Circle Theorems Introduction
Circle theorems establish relationships between angles, arcs, chords, and tangents in a circle. In Class 9, the focus is on the fundamental theorems involving chords, arcs, and angles subtended by chords.
These theorems form the foundation for geometric proofs and construction problems. Understanding circle theorems helps solve problems involving inscribed angles, cyclic quadrilaterals, and tangent properties studied in Class 10.
This topic is part of NCERT Class 9 Mathematics, Chapter: Circles. The key theorems covered here deal with the relationship between the angle at the centre and the angle at the circumference, equal chords, and perpendiculars from the centre.
What is Circle Theorems Introduction?
Key Terms:
- Circle: The set of all points in a plane equidistant from a fixed point (the centre).
- Radius: Distance from the centre to any point on the circle.
- Chord: A line segment whose endpoints lie on the circle.
- Diameter: A chord passing through the centre (longest chord).
- Arc: A continuous part of the circle. The shorter part is the minor arc; the longer part is the major arc.
- Segment: The region between a chord and its arc. Minor segment (smaller), major segment (larger).
- Sector: The region between two radii and their intercepted arc.
- Subtended angle: The angle formed at a point by two lines drawn from the endpoints of an arc or chord.
Central angle: The angle subtended by an arc at the centre.
Inscribed angle: The angle subtended by an arc at a point on the remaining part of the circle.
Circle Theorems Introduction Formula
Key Circle Theorems (Class 9):
Theorem 1:
Equal chords of a circle subtend equal angles at the centre.
Theorem 2 (Converse):
If angles subtended by two chords at the centre are equal, the chords are equal.
Theorem 3:
The perpendicular from the centre of a circle to a chord bisects the chord.
Theorem 4 (Converse):
The line drawn through the centre to bisect a chord is perpendicular to the chord.
Theorem 5:
Equal chords of a circle are equidistant from the centre.
Theorem 6:
The angle subtended by an arc at the centre is double the angle subtended at any point on the remaining circle.
Theorem 7:
Angles in the same segment of a circle are equal.
Theorem 8:
Angle in a semicircle is 90°.
Derivation and Proof
Proof of Theorem 6: The angle subtended by an arc at the centre is double the angle at any point on the remaining circle.
Given: A circle with centre O. Arc PQ subtends ∠POQ at the centre and ∠PRQ at point R on the remaining arc.
To prove: ∠POQ = 2 × ∠PRQ
Construction: Join RO and extend it to point S.
Proof:
- In triangle OPR: OP = OR (radii), so ∠OPR = ∠ORP (base angles of isosceles triangle). Let ∠OPR = ∠ORP = x.
- By exterior angle theorem: ∠POS = x + x = 2x.
- Similarly, in triangle OQR: OQ = OR (radii), so ∠OQR = ∠ORQ. Let ∠OQR = ∠ORQ = y.
- By exterior angle theorem: ∠QOS = y + y = 2y.
- ∠POQ = ∠POS + ∠QOS = 2x + 2y = 2(x + y).
- ∠PRQ = ∠ORP + ∠ORQ = x + y.
- Therefore: ∠POQ = 2 × ∠PRQ. Proved.
Corollary 1: Angles in the same segment are equal (since each equals half the central angle).
Corollary 2: Angle in a semicircle = 90° (since the central angle for a semicircle = 180°, so the inscribed angle = 90°).
Types and Properties
Categories of circle theorem problems:
1. Equal chords and equal angles
- Given equal chords, establish that central angles are equal.
- Or given equal central angles, establish that chords are equal.
2. Perpendicular from centre to chord
- Given the perpendicular distance and chord length, find the radius (using Pythagoras).
- Or given the radius and distance, find the chord length.
3. Central angle = 2 × inscribed angle
- Find the inscribed angle given the central angle, or vice versa.
- Two or more angles inscribed in the same segment are equal.
- Any angle inscribed in a semicircle is 90°.
- Used to establish right angles in geometric constructions.
6. Equal chords equidistant from centre
- If two chords are equal, they are the same distance from the centre.
- Conversely, chords equidistant from the centre are equal.
Solved Examples
Example 1: Example 1: Equal chords and equal angles
Problem: In a circle with centre O, chords AB and CD are equal. If ∠AOB = 70°, find ∠COD.
Solution:
Given:
- AB = CD (equal chords)
By Theorem 1:
- Equal chords subtend equal angles at the centre.
- ∠COD = ∠AOB = 70°
Answer: ∠COD = 70°.
Example 2: Example 2: Perpendicular from centre bisects chord
Problem: A chord of length 24 cm is at a distance of 5 cm from the centre. Find the radius of the circle.
Solution:
Given:
- Chord AB = 24 cm, distance from centre OM = 5 cm
Using the theorem:
- OM ⊥ AB, so M is the mid-point of AB.
- AM = 24/2 = 12 cm
In right triangle OMA:
- OA² = OM² + AM²
- OA² = 5² + 12² = 25 + 144 = 169
- OA = √169 = 13 cm
Answer: Radius = 13 cm.
Example 3: Example 3: Central angle double the inscribed angle
Problem: An arc PQ of a circle subtends an angle of 120° at the centre O. Find the angle subtended by the same arc at a point R on the remaining circle.
Solution:
Given:
- ∠POQ = 120° (central angle)
By Theorem 6:
- ∠POQ = 2 × ∠PRQ
- 120 = 2 × ∠PRQ
- ∠PRQ = 60°
Answer: ∠PRQ = 60°.
Example 4: Example 4: Angle in a semicircle
Problem: AB is a diameter of a circle with centre O. C is a point on the circle. If ∠OCA = 35°, find ∠ACB and ∠OAC.
Solution:
By Theorem 8:
- AB is a diameter, so ∠ACB = 90° (angle in semicircle).
In triangle OCA:
- OC = OA (radii), so triangle OCA is isosceles.
- ∠OCA = ∠OAC = 35°
Finding ∠OCB:
- ∠OCB = ∠ACB − ∠OCA = 90 − 35 = 55°
Answer: ∠ACB = 90°; ∠OAC = 35°; ∠OCB = 55°.
Example 5: Example 5: Angles in the same segment
Problem: In a circle, ∠BAC = 40° and ∠BDC = ? (where A and D are points on the same arc, and BC is a chord).
Solution:
By Theorem 7:
- A and D are in the same segment (same side of chord BC).
- Angles in the same segment are equal.
- ∠BDC = ∠BAC = 40°
Answer: ∠BDC = 40°.
Example 6: Example 6: Finding chord length from radius and distance
Problem: The radius of a circle is 17 cm and the distance from the centre to a chord is 8 cm. Find the length of the chord.
Solution:
Given:
- r = 17 cm, distance = 8 cm
Using Pythagoras theorem:
- Half chord = √(r² − d²) = √(289 − 64) = √225 = 15 cm
- Full chord = 2 × 15 = 30 cm
Answer: Chord length = 30 cm.
Example 7: Example 7: Equal chords equidistant from centre
Problem: Two chords AB and CD of a circle with radius 10 cm are equal and each is 12 cm long. Find their distance from the centre.
Solution:
Given:
- AB = CD = 12 cm, r = 10 cm
Finding distance:
- Half chord = 12/2 = 6 cm
- d = √(r² − (half chord)²) = √(100 − 36) = √64 = 8 cm
By Theorem 5: Equal chords are equidistant from the centre. Both chords are 8 cm from the centre.
Answer: Distance = 8 cm for both chords.
Example 8: Example 8: Finding central angle from inscribed angle
Problem: An inscribed angle ∠PRQ = 55°. Find the central angle ∠POQ.
Solution:
By Theorem 6:
- ∠POQ = 2 × ∠PRQ
- ∠POQ = 2 × 55 = 110°
Answer: ∠POQ = 110°.
Example 9: Example 9: Diameter and inscribed angle
Problem: PQ is a diameter. R is a point on the circle such that ∠RPQ = 28°. Find ∠PRQ and ∠PQR.
Solution:
Given:
- PQ is diameter, ∠RPQ = 28°
By Theorem 8:
- ∠PRQ = 90° (angle in semicircle)
In triangle PRQ:
- ∠RPQ + ∠PRQ + ∠PQR = 180°
- 28 + 90 + ∠PQR = 180
- ∠PQR = 62°
Answer: ∠PRQ = 90°; ∠PQR = 62°.
Example 10: Example 10: Comparing distances of unequal chords
Problem: Two chords of a circle of radius 15 cm are 18 cm and 24 cm long. Which chord is closer to the centre?
Solution:
Distance of chord 1 (18 cm):
- d₁ = √(15² − 9²) = √(225 − 81) = √144 = 12 cm
Distance of chord 2 (24 cm):
- d₂ = √(15² − 12²) = √(225 − 144) = √81 = 9 cm
Comparison: d₂ < d₁, so the 24 cm chord is closer to the centre.
Note: A longer chord is always closer to the centre. The diameter (longest chord) passes through the centre.
Real-World Applications
Applications of Circle Theorems:
- Engineering: Designing circular arches, gears, and pulleys using chord and arc relationships.
- Construction: Finding the centre of a circular object by using perpendicular bisectors of chords.
- Navigation: GPS positioning uses circle theorems (intersection of circles to determine location).
- Optics: Reflection and refraction in curved mirrors involve inscribed angle properties.
- Architecture: Design of domes, circular windows, and rotundas relies on arc and angle relationships.
- Surveying: Determining distances and angles using properties of circles inscribed around triangles.
Key Points to Remember
- Equal chords subtend equal angles at the centre, and vice versa.
- The perpendicular from the centre to a chord bisects the chord.
- The line from the centre bisecting a chord is perpendicular to the chord.
- Equal chords are equidistant from the centre.
- The central angle is double the inscribed angle subtended by the same arc.
- Angles in the same segment of a circle are equal.
- Angle in a semicircle is always 90°.
- The longer the chord, the closer it is to the centre.
- The diameter is the longest chord and passes through the centre.
- To find the radius when chord length and distance are given: r = √(d² + (half chord)²).
Practice Problems
- A chord of length 16 cm is at a distance of 6 cm from the centre. Find the radius.
- The radius of a circle is 25 cm. A chord is at a distance of 7 cm from the centre. Find the length of the chord.
- An arc subtends 140° at the centre. Find the angle it subtends at a point on the remaining circle.
- AB is a diameter. C is on the circle. ∠CAB = 50°. Find ∠ACB and ∠ABC.
- Two chords PQ and RS are equal. If the distance of PQ from the centre is 5 cm, find the distance of RS from the centre.
- Prove that the angle in a semicircle is 90°.
- Two chords of a circle of radius 20 cm are 24 cm and 32 cm. Find their distances from the centre.
- In a circle, ∠BAC = 45° (inscribed). Find ∠BOC (central angle).
Frequently Asked Questions
Q1. What are circle theorems?
Circle theorems are proven statements about relationships between angles, chords, arcs, and other elements of a circle. In Class 9, the key theorems cover equal chords, perpendicular from centre, central vs inscribed angles, same-segment angles, and angle in a semicircle.
Q2. What is the relationship between central angle and inscribed angle?
The angle subtended by an arc at the centre (central angle) is exactly double the angle subtended by the same arc at any point on the remaining circle (inscribed angle).
Q3. Why is the angle in a semicircle 90 degrees?
A semicircle corresponds to a diameter, which subtends a central angle of 180°. By the central-inscribed angle theorem, the inscribed angle = 180/2 = 90°.
Q4. What does 'angles in the same segment are equal' mean?
If two angles are inscribed in the same arc (both on the same side of a chord), they are equal. Both angles subtend the same arc at the centre, so both equal half the central angle.
Q5. How do you find the radius given chord length and distance from centre?
Use the formula: r = √(d² + (half chord)²), where d is the perpendicular distance from the centre to the chord. This follows from the Pythagoras theorem in the right triangle formed.
Q6. What is the longest chord of a circle?
The diameter is the longest chord. It passes through the centre and its length equals 2r.
Q7. Are circle theorems in the CBSE Class 9 syllabus?
Yes. Circle theorems covering chord properties, central and inscribed angles, same-segment angles, and angle in a semicircle are all part of CBSE Class 9 Mathematics, Chapter: Circles.
Q8. What is the difference between a chord and a diameter?
A chord is any line segment with both endpoints on the circle. A diameter is a special chord that passes through the centre. The diameter is the longest possible chord.
Related Topics
- Angle Subtended by a Chord
- Equal Chords and Equal Angles
- Perpendicular from Centre to Chord
- Cyclic Quadrilateral
- Angles in the Same Segment
- Angle in a Semicircle
- Tangent to a Circle
- Tangent is Perpendicular to Radius
- Number of Tangents from a Point
- Tangents from an External Point
- Tangent-Secant Relationship
- Angle Between Two Tangents
- Important Circle Theorem Problems
- Properties of Tangents - Summary
