Chapter Overview

What You'll Learn

Key Highlights

Chapter explores circle properties: chords, arcs, angles at center/circumference, and cyclic quadrilaterals. Theorems prove relationships like equal chords equidistant from center, arc angles double at circumference, and cyclic quads summing to 180° opposite angles. Applications include proving equal segments and concyclic points.

Comprehensive Chapter Summary

1. Angle Subtended by a Chord at a Point

• Recall: Circle parts from Class VI; angle subtended by segment PQ at R is ∠PRQ (Fig. 9.1).
• Angles: ∠POQ at center O; ∠PRQ, ∠PSQ at points R, S on major/minor arcs PQ (Fig. 9.2).
• Relationship: Longer chord subtends larger angle at center; draw chords to verify.
• Equal chords: Subtend equal angles at center (Fig. 9.3); measure to confirm.
• Theorem 9.1: Equal chords AB = CD subtend ∠AOB = ∠COD; proof via △AOB ≅ △COD (SSS: OA=OC, OB=OD radii, AB=CD) (Fig. 9.4), CPCT.
• Converse activity: Trace circle, cut equal angles AOB=POQ, segments ACB=PRQ congruent, so AB=PQ (Fig. 9.5).
• Theorem 9.2: Equal angles at center imply equal chords; △AOB ≅ △COD (SSS reverse).
• Exercise 9.1: Prove equal chords of congruent circles subtend equal angles; converse.
• Elaboration: Angles depend on arc size; minor arc subtends acute, major reflex; equal arcs from equal chords ensure symmetry.
• Application: Used in constructions, proving isosceles triangles in circle geometry.

2. Perpendicular from the Centre to a Chord

• Activity: Fold tracing paper circle along perpendicular from O to AB at M; ∠OMA=∠OMB=90°, MA=MB (Fig. 9.6).
• Proof: Join OA, OB; △OMA ≅ △OMB (hypotenuse-leg: OA=OB, AM=BM, right-angled).
• Theorem 9.3: Perpendicular from center bisects chord; general case via congruence.
• Converse: Line from center bisecting chord is perpendicular.
• Theorem 9.4: OM bisects AB at M; prove OM ⊥ AB; △OAM ≅ △OBM (SSS: OA=OB, AM=BM, OM common), CPCT ∠OMA=∠OMB=90° (Fig. 9.7).
• Verification: Test with various chords; always perpendicular.
• Elaboration: Bisector property aids in finding midpoints; useful in circle theorems for symmetry.
• Application: Proves equal segments in intersecting chords.

3. Equal Chords and their Distances from the Centre

• Distance: Perpendicular from point P to line AB is shortest PM (Fig. 9.8); zero if on line.
• Observation: Longer chords nearer center; diameter distance zero.
• Activity: Draw equal chords AB=CD, perpendiculars OM, ON; fold D to B, C to A; OM=ON (Fig. 9.9(i)).
• Repeat with congruent circles; superimpose, O=O', M=N; equal distances.
• Theorem 9.5: Equal chords (congruent circles) equidistant from center(s).
• Converse activity: Draw equal OL=OM inside circle; perpendicular chords PQ ⊥ OL, RS ⊥ OM; PQ=RS (Fig. 9.10).
• Theorem 9.6: Equidistant chords equal length.
• Example 1: Intersecting chords AB, CD at E; diameter PQ through E, ∠AEQ=∠DEQ; prove AB=CD.
• Solution: Perpendiculars OL, OM; ∠LOE=∠MOE (angle sum), △OLE ≅ △OME (AAS), OL=OM (CPCT), so AB=CD (Th. 9.6) (Fig. 9.11).
• Exercise 9.2: Common chord length; equal intersecting chords segments equal; angles with center; concentric circles AB=CD; distances in games; phone strings.
• Elaboration: Distances relate to chord length via Pythagoras: half-chord² + dist² = radius²; equal dist implies equal chords.
• Application: Solves real-world problems like park games, circle intersections.

4. Angle Subtended by an Arc of a Circle

• Arcs: Chord cuts minor/major arcs; equal chords congruent arcs (superimpose CD on AB, Fig. 9.13).
• Property: Equal chords congruent arcs; converse equal chords.
• Angle by arc: Minor ∠POQ, major reflex ∠POQ (Fig. 9.14).
• Result: Congruent arcs equal center angles (Th. 9.1).
• Theorem 9.7: Arc PQ at center ∠POQ = 2∠PAQ at circumference A (Fig. 9.15).
• Proof: Cases minor/semicircle/major; join AO to B; ∠BOQ=∠OAQ+∠AQO (ext. angle), OA=OQ isosceles ∠OAQ=∠OQA, so ∠BOQ=2∠OAQ; similarly ∠BOP=2∠OAP; sum ∠POQ=2∠PAQ.
• Remark: ∠PAQ in segment PAQP.
• Theorem 9.8: Angles in same segment equal (∠PCQ=∠PAQ, Fig. 9.16).
• Semicircle: ∠PAQ=90° (half 180°).
• Converse Th. 9.9: Equal angles same side imply concyclic (draw circle ACB, intersects at E; ∠ACB=∠AEB=∠ADB implies E=D, Fig. 9.17).
• Elaboration: Central angle twice inscribed; same segment equal due to shared arc; semicircle right angle from diameter theorem.
• Application: Proves concyclic points, angle chasing in diagrams.

5. Cyclic Quadrilaterals

• Definition: Quadrilateral ABCD cyclic if vertices on circle (Fig. 9.18).
• Activity: Draw cyclic quads, measure opposite angles; sum ∠A+∠C=180°, ∠B+∠D=180° (table).
• Theorem 9.10: Opposite angles sum 180°.
• Converse Th. 9.11: Sum 180° implies cyclic (similar to Th. 9.9).
• Example 2: AB diameter, CD=radius; AC, BD intersect E; prove ∠AEB=60° (Fig. 9.19).
• Solution: △ODC equilateral ∠COD=60°; ∠CBD=30° (Th. 9.7); ∠ACB=90° (semicircle); ∠BCE=90°, ∠CEB=60°.
• Example 3: Cyclic ABCD, diagonals AC,BD; ∠DBC=55°, ∠BAC=45°; find ∠BCD (Fig. 9.20).
• Solution: ∠CAD=55° (same segment); ∠DAB=100°; ∠BCD=80° (opposite 180°).
• Example 4: Circles intersect A,B; AD,AC diameters; prove B on DC (Fig. 9.21).
• Solution: ∠ABD=∠ABC=90° (semicircle); sum 180°, DBC straight line.
• Example 5: Internal bisectors form cyclic quad (Fig. 9.22).
• Solution: ∠FEH=180° - ½(∠A+∠B); ∠FGH=180° - ½(∠C+∠D); sum 180° (quad sum 360°), cyclic (Th. 9.11).
• Exercise 9.3: ∠ADC; chord=radius angles; ∠OPR; ∠BDC; ∠BAC; ∠BCD with AB=BC; diagonals diameters rectangle; isosceles trapezium cyclic; ∠ACP=∠QCD; circles on sides intersect on third; ∠CAD=∠CBD; cyclic parallelogram rectangle.
• Elaboration: Cyclic property from inscribed angles; converse for verification; applications in proving rectangles, concyclic tests.
• Application: Geometry problems, quadrilateral angle sums.

Key Concepts and Definitions

Important Facts

Questions and Answers from Chapter

Short Questions (1 Mark)

Medium Questions (3 Marks)

Long Questions (6 Marks)

Quick Revision Notes