# The set of points in 2-space that satisfy a condition

Let u= $\langle X,Y \rangle$ and v= $\langle X_1,Y_1 \rangle$. Describe the set of points $(X,Y)$ in 2-space that satisfy the stated conditions:

$(a)$ ||u - v||$=1$

$(b)$ ||u - v||$≤1$

$(c)$ ||u - v||$>1$

I don't know how to answer these questions. I see that the answer to $(a)$ would be two concentric circles where the difference between their respective radii would be 1, but I don't know how to answer the question. I am even more lost on $(b)$ and $(c)$. Any help would be appreciated. Thank you.

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I'm not sure if you meant for subscripts for v? If my edits are incorrect with respect to what you intended, feel free to undo them. For angle brackets use "\langle X, Y \rangle" (enclosed in dollar signs, of course). Your coordinates inside "< >" were rendered invisible, for some reason. Angle brackets per code: $\langle X, Y \rangle$ vs. $<X, Y>$...see the difference? – amWhy Jul 5 '11 at 21:57
@Theo: Hello!! Good suggestion: using backticks ... Sometimes I think they're overused, but in cases like this, when pointing out (La)TeX)-issues, etc., they're particularly helpful. Thanks! – amWhy Jul 6 '11 at 1:09
I assume $X_1$, $Y_1$ are fixed numbers. Then (a) is tthe circle with center $(X_1,Y_1)$, radius $1$. (b) is same circle plus its interior. (c) is all points outside the same circle. Can't say much more, hard to type in TeX without feedback. – André Nicolas Jul 6 '11 at 5:06

The question is not entirely clear. I will take it to mean, $\bf v$ is given, describe the set of $\bf u$ such that $\|{\bf u}-{\bf v}\|=1$, etc. If my interpretation is incorrect, perhaps OP will clarify and we can make some progress.
So anyway the set of $\bf u$ such that $\|{\bf u}-{\bf v}\|=1$ is a circle of radius 1, centered at $\bf v$. In terms of the $x$, $y$ variables, it's the set of $(x,y)$ such that $(x-x_1)^2+(y-y_1)^2=1$.
I don't blame you for the unclear wording. Do you understand why the set of $\bf u$ such that $\|{\bf u}-{\bf v}\|=1$ is a circle centered at $\bf v$? Do you understand why that leads to the equation $(x-x_1)^2+(y-y_1)^2=1$? Did the comment from user6312 do anything for you? – Gerry Myerson Jul 7 '11 at 1:15