Answer:
B. x= 4
Step-by-step explanation:
We know that,
The result for the intersection of the secant and the tangent is given by by the figure below,
[tex]UV^{2}=UX\times UY[/tex]
According to the question, we have,
[tex]8^{2}=x(12+x)[/tex]
i.e. [tex]64=12x+x^2[/tex]
i.e. [tex]x^2+12x-64=0[/tex]
Now, the solution of a quadratic equation [tex]ax^2+bx+c=0[/tex] is given by, [tex]x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}[/tex]
So, we get,
[tex]x^2+12x-64=0[/tex] implies a= 1, b= 12 and c= -64.
Thus, the solution is given by,
[tex]x=\frac{-12\pm \sqrt{12^{2}-4\times 1\times (-64)}}{2\times 1}[/tex]
i.e. [tex]x=\frac{-12\pm \sqrt{144+256}}{2}[/tex]
i.e. [tex]x=\frac{-12\pm \sqrt{400}}{2}[/tex]
i.e. [tex]x=\frac{-12\pm 20}{2}[/tex]
i.e. [tex]x=\frac{-12-20}{2}[/tex] and i.e. [tex]x=\frac{-12+20}{2}[/tex]
i.e. [tex]x=\frac{-32}{2}[/tex] and i.e. [tex]x=\frac{8}{2}[/tex]
i.e x= -16 and x= 4.
Since, length cannot be negative.
So, we have, x= 4.
