Review Solutions — Circle Geometry

1. Circle theorem angles. Fluency

   • (a) Central 112°, inscribed on major arc: 56°.
   • (b) Inscribed 47°, find central: 94°.
   • (c) AB diameter, angle BAC = 28°, find ABC: ACB = 90° ⇒ ABC = 62°.
   • (d) Cyclic ABCD, A = 73°, B = 88°; find C and D: C = 107°, D = 92°.

2. Chords and tangents. Fluency

   • (a) r=15, chord 9 cm from centre: half-chord = √144 = 12. Chord = 24 cm.
   • (b) One tangent = 11 cm; find other: 11 cm (equal tangents).
   • (c) ET=9, EA=3; find EB: 81 = 3×EB ⇒ EB = 27 cm.
   • (d) Chords intersect: AX=4, XC=9, BX=6; find XD: 36 = 6×XD ⇒ XD = 6.

3. Arc length and sector area. Fluency

   • (a) r=9, θ=80°; arc length: 4π ≈ 12.57 cm.
   • (b) r=7, θ=270°; sector area: (147/4)π ≈ 115.45 m².
   • (c) Area=25π, r=10; find θ: 90°.
   • (d) r=6, θ=120°; minor segment area: 12π − 9√3 ≈ 22.11 cm².

4. Mixed calculations. Fluency

   • (a) OP=17, r=8; tangent PT: √225 = 15 cm.
   • (b) r=13, chord 5 cm from centre: 24 cm.
   • (c) Arc=8π, θ=160°; find r: 9 cm.
   • (d) Alternate segment, tangent-chord = 52°: 52°.

5. Angles in a circle (diagram, AOB = 100°). Understanding

   • (a) Angle ACB (C on major arc): 100/2 = 50°. (Inscribed angle = half central angle.)
   • (b) Angle ADB (D on major arc): 50°. (Angles in same segment are equal.)
   • (c) Angle AEB (E on minor arc): Reflex AOB = 260°. AEB = 130°.
   • (d) Angle ADB + angle ACB: Both = 50° (same segment). Sum = 100°.

6. Tangent and chord combined (PA=5, AB=20). Understanding

   • (a) Find PT: √(5×25) = 5√5 ≈ 11.18 cm.
   • (b) Radius (OP=15): √(225−125) = 10 cm.
   • (c) Angle TPO: sin²¹(10/15) ≈ 41.8°.
   • (d) Power of point M (midpoint OP, OM=7.5): r²−OM² = 100−56.25 = 43.75 cm².

7. Sector perimeter and area. Understanding

   • (a) Perimeter 30, r=9; find θ: θ ≈ 76.4°.
   • (b) Area 40, r=8; find θ: θ ≈ 71.6°.
   • (c) Arc=10, area=40; find r and θ: r = 8 cm, θ ≈ 71.6°.
   • (d) Ratio sector to triangle (r=10, θ=60°): 2π/(3√3) ≈ 1.21.

8. Cyclic quadrilateral angles. Understanding

   • (a) A=(x+20)°, C=(2x−5)°; find x: 3x+15=180 ⇒ x = 55. A=75°, C=105°.
   • (b) P:R = 2:3; find both: P = 72°, R = 108°.
   • (c) Opposite angles both 90°; what shape?: Rectangle.
   • (d) Tangent-chord = 65° at D; find ABC: Insufficient information; cannot be determined uniquely from this alone.

9. Radar sweep (r=80 km, θ=140°). Problem Solving

   • (a) Arc length: (560/9)π ≈ 195.48 km.
   • (b) Area one sweep: (44800/18)π ≈ 7817.6 km².
   • (c) Total area scanned 1 min (back-and-forth): Same sector rescanned. Area = 7817.6 km².
   • (d) Aircraft 60 km from A, flying toward A at 900 km/h: Time to reach A = 60/900 h ≈ 4 min.

10. Garden fountain (r=3 m, PT=4 m, PA=2 m). Problem Solving

    • (a) Distance OP: √(9+16) = 5 m.
    • (b) Find PB: 16 = 2×PB ⇒ PB = 8 m.
    • (c) Chord AB: PB − PA = 6 m.
    • (d) Distance from O to chord: √(9−9) = 0 m. The chord is a diameter!

11. Two equal circles (r=5, centres 13 cm apart). Problem Solving

    • (a) Why tangent ⊥ radius: By the tangent-radius theorem, the radius to a tangent point is always perpendicular to the tangent.
    • (b) Common external tangent length (T⊂1;T⊂2;): √(13²−(5−5)²) = √169 = 13 cm.
    • (c) Confirm T⊂1;T⊂2;: 13 cm (equal radii, so difference = 0).
    • (d) Common internal tangent length: √(169−100) = √69 ≈ 8.31 cm.

12. Sector optimisation (P=20 cm). Problem Solving

    • (a) Perimeter formula: 2r + πrθ/180 = 20.
    • (b) θ in terms of r: θ = 180(20−2r)/(πr).
    • (c) Area as function of r: A = 10r − r².
    • (d) Optimal radius: r=5 gives A=25. Maximum at r = 5 cm. θ ≈ 114.6°.