Question 198771
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The radius of the smaller circle is *[tex \Large r]


The radius of the larger circle is *[tex \Large r + 3]


The circumference of the smaller circle is *[tex \Large 2\pi r]


The circumference of the larger circle is *[tex \Large 2\pi(r + 3) = 2\pi r + 6\pi]


So subtract *[tex \Large 2\pi r + 6\pi - 2\pi r]



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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