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For part b, I'm getting a little stuck. So I'm trying to show that if $(x,y,r)R(a,b,s)$ and $(a,b,s)R(x,y,r)$, then $(x,y,r)=(a,b,s)$

And so $(x,y,r)R(a,b,s)$ implies $\sqrt{(x-a)^2+(y-b)^2} \leq s-r$ and $(a,b,s)R(x,y,r)$ implies $\sqrt{(a-x)^2+(b-y)^2} \leq r-s$

After some algebra, here's what I get: For the first one, $x^2+y^2-s^2-2xa-2yb+2sr+b^2+a^2 \leq r^2$ and for the second one, $x^2+y^2-s^2-2ax-2by+2sr+b^2+a^2 \leq r^2$

and so they come out to be the same, but in an example from the notes, I think I should have came up with something like: $a \leq c$ and $c \leq a$, so $c=a$, and I didnt get anything of that form, so I'm wondering if my steps above are also correct. Have I proven anything, or if not, what have I done/assumed wrong?

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    Prime: Tags are meant to be (i) informative and to (ii) call attention to your question from people who are interested or knowledgeable in the topic. If you are *inventing* a tag, then nobody will be "watching" for the tag, so you will fail to achieve the second purpose of tags. Of course, if there are *no* tags that adequately describe your topic, you should create one that is informative; but here, you *did* have a tag ("relations").2010-12-02

3 Answers 3

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The relation is saying that the first circle is completely included inside the second circle. If two circles are completely included in one another, they must be the same circle.

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    @f-Prime Now that I have claimed it, you can try to prove it. It's no well-known property, just something that can be deduced from the definition of circle (all points at distance at most the radius from the center).2010-12-02
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Your two square roots are the same, so $r=s$ follows from the antisymmetry of $\leq$

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Part (a) would give you the intuition. But the first thing to notice here (even before you notice that the square roots have the same value) is that since the square root is always nonnegative, the first condition gives you that $s-r\geq 0$ and the second gives you that $r-s\geq 0$. Those two together tell you what about $s$ and $r$?

Once you have that, you know something about the square root; and since the quantity inside is a sum of squares, the only way that something can happen is if both squares are -fill in the blank-. Which tells you something about $x$, $y$, $a$, and $b$.

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    Oh ok so with all we've done above, we've shown that x=a, y=b, and$r=s$huh?2010-12-02