Answer:
x = 4[tex]\sqrt{3}[/tex]
Step-by-step explanation:
Given a tangent and secant 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
x² = 4(4 + 8) = 4 × 12 = 48 ( take the square root of both sides )
x = [tex]\sqrt{48}[/tex] = [tex]\sqrt{16(3)}[/tex] = [tex]\sqrt{16}[/tex] × [tex]\sqrt{3}[/tex] = 4[tex]\sqrt{3}[/tex]
