<PRE><B>Question 657375</B>
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If the original circle had a radius of *[tex \LARGE x], then the area of the original circle was *[tex \LARGE \pi{x^2}].


The new circle then has a radius of *[tex \LARGE x\ +\ 2], so the area of the new circle is *[tex \LARGE \pi(x\ +\ 2)^2].


We are given that the difference between these two areas is *[tex \LARGE 16\pi], so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \pi(x\ +\ 2)^2\ -\ \pi{x^2}\ =\ 16\pi]


Solve for *[tex \LARGE x]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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