直线 PAT 切圆于点 A，弦 AB 将圆分隔，则 ∠BAT = ∠BCA（弦切角等于其交错弓形上的圆周角）。

This theorem often appears in CCEA questions involving tangents and chords. Identify the chord from the point of tangency, then look for an angle sitting on the opposite arc inside the circle.

该定理常出现在 CCEA 试题中涉及切线和弦的题目里。从切点出发找到弦，再在圆的对面弧上寻找圆周角即可。

8. Perpendicular from Centre to a Chord | 圆心到弦的垂线

The perpendicular from the centre of a circle to a chord bisects the chord. If O is the centre, and OM is perpendicular to chord AB, then AM = MB.

圆心到弦的垂线平分这条弦。若 O 为圆心，OM 垂直于弦 AB，则 AM = MB。

Conversely, the line joining the centre to the midpoint of a chord is perpendicular to the chord. This property links chords with right triangles and can be used to find distances or chord lengths via Pythagoras.

反之，连接圆心与弦的中点的直线垂直于该弦。该性质将弦与直角三角形联系起来，可借助毕达哥拉斯定理求距离或弦长。

9. Equal Chords and Distance from Centre | 等弦与圆心距离

Equal chords are equidistant from the centre. If two chords AB and CD are equal in length, then the perpendicular distances from the centre O to these chords (OM and ON) are equal.

相等的弦到圆心的距离相等。若两弦 AB 与 CD 长度相等，则圆心 O 到这两条弦的垂直距离 OM 和 ON 相等。

Conversely, chords that are equidistant from the centre are equal in length. CCEA questions may ask you to prove chord equality by showing that the perpendicular distances are equal, or vice versa.

反之，到圆心距离相等的弦长度相等。CCEA 考题可能要求你通过证明垂直距离相等来证明两条弦相等，或反过来。

10. Problem-Solving Strategy | 综合解题策略

Typical CCEA circle geometry questions involve multiple theorems in one diagram. Start by marking all clearly stated angles and right angles (tangent-radii, semicircle). Use the angle at the centre, cyclic quadrilateral or same segment theorems to systematically fill in missing angles. Always state the theorem used as a reason in a proof or ‘show that’ question.

典型的 CCEA 圆几何题目会在一张图中涉及多个定理。先从标出所有明确给出的角和直角（切线-半径、半圆）开始。然后利用圆心角、圆内接四边形或同弧定理系统性地推导出未知角。在证明题或“求证”题型中，务必把你所用的定理作为理由写出来。

For problems involving tangents, exploit the 90° radius-tangent angle and the equal tangents from a point to create isosceles or right triangles. When chords are involved, draw the perpendicular from the centre – it always creates symmetry and right triangles.

涉及切线的问题，要充分利用半径与切线的 90° 关系以及等长切线来构造等腰或直角三角形。涉及弦的问题，记得添加圆心到弦的垂线——它总能带来对称性和直角三角形。

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Circle Geometry – Key Theorems for CCEA IGCSE | IGCSE CCEA 数学：圆周运动 考点精讲