IGCSE Maths Topic 4 Geometry: Circle Theorems & Angle Properties | IGCSE 数学主题 4：圆定理与角性质

📚 IGCSE Maths Topic 4 Geometry: Circle Theorems & Angle Properties | IGCSE 数学主题 4：圆定理与角性质

Circle theorems are a core part of IGCSE Mathematics Topic 4. They describe the special angle and length relationships that appear whenever points lie on the circumference of a circle, whenever tangents are drawn, and whenever chords intersect. In this revision article, you will work through every theorem you need for the exam, learn how to spot the correct diagram pattern, and practise applying the rules to multi-step problems.

圆定理是 IGCSE 数学主题 4 的核心内容。它们描述了当点位于圆周上、当画出切线、当弦相交时出现的特殊角度和长度关系。在这篇复习文章中，你将逐步掌握考试所需的每一个圆定理，学会识别正确的图形模式，并练习将规则应用到多步骤问题中。

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1. Circle Language and Notation | 圆的基本语言与记号

Before using circle theorems, you must be confident with the basic vocabulary: the radius is a line from the centre to the circumference; a chord is a line segment joining two points on the circumference; a tangent is a straight line that touches the circle at exactly one point; an arc is a part of the circumference between two points; a segment is the region between a chord and an arc; and a sector is the region between two radii and an arc.

在使用圆定理之前，你必须熟悉基本术语：半径是从圆心到圆周的线段；弦是连接圆周上两点的线段；切线是与圆恰好相交于一点的直线；弧是圆周上两点之间的部分；弓形是弦与弧之间的区域；扇形是两条半径与一条弧之间的区域。

• Radius: line from centre O to point A on the circle, written OA.
• Chord: line joining two points A and B on the circle, written AB.
• Tangent: line that touches the circle at exactly one point T.
• Arc: curved part of the circumference between two points, such as arc AB.

• 半径：从圆心 O 到圆上一点 A 的线段，记作 OA。
• 弦：连接圆上两点 A 和 B 的线段，记作 AB。
• 切线：与圆恰好相切于一点 T 的直线。
• 弧：圆周上两点之间的弯曲部分，例如弧 AB。

In IGCSE diagrams, the centre is usually labelled O, and points on the circumference are labelled with capital letters such as A, B, C, and P. Always mark the centre clearly before applying any theorem, because several theorems depend on comparing an angle at the centre with an angle at the circumference.

在 IGCSE 图形中，圆心通常标记为 O，圆周上的点用大写字母如 A、B、C 和 P 标记。应用任何定理之前，一定要清楚地标出圆心，因为多个定理都依赖于将圆心角与圆周角进行比较。

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2. Angle at the Centre Is Twice the Angle at the Circumference | 圆心角等于圆周角的二倍

The first major circle theorem states that 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 circumference. In diagram form, if arc AB subtends angle ∠AOB at the centre and angle ∠APB at a point P on the circumference, then ∠AOB = 2 × ∠APB.

第一个重要圆定理指出，一条弧在圆心处所对的角是该弧在圆周上任意一点处所对角的二倍。用图形表示，如果弧 AB 在圆心处所对的角为 ∠AOB，在圆周上一点 P 处所对的角为 ∠APB，那么 ∠AOB = 2 × ∠APB。

This theorem works whether the centre lies inside the angle at the circumference or outside it. It also tells you that if you know the angle at the centre, you can find the angle at the circumference by dividing by 2, and if you know the angle at the circumference, you can find the angle at the centre by multiplying by 2.

无论圆心是在圆周角的内部还是外部，这个定理都成立。它还告诉你，如果知道圆心角，可以通过除以 2 求出圆周角；如果知道圆周角，可以通过乘以 2 求出圆心角。

∠AOB = 2 × ∠APB

Worked example: In a circle with centre O, points A, B and P lie on the circumference. Given that ∠APB = 35°, find ∠AOB.

例题：在圆心为 O 的圆中，点 A、B 和 P 位于圆周上。已知 ∠APB = 35°，求 ∠AOB。

Solution: Since arc AB subtends ∠APB at the circumference and ∠AOB at the centre, use the centre-angle theorem: ∠AOB = 2 × 35° = 70°.

解答：由于弧 AB 在圆周上对着 ∠APB，在圆心处对着 ∠AOB，使用圆心角定理：∠AOB = 2 × 35° = 70°。

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3. Angle in a Semicircle Is 90° | 半圆内的角是 90°

The angle in a semicircle theorem is a special case of the centre-angle theorem. If AB is a diameter of the circle, then arc AB is a semicircle, so the central angle ∠AOB is 180°. Therefore, any angle ∠APB subtended by the diameter at the circumference must be half of 180°, which is 90°.

半圆内角定理是圆心角定理的一个特例。如果 AB 是圆的直径，那么弧 AB 是一个半圆，因此圆心角 ∠AOB 为 180°。所以，直径在圆周上任意一点所对的角 ∠APB 必定是 180° 的一半，即 90°。

If AB is a diameter, then ∠APB = 90°

This theorem is extremely useful for proving that a triangle is right-angled when all three vertices lie on a circle and one side is a diameter. It is also a quick way to find missing angles in diagrams that contain a semicircle.

这个定理在证明一个三角形是直角三角形时非常有用，当三角形的三个顶点都在圆上且有一条边是直径时。它也是在包含半圆的图形中快速找出缺失角度的一种方法。

Worked example: A, B and C lie on a circle. AB is a diameter. ∠CAB = 28°. Find ∠ABC.

例题：A、B 和 C 位于圆上。AB 是直径。∠CAB = 28°。求 ∠ABC。

Solution: Since AB is a diameter, ∠ACB = 90°. In triangle ABC, the sum of angles is 180°, so ∠ABC = 180° – 90° – 28° = 62°.

解答：因为 AB 是直径，所以 ∠ACB = 90°。在三角形 ABC 中，内角和为 180°，所以 ∠ABC = 180° – 90° – 28° = 62°。

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4. Angles in the Same Segment Are Equal | 同一弓形上的圆周角相等

Angles subtended by the same chord or arc in the same segment of a circle are equal. In diagram form, if points P and Q lie on the same side of chord AB and both angles ∠APB and ∠AQB are subtended by arc AB, then ∠APB = ∠AQB.

同一条弦或同一条弧在同一弓形中所对的圆周角相等。用图形表示，如果点 P 和 Q 位于弦 AB 的同一侧，并且 ∠APB 和 ∠AQB 都由弧 AB 所对，那么 ∠APB = ∠AQB。

∠APB = ∠AQB

This theorem often appears in diagrams with two triangles sharing the same base chord. It is especially helpful when you need to transfer an angle from one part of the circle to another part. Remember that the angles must be in the same segment, meaning they must be on the same side of the chord.

这个定理经常出现在两个三角形共用同一条底弦的图形中。当你需要把一个角从圆的一个部分转移到另一个部分时，它特别有用。请记住，这些角必须位于同一弓形中，也就是说它们必须位于弦的同一侧。

Worked example: Chord AB is drawn in a circle. Point P lies on the major arc AB, and point Q also lies on the major arc AB. If ∠APB = 47°, state the size of ∠AQB and explain why.

例题：在圆中画出弦 AB。点 P 位于优弧 AB 上，点 Q 也位于优弧 AB 上。如果 ∠APB = 47°，说明 ∠AQB 的大小并解释原因。

Solution: ∠AQB = 47° because angles in the same segment are equal. Both angles are subtended by chord AB and lie on the same side of that chord.

解答：∠AQB = 47°，因为同一弓形上的圆周角相等。这两个角都由弦 AB 所对，并且位于该弦的同一侧。

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5. Opposite Angles in a Cyclic Quadrilateral Sum to 180° | 圆内接四边形对角互补

A cyclic quadrilateral is a quadrilateral whose four vertices all lie on the circumference of a circle. The key theorem states that the sum of each pair of opposite angles in a cyclic quadrilateral is 180°. If vertices are labelled A, B, C and D in order on the circle, then ∠A + ∠C = 180° and ∠B + ∠D = 180°.

圆内接四边形是指四个顶点都在同一个圆上的四边形。关键定理指出，圆内接四边形中每一组对角的和都为 180°。如果顶点按顺序标记为 A、B、C 和 D，那么 ∠A + ∠C = 180°，且 ∠B + ∠D = 180°。

∠A + ∠C = 180° and ∠B + ∠D = 180°

This result follows from the centre-angle theorem: each pair of opposite angles is subtended by two arcs that together make up the whole circumference. It is very common in IGCSE questions that ask you to find a missing angle in a four-point circle diagram.

这个结果可以从圆心角定理推出：每一组对角由两段弧所对，而这两段弧合起来正好构成整个圆周。在 IGCSE 题目中，经常会出现要求你找出四点共圆图形中缺失角度的情况。

Worked example: A, B, C and D lie on a circle. ∠A = 105° and ∠C = 75°. State whether ABCD could be a cyclic quadrilateral.

例题：A、B、C 和 D 位于圆上。∠A = 105°，∠C = 75°。判断 ABCD 是否可能是圆内接四边形。

Solution: Yes, because ∠A + ∠C = 105° + 75° = 180°. The opposite angles add to 180°, so the quadrilateral can be cyclic.

解答：可以，因为 ∠A + ∠C = 105° + 75° = 180°。对角之和为 180°，所以这个四边形可以是圆内接四边形。

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6. Tangent and Radius Are Perpendicular | 切线与半径垂直

A tangent to a circle is a straight line that touches the circle at exactly one point. The tangent-radius theorem states that the tangent at any point on the circle is perpendicular to the radius drawn to that point. If T is the point of contact and O is the centre, then OT ⟂ tangent line.

圆的切线是与圆恰好相交于一点的直线。切线-半径定理指出，圆上任意一点处的切线都垂直于经过该点的半径。如果 T 是切点，O 是圆心，那么 OT 垂直于切线。

OT ⟂ tangent at T

This theorem creates right angles that allow you to use Pythagoras’ theorem or trigonometry in problems involving tangents and radii. It also helps to identify right-angled triangles when a tangent is drawn from an external point.

这个定理构造出直角，使你在涉及切线和半径的问题中可以使用勾股定理或三角函数。当从圆外一点画切线时，它也有助于识别直角三角形。

Worked example: A tangent at T touches a circle with centre O. OT = 6 cm and a tangent segment from an external point P to T is 8 cm. Find OP.

例题：一条切线在 T 点与圆心为 O 的圆相切。OT = 6 cm，从圆外一点 P 到 T 的切线长为 8 cm。求 OP。

Solution: Since radius OT is perpendicular to tangent PT, triangle OTP is right-angled at T. Use Pythagoras’ theorem: OP² = OT² + PT² = 6² + 8² = 36 + 64 = 100, so OP = √100 = 10 cm.

解答：因为半径 OT 垂直于切线 PT，所以三角形 OTP 在 T 处为直角。使用勾股定理：OP² = OT² + PT² = 6² + 8² = 36 + 64 = 100，所以 OP = √100 = 10 cm。

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7. Tangent Lengths from the Same External Point Are Equal | 同一点引出的两条切线长相等

If two tangents are drawn from the same external point P to a circle, touching the circle at points T and U, then the lengths of the tangent segments PT and PU are equal. The line PO from the external point to the centre also bisects the angle between the two tangents, so ∠TPO = ∠OPU.

如果从同一个圆外点 P 向圆引两条切线，分别切圆于 T 点和 U 点，那么切线长 PT 和 PU 相等。从圆外点到圆心的线段 PO 还平分两条切线之间的夹角，因此 ∠TPO = ∠OPU。

PT = PU and ∠TPO = ∠OPU

This theorem is often used to form isosceles triangles. Once you know that two tangent segments are equal, you can use base angles in an isosceles triangle to find unknown angles or use lengths in algebraic problems.

这个定理经常用于构造等腰三角形。一旦你知道两条切线长相等，你就可以利用等腰三角形的底角相等来求未知角，或在代数问题中使用长度关系。

Worked example: From point P, two tangents PT and PU touch a circle at T and U. ∠TPU = 52°. Find ∠PTU.

例题：从点 P 引两条切线 PT 和 PU，分别切圆于 T 和 U。∠TPU = 52°。求 ∠PTU。

Solution: PT = PU, so triangle PTU is isosceles with PT = PU. The sum of angles in triangle PTU is 180°, so the two base angles are equal: ∠PTU = ∠PUT = (180° – 52°) ÷ 2 = 128° ÷ 2 = 64°.

解答：PT = PU，所以三角形 PTU 是以 PT = PU 为腰的等腰三角形。三角形 PTU 的内角和为 180°，两个底角相等：∠PTU = ∠PUT = (180° – 52°) ÷ 2 = 128° ÷ 2 = 64°。

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8. Alternate Segment Theorem | 弦切角定理

The alternate segment theorem connects a tangent and a chord drawn at the point of contact. It states that the angle between the tangent and a chord through the point of contact is equal to the angle in the alternate segment of the circle. In diagram form, if tangent PT touches the circle at T and chord TA is drawn, then the angle between tangent PT and chord TA equals the angle subtended by chord TA in the alternate segment, such as ∠TBA where B lies on the other side of chord TA.

弦切角定理将切线与过切点的弦联系起来。它指出，切线与经过切点的弦之间的夹角，等于圆内交替弓形中的角。用图形表示，如果切线 PT 在 T 点与圆相切，并画出弦 TA，那么切线 PT 与弦 TA 之间的夹角等于弦 TA 在交替弓形中所对的角，例如 ∠TBA，其中 B 位于弦 TA 的另一侧。

∠PTA = ∠TBA

This theorem is very powerful but also the most commonly misunderstood. Always check that the angle is formed by a tangent and a chord, and then match it with the angle in the opposite segment on the other side of the chord.

这个定理非常有用，但也是最常被误解的定理。一定要检查这个角是由切线和一条弦构成的，然后将它与弦另一侧相对弓形中的角对应起来。

Worked example: Tangent PT touches a circle at T. Chord TA is drawn. B is a point on the circle such that ∠TBA = 41°. Find the angle between tangent PT and chord TA.

例题：切线 PT 在 T 点与圆相切。画出弦 TA。B 是圆上一点，使 ∠TBA = 41°。求切线 PT 与弦 TA 之间的夹角。

Solution: By the alternate segment theorem, the angle between tangent PT and chord TA is equal to the angle in the alternate segment ∠TBA. Therefore the required angle is 41°.

解答：根据弦切角定理，切线 PT 与弦 TA 之间的夹角等于交替弓形中的角 ∠TBA。因此所求角为 41°。

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9. Chords and Perpendicular Bisectors | 弦与垂直平分线

A line from the centre of a circle that is perpendicular to a chord bisects that chord. Equally, the perpendicular bisector of any chord passes through the centre of the circle. This means that if OM is drawn from centre O to chord AB and OM is perpendicular to AB, then AM = MB.

从圆心出发且垂直于一条弦的直线平分该弦。同样，任何弦的垂直平分线都经过圆心。这意味着如果从圆心 O 到弦 AB 画出 OM，并且 OM 垂直于 AB，那么 AM = MB。

If OM ⟂ AB, then AM = MB

This theorem is extremely useful in problems involving lengths and distances inside a circle. It allows you to create right-angled triangles with the radius as the hypotenuse, which can then be solved by Pythagoras’ theorem.

这个定理在涉及圆内长度和距离的问题中非常有用。它可以让你构造出以半径为斜边的直角三角形，然后利用勾股定理进行求解。

Worked example: A circle has centre O and radius 10 cm. Chord AB is 16 cm long. Find the perpendicular distance from O to AB.

例题：一个圆的圆心为 O，半径为 10 cm。弦 AB 长 16 cm。求从 O 到 AB 的垂直距离。

Solution: Let M be the midpoint of AB. Since OM is perpendicular to AB, AM = MB = 16 ÷ 2 = 8 cm. In right-angled triangle OMA, OA = 10 cm and AM = 8 cm, so OM² = OA² – AM² = 10² – 8² = 100 – 64 = 36, therefore OM = √36 = 6 cm.

解答：设 M 是 AB 的中点。因为 OM 垂直于 AB，所以 AM = MB = 16 ÷ 2 = 8 cm。在直角三角形 OMA 中，OA = 10 cm，AM = 8 cm，所以 OM² = OA² – AM² = 10² – 8² = 100 – 64 = 36，因此 OM = √36 = 6 cm。

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10. Recognising Multi-Step Circle Theorem Problems | 识别多步骤圆定理问题

Most IGCSE exam questions do not test a single circle theorem in isolation. Instead, they combine two or three theorems in one diagram. Your first step should be to mark all given angles on the diagram and label the centre, tangents, diameters, and chords. Then look for which theorem matches the angle you are trying to find.

大多数 IGCSE 考试题目不会孤立地考查单个圆定理。相反，它们在一个图形中综合考查两个或三个定理。你的第一步应该是在图上标出所有给定角度，并标出圆心、切线、直径和弦。然后寻找与所求角相匹配的定理。

Diagram feature	Theorem to use
Diameter present	Angle in a semicircle is 90°
Tangent and radius meet	Radius is perpendicular to tangent
Two tangents from one point	Tangent lengths are equal
Cyclic quadrilateral marked	Opposite angles sum to 180°
Angles in same segment	Equal angles
Tangent and chord meet	Alternate segment theorem

Worked example: In a circle with centre O, A and B lie on the circumference. P is an external point. Tangent PT touches the circle at T, and chord TA is drawn. The angle between tangent PT and chord TA is 38°. Also, AB is a diameter. Find ∠ATB.

例题：在圆心为 O 的圆中，A 和 B 位于圆周上。P 是圆外一点。切线 PT 在 T 点与圆相切，并画出弦 TA。切线 PT 与弦 TA 之间的夹角为 38°。另外，AB 是直径。求 ∠ATB。

Solution: By the alternate segment theorem, ∠TBA = 38° because it is the angle in the alternate segment. Since AB is a diameter, ∠ATB = 90° by the angle in a semicircle theorem. If you also needed a third angle in triangle ABT, the sum would give ∠TAB = 180° – 90° – 38° = 52°.

解答：根据弦切角定理，∠TBA = 38°，因为它是交替弓形中的角。由于 AB 是直径，根据半圆内角定理，∠ATB = 90°。如果你还需要三角形 ABT 中的第三个角，内角和将给出 ∠TAB = 180° – 90° – 38° = 52°。

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11. Common Mistakes and How to Avoid Them | 常见错误与如何避免

The most common error is confusing the angle at the centre theorem with the angle in the same segment theorem. In the centre theorem, one angle is at the centre and the other is at the circumference, so one angle is double the other. In the same segment theorem, both angles are at the circumference and they are equal. Always check whether the vertex of each angle is at the centre or on the circumference before applying a rule.

最常见的错误是将圆心角定理与同一弓形上的圆周角定理混淆。在圆心角定理中，一个角在圆心，另一个角在圆周上，所以一个角是另一个角的两倍。在同一弓形定理中，两个角都在圆周上，并且它们相等。在应用规则之前，一定要检查每个角的顶点是在圆心还是在圆周上。

Another common mistake is assuming that any quadrilateral inside a circle is cyclic. It is only cyclic if all four vertices lie on the circumference. If the centre is inside the quadrilateral but the vertices are not on the circle, the cyclic quadrilateral theorem does not apply.

另一个常见错误是假设圆内的任何四边形都是圆内接四边形。只有当四个顶点都位于圆周上时，它才是

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