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I have two circles, one of which is completely within the other. They do not touch, but are not necessarily concentric. I am given the sum of their circumferences, and the difference in their areas (ie, the area of the space inside the outer circle and outside the inner circle). I need to find the average distance between the circles or, equivalently, the radii of the circles.

In other words, given $C = C_o + C_i$, and $A = A_o - A_i$, find $R_o$ and $R_i$.

I've gotten as far as expressing $A$ in terms of $C_o$ and $C_i$, but then my long-unused algebra fails me. Any help? Thanks...

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What is the average distance between two non-concentric shapes, anyway? – Rahul Jul 31 '12 at 2:51
up vote 3 down vote accepted

Recall the area is $\pi r^2$ and circumference is $2\pi r.$ The sum of circumferences gives you equation: $$ r_1 + r_2 = \frac{C}{2\pi}. \tag{1}$$ The difference of areas gives a second equation: $$ r_1^2 - r_2^2 = \frac{A}{\pi}. \tag{2}$$ Factor LHS of $(2)$ into $(r_1 - r_2)(r_1 + r_2),$ substitute the value of $r_1 + r_2$ from $(1),$ and you will then have 2 linear equations in two unknowns, which you can solve.

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Hint, the two equations are $$ r_1 + r_2 = \frac{C}{2\pi} \\ r_1 - r_2 = \frac{2A}{C}. $$ Add them up to solve for $r_1,$ then subtract or substitute to solve for $r_2.$ – user2468 Jul 31 '12 at 2:45
Finally, it's $$ R_o = \frac{1}{2}\left(\frac{C}{2\pi} + \frac{2A}{C} \right)\\ R_i = \frac{1}{2}\left(\frac{C}{2\pi} - \frac{2A}{C} \right). $$ – user2468 Jul 31 '12 at 2:51
Thanks; it's been a long time since I've used my algebra, but I was able to follow your answer for the most part. I got lost on the equation for r_1 - r_2 (which is really all I need). You came up with 2A/C but, following your instructions (factor LHS of (2), substitute RHS of (1)), I came up with (pi X A )/C. Could you double-check? – user36864 Jul 31 '12 at 3:31
We have $$r_1^2 - r_2^2 = (r_1 + r_2)(r_1 - r_2) = \frac{A}{\pi}$$ and $$r_1 + r_2 = \frac{C}{2\pi}.$$ So $$\frac{C}{2\pi}(r_1 - r_2) = \frac{A}{\pi}.$$ Cancel the $\pi$'s from both sides, and move $\frac{C}{2}$ into the other side, we get $$r_1 - r_2 = \frac{2A}{C}.$$ – user2468 Jul 31 '12 at 4:13
@user36864 feel free to ask follow up questions. – user2468 Jul 31 '12 at 4:15

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