Answer:
x = 4, x = 8
Explanation:
(9)
Given 2 secants from an external point to the circle, then
The product of the external part and the whole of one secant is equal to the product of the external part and the whole of the other secant, that is
x(x + 6) = 5(5 + 3)
x² + 6x = 5 × 8 = 40 ( subtract 40 from both sides )
x² + 6x - 40 = 0 ← in standard form
(x + 10)(x - 4) = 0 ← in factored form
Equate each factor to zero and solve for x
x + 10 = 0 ⇒ x = - 10
x - 4 = 0 ⇒ x = 4
x > 0 then x = 4
(11)
Given a secant and a tangent from an external point to the circle, then
The square of the tangent is equal to the product of the external part and the whole of the secant, that is
x(x + 10) = 12²
x² + 10x = 144 ( subtract 144 from both sides )
x² + 10x - 144 = 0 ← in standard form
(x + 18)(x - 8) = 0 ← in factored form
Equate each factor to zero and solve for x
x + 18 = 0 ⇒ x = - 18
x - 8 = 0 ⇒ x = 8
x > 0 then x = 8