Solutions — Chords and Tangents

1. Perpendicular from centre to chord. Fluency

   • (a) r=10, d=6: half-chord = √(100−36) = √64 = 8. Chord = 16 cm.
   • (b) r=17, chord=16 (half=8): d = √(289−64) = √225 = 15 cm.
   • (c) r=25, chord=14 (half=7): d = √(625−49) = √576 = 24 cm.
   • (d) r=25, d=24 for equal chords: half-chord = √(625−576) = √49 = 7. Chord = 14 cm.

2. Intersecting chords. Fluency

   • (a) AX=3, XC=12, BX=4: 3×12 = 4×XD ⇒ XD = 36/4 = 9.
   • (b) PX=6, XR=8, QX=4: 6×8 = 4×XS ⇒ XS = 48/4 = 12.
   • (c) First chord: 5×9=45. Second: equal split: Let the second chord be split into a and b where a×b = 45. Since the chords are equal in total length (5+9=14), a+b=14 and a×b=45. Solving: a+b=14, ab=45 ⇒ quadratic t²−14t+45=0 ⇒ t=(14±√(196−180))/2=(14±4)/2. So splits: 5 and 9 (same split by symmetry).
   • (d) x(4) = (x+2)(3): 4x = 3x+6 ⇒ x = 6.

3. Tangents from an external point. Fluency

   • (a) Two tangents from P, one=15: Both tangents are equal: other = 15 cm.
   • (b) ET=8, EA=4: 64 = 4×EB ⇒ EB = 16 cm.
   • (c) EA=5, EB=20: ET = √(5×20) = √100 = 10 cm.
   • (d) OP=13, r=5: PT = √(OP²−r²) = √(169−25) = √144 = 12 cm.

4. Radius perpendicular to tangent. Fluency

   • (a) r=8, PT=15: OP = √(8²+15²) = √(64+225) = √289 = 17 cm.
   • (b) OT=6, OP=10: PT = √(100−36) = √64 = 8 cm.
   • (c) Angle 30°, r=5: sin 30° = r/OE ⇒ OE = 5/sin30° = 5/0.5 = 10 cm. PT = √(100−25) = √75 = 5√3 ≈ 8.66 cm.
   • (d) Angle APB=60°, r=7: The angle OAP = 90° (radius ⊥ tangent). In triangle OAP: angle AOP = 90°−30° = 60° (half of angle APB since OA bisects by symmetry). OA/OP = sin30° ⇒ OP = 7/sin30° = 14 cm. PA = √(196−49) = √147 = 7√3 ≈ 12.12 cm. OA = 7 cm (radius).

5. Chord and tangent from a diagram. Understanding

   • (a) Chord AB (r=10, OM=6): Half-chord AM = √(10²−6²) = √(100−36) = √64 = 8 cm. AB = 2×8 = 16 cm.
   • (b) Tangent PT (OP=26, r=10): PT = √(26²−10²) = √(676−100) = √576 = 24 cm.
   • (c) Angle OAM: In right triangle OAM: OA=10 (radius), AM=8 (half-chord), OM=6. sin(OAM) = OM/OA = 6/10 = 0.6 ⇒ angle OAM ≈ 36.9°. Angle AOM = sin²¹(AM/OA)... wait: cos(OAM)=AM/OA gives angle OAB = cos²¹(8/10) = cos²¹(0.8) ≈ 36.9°.
   • (d) Verify PT² = PX × PY (PX=2, XY=23 so PY=25): PT² = 24² = 576. PX×PY = 2×25 = 50. Hmm, 576 ≠ 50 — this would mean the secant is not from the same external point at distance 26. Let us recheck: if OP=26, PT=24, and the secant goes through the circle with near point X and far point Y satisfying PX×PY = PT² = 576. With PX=2: PY=288. XY=286 cm. For PX=2, PY=288 (not 25). The question’s secant values (PX=2, XY=23) are separate illustrative numbers; verify: 2×25=50≠576. Choose consistent values: PX=6, PY=576/6=96, XY=90. Or PX=24 (tangent as degenerate secant): 24×24=576 ✓.

6. Chord distance and equal chords. Understanding

   • (a) Parallel chords 16 and 12 (r=10), same side: d⊂1; = √(100−64) = 6 cm. d⊂2; = √(100−36) = 8 cm. Distance = 8−6 = 2 cm.
   • (b) Opposite sides: Distance = 6 + 8 = 14 cm.
   • (c) Chord 24 (half=12), d=5: r = √(144+25) = √169 = 13 cm.
   • (d) Equal chords 18 (half=9), r=15: d = √(225−81) = √144 = 12 cm.

7. Tangent angle problems. Understanding

   • (a) r=5, OP=13. Angle APB: PA = √(169−25) = 12 cm. In triangle OAP: sin(OPA) = 5/13. Angle OPA = sin²¹(5/13) ≈ 22.6°. Angle APB = 2×22.6 ≈ 45.2°.
   • (b) Angle AOB: Angle AOB + angle APB = 180° (they are supplementary since OAPB is a kite with right angles at A and B). AOB = 180−45.2 ≈ 134.8°.
   • (c) Secant through O: PA=4, AB=12 (so PB=16): PT = √(PA×PB) = √(4×16) = √64 = 8 cm.
   • (d) PT=24, angle=28°: tan 28° = OT/PT... actually angle between tangent and OP: sin28° = r/OP ⇒ OP = r/sin28°. And tan28° = PT/OP... No: angle at P between OP and PT is 28°; OTP = 90°. So sin28° = OT/OP and cos28° = PT/OP. OP = PT/cos28° = 24/cos28° = 24/0.8829 ≈ 27.18 cm. r = OT = OP×sin28° ≈ 27.18×0.4695 ≈ 12.76 cm.

8. Intersecting secants from external point. Understanding

   • (a) EA=4, EB=9, EC=3: 4×9 = 3×ED ⇒ ED = 36/3 = 12.
   • (b) EA=6, EB=24, EC=8: 6×24 = 8×ED ⇒ ED = 144/8 = 18.
   • (c) EA=5, EB=20 (tangent case EC=ED): ET² = 5×20 = 100 ⇒ ET = 10.
   • (d) Same relationship: Both rules express the same power-of-a-point theorem: for any point P and circle, the product PA×PB is constant for all lines through P intersecting the circle at A and B. When P is inside, the products of the two chord segments are equal. When P is outside, the products of the two secant segments (near×far) are equal. The tangent is the limiting case of a secant where both intersection points coincide.

9. Combined chord and tangent. Problem Solving

   • (a) PT in terms of r and OP: PT = √(OP² − r²) = √((d+h)² − r²).
   • (b) Substitute r²=a²+d²: PT = √((d+h)² − (a²+d²)) = √(d²+2dh+h²−a²−d²) = √(2dh+h²−a²).
   • (c) r=5, a=4, h=7: d = √(r²−a²) = √(25−16) = 3 cm. PT = √(2×3×7 + 49 − 16) = √(42+49−16) = √75 = 5√3 ≈ 8.66 cm.
   • (d) PT=a: a² = 2dh + h² − a² ⇒ 2a² = 2dh+h² ⇒ h²+2dh−2a² = 0 ⇒ h = (−2d + √(4d²+8a²))/2 = −d + √(d²+2a²).

10. Cyclic quadrilateral and tangent. Problem Solving

    • (a) Tangent at A makes 40° with AB. Find angle ACB: By the alternate segment theorem: angle ACB = 40°.
    • (b) Central angle AOB: AOB = 2 × ACB = 2 × 40 = 80°.
    • (c) Angle ABC=85°, find angle ADB: ADB = 85°. (Angles in the same segment subtending arc AC are equal.)
    • (d) Angle ADC (opposite to ABC in cyclic quad): ADC = 180 − 85 = 95°.