In the figure, AB=12, BC=8, DE=6, PD=4, and A is the point of tangency. Find the radius of circle P.

Answer:The radius of the circle P = 2√10 = 6.325Step-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 = 4Join PC and PE (radii)In ΔBDC and ΔPDE ⇒ ∵ ∠PDC = Ф , ∴ ∠PDE = 180 - ФUse 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