Question

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 1

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