F5 basic properties of circle

(1)(a)

∵ ∠PQR + ∠SRQ
= 90° + 90° (tangent ⊥ radius)
= 180°
∴ PQ // SR (int. ∠s. supp.)

(1)(b)

Let S' be the projection of S on PQ
SS' ² + PS' ² = PS ² (Pyth. theorem)
SS' ² + (8 - 6)² = (8 + 6)² (tangent properties)
SS' ² = 192
SS' = 8√3 cm
thus radius = 8√3 / 2 = 4√3 cm

(1)(c)

Since ∆PBQ ~ ∆RBS (AAA)
QB : BS = 8 : 6 = 4 : 3
QB = 4BS / 3

QS ² = QR ² + SR ² (Pyth. theorem)
QS ² = (8√3)² + 6² = 228
QS = 2√57 cm
QB + BS = 2√57
4BS / 3 + BS = 2√57
BS = 6√57 / 7 cm

In ∆SAB and ∆SPQ,
SA : SP = 6 : 14 = 3 : 7
SB : SQ = (6√57 / 7) : 2√57 = 3 : 7
∠ASB = ∠PSQ (common ∠)
∴∆SAB ~ ∆SPQ (2 sides proportional, int. ∠ equal)

∴∠SAB = ∠SPQ (corr. ∠s of ~∆s)
∴ AB // PQ (corr. ∠s equal)

(1)(d)

AB : 8 = 6 : 14 (corr. sides of ~∆s)
AB = 24 / 7 cm

(2)

Let TA touches O₂ and O₁ at A and E respectively,
and TB touches O₂ and O₁ at B and F respectively,
Let r₁ cm and r₂ cm be the radii of O₁ and O₂ respectively,

O₁E = O₁F = O₁D = DQ = FQ = r₁
O₂A = O₂B = O₂C = CQ = QB = r₂
PC = PA = PQ - CQ = 6 - r₂ (tangent properties)
PE = PD = PQ - DQ = 6 - r₁ (tangent properties)

PT ² = PQ ² + QT ² (Pyth. theorem)
PT ² = 6 ² + 8 ² = 100
PT = 10 cm

TE = TF = PT - PE = 10 - (6 - r₁) = 4 + r₁ (tangent properties)
TQ = TF + FQ => 8 = 4 + r₁ + r₁ => r₁ = 2

O₁F : O₂B = TF : TB (corr. sides of ~∆s)
2 : r₂ = 6 : 8 + r₂
16 + 2r₂ = 6r₂
r₂ = 4

therefore, the radii of O₁ and O₂ are 2cm and 4cm respectively.