Answer:
The radius of the circle P = 2√10 = 6.325
Step-by-step explanation:
∵ AB is a tangent to circle P at A
∴ (AB)² = BC × BE
∵ BC = 8 , AB = 12 , ED = 6
∵ BE = ED + DC + CB
∴ BE = 6 + CD + 8 = 14 + CD
∴ (12)² = 8 × (14 + DC) ⇒ (12)²/8 = 14 + CD ⇒ CD = (12)²/8 - 14
∴ CD = 4
Join PC and PE (radii)
In ΔBDC and ΔPDE ⇒ ∵ ∠PDC = Ф , ∴ ∠PDE = 180 - Ф
Use cos Rule:
∵ r² = (PD)² + (DC)² - 2(PD)(DC)cosФ
∴ r² = 16 + 16 - 32cosФ = 32 - 32cosФ ⇒ (1)
∵ r² = (PD)² + (DE)² - 2(PD)(DE)cos(180 - Ф) ⇒ cos(180 - Ф) = -cosФ
∴ r² = 16 + 36 + 48cosФ = 52 + 48cosФ ⇒ (2)
∵ (1) = (2)
∴ 32 - 32 cosФ = 52 + 48cosФ
∴ 32 - 52 = 48cosФ + 32cosФ
∴ -20 = 80cosФ
∴ cosФ = -20/80 = -1/4
∴ r² = 32 - 32(-1/4) = 32 + 8 = 40
∴ r = √40 = 2√10 = 6.325
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