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The Villarceau circles are things whose existence is surprising. To find radii of Villarceau circles, I stupidly went through a bit of trigonometry and got a much simpler result than I expected, and then realized there was a glaringly obvious way to do it that I hadn't thought of.

In the $xyz$-space imagine a circle of radius $r>0$ in the $xz$-plane, whose center is at a distance $R>r$ from the $z$-axis, and revolve it about the $z$-axis, getting a torus embedded in $\mathbb R^3$. The intersection of that surface with the $xz$-plane is two circles not crossing each other. A line $\ell_1$ touches one of those circles on one side and another on the other side, and that line is in a plane parallel to the $y$-axis, and the intersection of that plane with the torus is the union of two Villarceau circles. So I thought: let's draw a line $\ell_2$ touching both circles on the same side, and the other line $\ell_3$ touching both circles on the same side, and the distance from the intersection of $\ell_1$ with $\ell_2$ to the intersection of $\ell_1$ with $\ell_3$ is the diameter of the Villarceau circle. So I thought: first, the distance from the center to the point of tangency of $\ell_1$ with one circle, is $$\sqrt{R^2-r^2}.\tag{1}$$ Add to that the distance the distance from that point of tangency to the point of intersection of $\ell_1$ with the nearest of those two parallel $\ell$s, and that distance is $$ \frac{r^2}{R+\sqrt{R^2 - r^2}}.\tag{2} $$ So the sum of $(1)$ and $(2)$ involves finding a common denominator, doing some routine cancelations, getting $$ \frac{R\left(R+\sqrt{R^2-r^2}\,\right)}{R+\sqrt{R^2-r^2}} $$ and one more cancellation gives you $R$. Then I realized that the obvious way to see that the radius is $R$ doesn't involve doing any of that.

But I recognized that bit of trivial algebra from a routine calculus problem: $$ \frac{d}{dx} \log\left(x+\sqrt{x^2-1}\,\right). $$ Going through the usual algorithms, you get this down to $$ \frac{x+\sqrt{x^2-1}}{\sqrt{x^2-1}\left(x+\sqrt{x^2-1}\,\right)} $$ and again to one last cancelation to get $\dfrac{1}{\sqrt{x^2-1}}$.

SO MY QUESTION IS: What particular commonality between these two problems causes this same bit of algebra to occur in both places?

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    *Psst. The ([tag:algebra]) tag is no longer being used.*2012-10-14
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    @JenniferDylan : Would you suggest some other tag, or do you prefer to delete it? (And apparently it _is_ still being used, since, at least for now, this question bears it.)2012-10-15
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    +1 for using the algebra tag and taking a stand on the fiasco by which algebra-meaninglesslabels are used as the permanent replacement.2012-10-19
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    I think [tag:algebra-precalculus] is close enough - algebraic manipulations with various expression fall under this tag nicely. But I did not retag, since I don't want to go into retagging war.2012-10-23
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    @zyx Meta is a better place for such discussions - and you can find several discussions about this tag there. I just wanted to say that what you call *fiasco* was done based on community consensus.2012-10-23
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    @MartinSleziak, there was no community consensus, but a series of unilateral actions leading to the *fiasco*. As you say, it is for the meta.2012-10-23
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    @MartinSleziak : I'd rather abolish the "algebra-precalculus" tag. However, supposing its use to be appropriate in some cases, it does not seem appropriate here. If you think otherwise, demonstrate that some methods taught in courses that prepare students to take calculus can suffice to answer this question.2012-10-23
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    I am not sure continuing this discussion can lead to something constructive. (Other than saying that we disagree about the matter.) But in case some of you have something more to say, let's do it on meta or in [chat](http://chat.stackexchange.com/transcript/message/6613076#6613076). They are more suitable for discussions of this type, which are irrelevant to the question.2012-10-24
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    The similarity between the two examples seems really tenuous, and it is unclear why whatever perceived pattern is special and not just something that happens thousands of unrealated places. Is there some similarity that Is supposed to be striking, here?2015-11-23
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    Attempt to integrate the derivative you found at the end. You'll be astonished.2015-12-17
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    @SimpleArt : I don't think I will be astonished since I've seen this done many times.2015-12-17
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    My point is that since a logarithmic function like yours has the same derivative as a trig. function, then that is the connection you probably seek.2015-12-17
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    ok${}\,\ldots\ldots\qquad{}$2015-12-18

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