Complete notes with theorems, properties, and formula sheet
A circle is a collection of all points in a plane that are at a fixed distance from a fixed point. Circles are one of the most important shapes in geometry with numerous real-life applications.
Circle: Set of all points in a plane at a fixed distance from a fixed point.
Centre: The fixed point from which all points on the circle are equidistant.
Radius: The fixed distance from the centre to any point on the circle.
Diameter: A line segment passing through the centre with endpoints on the circle.
Diameter = 2 × Radius
Equal chords of a circle are equidistant from the centre.
Converse: Chords equidistant from the centre are equal in length.
The perpendicular from the centre of a circle to a chord bisects the chord.
Converse: The line joining the centre to the midpoint of a chord is perpendicular to the chord.
Equal chords of a circle subtend equal angles at the centre.
Converse: If angles subtended by chords at the centre are equal, the chords are equal.
Q: In a circle with centre O, chord AB = 12 cm. The perpendicular distance from O to AB is 8 cm. Find the radius.
Solution:
Let M be the foot of perpendicular from O to AB
OM ⊥ AB, so M bisects AB (by theorem)
AM = ½ × AB = ½ × 12 = 6 cm
In right triangle OMA:
OA² = OM² + AM² (Pythagoras)
r² = 8² + 6²
r² = 64 + 36 = 100
r = 10 cm
The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
∠AOB = 2 × ∠ACB (where C is on the circle)
Angles in the same segment of a circle are equal.
If points P and Q are on the same arc, then ∠APB = ∠AQB
The angle in a semicircle is a right angle (90°).
If AB is a diameter and C is any point on the circle, then ∠ACB = 90°
Q: In a circle with centre O, arc AB subtends 60° at the centre. Find the angle subtended at any point C on the major arc.
Solution:
Angle at centre = 60°
By theorem: Angle at circumference = ½ × angle at centre
∠ACB = ½ × 60°
∠ACB = 30°
A quadrilateral is called cyclic if all four vertices lie on a circle.
The circle is called the circumcircle of the quadrilateral.
The sum of opposite angles of a cyclic quadrilateral is 180°.
∠A + ∠C = 180° and ∠B + ∠D = 180°
Converse: If opposite angles of a quadrilateral sum to 180°, it's cyclic.
The exterior angle of a cyclic quadrilateral equals the opposite interior angle.
Q: ABCD is a cyclic quadrilateral where ∠A = 70°, ∠B = 100°, ∠C = 110°. Find ∠D.
Solution:
In cyclic quadrilateral: opposite angles sum to 180°
∠A + ∠C = 180°
70° + 110° = 180° ✓ (verified)
∠B + ∠D = 180°
100° + ∠D = 180°
∠D = 80°
| Concept | Property | Use |
|---|---|---|
| Equal chords | Equidistant from centre | Finding distances |
| Perpendicular to chord | Bisects the chord | Finding chord length |
| Angle at centre | Double of angle at circumference | Finding angles |
| Semicircle | Angle = 90° | Right angle problems |
| Cyclic quadrilateral | Opposite angles = 180° | Angle calculations |
Diameter = 2r
Circumference = 2πr
Area = πr²
r = radius
Equal chords → Equidistant from centre
Perpendicular from centre → Bisects chord
Equal chords → Equal angles at centre
Fundamental theorems
Angle at centre = 2 × angle at circumference
∠AOB = 2∠ACB
Same arc AB
All angles in same segment are equal
∠APB = ∠AQB = ∠ARB
P, Q, R on same arc
Angle in semicircle = 90°
If AB is diameter, ∠ACB = 90°
Very important!
∠A + ∠C = 180°
∠B + ∠D = 180°
Opposite angles supplementary
Key property
Arc length = (θ/360°) × 2πr
Sector area = (θ/360°) × πr²
θ in degrees
• Draw clear diagrams
• Mark centre and radii
• Use Pythagoras for chords
• Remember 90° in semicircle!
• Always draw a clear diagram with centre marked.
• Mark all radii - they're always equal!
• Look for right triangles formed by perpendiculars to chords.
• Remember: angle at centre = 2 × angle at circumference (most used theorem!).
• Semicircle always gives 90° - memorize this!
• For cyclic quadrilaterals, opposite angles always sum to 180°.
• Use Pythagoras theorem frequently with radius, chord, and perpendicular distance.
• Practice identifying same segments for equal angles.
Q1: A chord of length 16 cm is at distance 6 cm from centre. Find radius.
Solution: Half chord = 8 cm, distance = 6 cm
r² = 8² + 6² = 64 + 36 = 100 → r = 10 cm
Q2: Arc subtends 80° at centre. Find angle at circumference.
Solution: Angle at circumference = ½ × 80° = 40°
Q3: In cyclic quadrilateral, ∠P = 85°, ∠Q = 70°. Find ∠R.
Solution: ∠P + ∠R = 180° → 85° + ∠R = 180° → ∠R = 95°