Circles — Angle and Cyclic Properties
Introduction
Circle geometry is a major component of ICSE Class 10 mathematics. This chapter covers the angle properties of circles, cyclic quadrilaterals, and the relationship between tangents and chords.
Key Theorems
Theorem 1: Angle at the Centre
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 part of the circle.
∠AOB = 2 × ∠APB (where O is the centre and P is any point on the circumference on the opposite side of AB)
Theorem 2: Angle in a Semicircle
The angle subtended by a diameter at the circumference is a right angle.
If AB is a diameter, then ∠APB = 90° for any point P on the circle.
Theorem 3: Angles in the Same Segment
Angles subtended by the same chord (or equal chords) in the same segment of a circle are equal.
∠APB = ∠AQB = ∠ARB (if P, Q, R all lie on the same segment of chord AB)
Cyclic Quadrilateral
A quadrilateral whose vertices all lie on a circle is called a cyclic quadrilateral.
Theorem 4: Opposite Angles of a Cyclic Quadrilateral
The sum of opposite angles of a cyclic quadrilateral is 180°.
∠A + ∠C = 180° and ∠B + ∠D = 180°
Theorem 5: Exterior Angle of a Cyclic Quadrilateral
The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.
∠DCE = ∠BAD (where CE is the extension of side BC)
Tangent-Chord Theorem (Alternate Segment Theorem)
The angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment.
If AT is the tangent at A, then ∠BAT = ∠ACB (angle in the alternate segment).
Worked Examples
Example 1: Angle at Centre
In a circle with centre O, chord AB subtends ∠AOB = 80° at the centre. Find ∠APB where P is on the circumference on the opposite side of AB.
Solution: ∠APB = ¹/₂ × ∠AOB = ¹/₂ × 80° = 40°
Example 2: Angle in a Semicircle
AB is a diameter and C is a point on the circle such that ∠CAB = 35°. Find ∠CBA.
Solution: ∠ACB = 90° (angle in a semicircle) In ΔABC: ∠CAB + ∠CBA + ∠ACB = 180° 35° + ∠CBA + 90° = 180° ∠CBA = 55°
Example 3: Cyclic Quadrilateral
In a cyclic quadrilateral ABCD, ∠A = 70° and ∠B = 110°. Find ∠C and ∠D.
Solution: In a cyclic quadrilateral: ∠A + ∠C = 180° → 70° + ∠C = 180° → ∠C = 110° ∠B + ∠D = 180° → 110° + ∠D = 180° → ∠D = 70°
∠C = 110°, ∠D = 70°
Example 4: Tangent-Chord Theorem
A tangent at A touches the circle at A and ∠ACB = 40°. Find ∠BAT.
Solution: By the alternate segment theorem: ∠BAT = ∠ACB = 40°
Example 5: Complex Cyclic Quadrilateral
In a circle, AB ∥ CD and ABCD is a cyclic quadrilateral. Prove that AC = BD.
Solution: Since AB ∥ CD, ∠BAD + ∠CDA = 180° (co-interior angles) ...(i) In cyclic quadrilateral ABCD, ∠BAD + ∠BCD = 180° ...(ii)
From (i) and (ii): ∠CDA = ∠BCD Since these are equal angles in the same segment, arc BC = arc AD Therefore chord BC = chord AD, i.e., AC = BD. ✓
Comparison Table
| Property | Theorem | Statement |
|---|---|---|
| Angle at centre | ∠AOB = 2∠APB | Central angle is double the inscribed angle |
| Semicircle | ∠APB = 90° | Any angle in a semicircle is right |
| Same segment | ∠APB = ∠AQB | Angles in same segment are equal |
| Cyclic quad | ∠A + ∠C = 180° | Opposite angles sum to 180° |
| Tangent-chord | ∠BAT = ∠ACB | Angle between tangent and chord equals angle in alternate segment |
Common Mistakes and Fixes
| Mistake | Fix |
|---|---|
| Confusing angle at centre and circumference | Centre angle = 2 × circumference angle (for same arc) |
| Forgetting the 180° condition for cyclic quadrilaterals | Opposite angles sum to 180°, NOT adjacent |
| Applying angle in semicircle without diameter condition | Only works if AB is a diameter |
| Misidentifying the alternate segment | The alternate segment is opposite to the tangent-chord angle |
ICSE Exam Focus
This is a high-weightage topic carrying 12–16 marks in ICSE exams. Questions include:
- Direct application of angle properties.
- Cyclic quadrilateral problems.
- Proof-based questions (5–6 marks).
- Combined problems with tangents and chords.
Marks Blueprint:
| Topic | Marks |
|---|---|
| Angle at centre / semicircle | 3 |
| Cyclic quadrilateral properties | 4 |
| Tangent-chord theorem | 3 |
| Proof-based questions | 4–6 |
| Combined problems | 4 |
Self-Test Questions
-
In a circle, chord AB subtends 60° at a point on the circumference. Find the angle subtended at the centre.
-
In a cyclic quadrilateral PQRS, ∠P = 85° and ∠Q = 95°. Find ∠R and ∠S.
-
Prove that the angle in a semicircle is a right angle.
-
A tangent at T touches the circle at T and chord TP subtends ∠TQP = 50° in the alternate segment. Find ∠PTQ.
-
In the given figure, O is the centre and ∠AOB = 100°. Find ∠APB and ∠AQB where P and Q are points on the circle on opposite arcs.
-
Show that a parallelogram inscribed in a circle must be a rectangle.
In ICSE, circle theorems are best learned by drawing clear diagrams. Mark all known angles and use the theorems step-by-step to derive unknown angles.
