Here is Prob. 29, in the Additional Problems For Chapter 1, in the book Calculus With Analytic Geometry by George F. Simmons, 2nd edition:  A point moves in such a way that the ratio of its distances from two fixed points is a constant $k \neq 1$. Show that the locus is a circle.  My Attempt:  Let the two fixed points be $Q = (a, b)$ and $R = (c, d)$. If $P = (x, y)$ is any point such that $$ \frac{PQ}{PR} = k, $$ then upon squaring and multiplying out by the denominator we obtain $$ k^2 (x-c)^2 + k^2 (y-d)^2 = (x-a)^2 + (y-b)^2, $$ which we can also rewrite as $$ \left( k^2-1\right) x^2 + \left( k^2-1\right) y^2 - 2\left( k^2 c - a \right) x - 2 \left( k^2 d - b \right) y = a^2 + b^2 -  k^2 \left( c^2 + d^2 \right), $$ and since $k &gt; 0$ and $k \neq 1$, therefore we have $k^2 -1 \neq 0$, and thus upon dividing both sides by $k^2-1$, we obtain $$ x^2 + y^2 - \frac{ 2\left( k^2 c - a \right)  }{ k^2 - 1} x - \frac{ 2 \left( k^2 d - b \right)  }{k^2-1} y = \frac{ a^2 + b^2 -  k^2 \left( c^2 + d^2 \right) }{ k^2-1},  $$ and upon collecting together the terms in $x$ and the terms in $y$, we get $$ x^2 - 2 \frac{ \left( k^2 c - a \right)  }{ k^2 - 1} x  + y^2 - 2 \frac{  \left( k^2 d - b \right)  }{k^2-1} y = \frac{ a^2 + b^2 -  k^2 \left( c^2 + d^2 \right) }{ k^2-1},  $$ and upon completing the two squares on the left-hand-side, we get $$ \begin{align}  &amp; \ \ \ \left( x - \frac{ \left( k^2 c - a \right)  }{ k^2 - 1}  \right)^2 + \left( y - \frac{  \left( k^2 d - b \right)  }{k^2-1} \right)^2 \\  &amp;= \frac{ \left( k^2 c - a \right)^2  }{ \left( k^2 - 1 \right)^2 }  + \frac{  \left( k^2 d - b \right)^2   }{ \left( k^2-1 \right)^2 } + \frac{ a^2 + b^2 -  k^2 \left( c^2 + d^2 \right) }{ k^2-1} \\ &amp;= \frac{ \left( k^2 c - a \right)^2  +  \left( k^2 d - b \right)^2   + \left( k^2-1 \right) \left( a^2 + b^2 \right) -  k^2 \left( k^2-1 \right) \left( c^2 + d^2 \right)   }{ \left( k^2 - 1 \right)^2 } \\ &amp;= \frac{  k^2 \left( 2c^2 + 2d^2 -ac - bd + a^2 + b^2  \right) - k^4 \left( c^2 + d^2 \right) }{ \left( k^2 - 1 \right)^2 } \\ &amp;= \ldots. \end{align} $$ Now the locus of the point $P = (x, y)$ is a circle if and only if $$ \frac{  k^2 \left( 2c^2 + 2d^2 -ac - bd + a^2 + b^2  \right) - k^4 \left( c^2 + d^2 \right) }{ \left( k^2 - 1 \right)^2 } &gt; 0, $$ which is the case if and only if $$ k^2 \left( 2c^2 + 2d^2 -ac - bd + a^2 + b^2  \right) - k^4 \left( c^2 + d^2 \right) &gt; 0. $$  Is what I have done so far correct in each and every detail? If so, how can we show that the last inequality always holds? Or, can we show this for all possible $a, b, c, d, k \in \mathbb{R}$ such that (i) $(a, b) \neq (c, d)$,  and (ii) $k &gt; 0$ and $k \neq 1$? Or, are there any exceptions?
Say I have two points. One at $A = (0, 0)$ and another at $B = (0, 10)$. I wish to derive an equation for a line that, for any point $P$ on the line, would equal a set ratio for the distance $P$ to $A$ and $P$ to $B$.  This is easy if the predetermined ratio for $A:B$ is $1:1$. I would create a line from $A$ to $B$, find its midpoint, then create an orthogonal line at the midpoint. Then at any $P$ on the line I know that the ratio distances from $P$ to $A$ and $B$ will be $1:1$.  But what if the ratio for $A:B$ is something like $11:18$?  This is for a positioning algorithm I am implementing - not for homework :)  EDIT: Note - If an exact formula is not possible, an approximation of the line out to $\pm 200$ from $A$ or $B$ is acceptable as values further from $A$ and $B$ will not be utilized anyway. A linear equation for the final result/approximation is ideal for speed in running multiple computations.
Firefox browser opens but URL is not launched How to prove positive semidefiniteness of two matrices through Schur Complement? Let matrices $X$ and $Y$ be positive semideﬁnite. Show that $X \succeq Y \succ 0$ is equivalent to $Y^{-1} \succeq X^{-1}$.   The teacher tells me that it is easy to prove via the Schur complement. But I could not get his idea. Could anyone help me and give me some intructions? Thanks in advance! I am having a lot of trouble with these series questions. Up until this point, I had relatively little trouble with all the questions in the book. These seem to require knowledge about approximations of functions and other external experience-based knowledge, which I just don't have yet.  Determine convergence or divergence of the given series. In the case of convergence, determine whether the series converges absolutely or conditionally.  $$\sum_{n=1}^\infty (-1)^n\left[e-\left(1+\frac 1 n \right)^n\right]$$  It's easy to see that  $$\lim_{n\to\infty}\left[e-\left(1+\frac 1 n\right)^n\right]=0$$  however, in order to apply Leibniz's Rule and show conditional convergence I need to show that the sequence is monotonically decreasing. This doesn't seem doable with straight inequalities, so I tried taking the derivative, which just resulted in an uninterpretable mess. This doesn't even begin to address the question of absolute convergence/divergence.  There are 54 of these questions... I must be missing something really fundamental if they all take this long. I was having a look at an old question of mine when I browsed the  from the timeline. I scrolled downwards and realized the Review completed panel comes along for the ride. While that might be , tags flowing over top of it as pictured below probably isn't.  Here's a .gif of the Question and Duplicate tabs flowing over it as well. Note that the timeline button also flows over. If vote buttons are present, they will also flow over it.  Reproduced on Chrome and Microsoft Edge. Update Per  by Tiago Martins Peres, this issue also happens in other Review types, namely Late Answers, First Posts and Low Quality Posts. In addition to that, it is causing the action buttons on reviews to become unclickable if you scroll low enough.  As a temporary workaround, you can use 's , which  this issue. I've got a bunch of files named with a the pattern 99 - DescriptiveName.txt and I'd like to remove the number from the front so I just have DescriptiveName.txt.  How can I do this? Can I do it from the command line or is there a utility that can do this? What does `invalid initialization of non-const reference` mean? In Java, are enum types inside a class static? What's the difference between "speak" and "talk", grammatically speaking? There are a number of questions e.g.  and  that deal with the slightly different connotations of the words "speak" and "talk". However, there also seem to be some grammatical differences between the two words. This question is about whether there's a way to formally pin down these differences, or whether they're just contingent features of the two words having different histories.  Some examples of grammatical differences are as follows:   be talkative ... be speakative speak French ... talk French give a talk ... give a speech (as opposed to a speak) speak up ... talk up (the meanings of the two phrases being completely different, with "up" being a preposition only in the latter case) speaking of which ... talking of which grammatically speaking ... grammatically talking   On the other hand, many other constructions work just as well with either (though they might have subtly different meanings), for example   speak to ... talk to speak with ... talk with speak about ... talk about   I'm interested in whether there's a way to pin down these differences (e.g. are the two words classified as different types of verb in some way?), or whether they're essentially just arbitrary. I'm also interested in why we have these two different words with subtly different meanings. Is the distinction between “speak” and “talk” a feature of many languages, or is it just a peculiarity of English?  Etymologically, both words are from Germanic origins. “Talk” seems to have been formed from the Middle English tale, even though “speak” already existed in the English language by then (as far as I can tell). This makes it even more mysterious: Why did we form a new word as a synonym of one that was already established? Or were the meanings different at that time? Does $IJ=IK\implies J=K$ always hold for integral domain and finitely generated nonzero ideal $I$? Let $R$ be a commutative integral domain, $I,J,K$ three ideals of $R$ with $I\neq (0)$ being finitely generated. Then does $IJ=IK$ imply $J=K$?   With Nakayama lemma, I can prove it if one of $J$ and $K$ equals to $R$. And I also know it holds when $R$ is a Prüfer domain or $I$ is singly generated. (Built-in) way in JavaScript to check if a string is a valid number Bug: biblatex can not deal name prefix like "van der ", Miktex updated on 2016.01.16 Firstly, hello everyone. I am new here : )  I am writing my phd thesis using MikTeX + TeXnicCenter and everything goes well, until I have updated MikTeX today.  My Problem:  The program gives an error message to the "same old" command:  \printbibliography[prefixnumbers={},notkeyword={Eigene}, notkeyword={Datenblatt}, title={Fremdliteratur}, heading=subbibliography]   The error message looks like this:  Undefined control sequence. &lt;argument&gt; ...nameprefix {van~der}\isdot }\ifprefchar    Investigation  I have found out that this error comes from the name prefix "van der". If I write them together without space in *.bib file and compile it, it works again.  What's the best way to fix this or work around it? Is it a bug? 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From the depths of space comes these microbes that latch onto electromagnetic fields. Anything electronic is rendered useless. There's the usual stuff where governments try to solve the issue but make it ultimately worse. It's only when they realise that the microbes surround the planet that they realise it is futile. The story closes with people having returned to a simpler life, using medieval farming methods, not using electricity in any form, getting about by horse, etc. I think it was a short story. horizontal alignment of dot over hat over bold greek letter! For some strange reason, this:   \hat{\dot{\bm{\phi}}}   prints the dot and the hat slightly off the left of the phi letter. Any ideas if this can be fixed? thanks. Using bm package for bold symbols. Is it possible to follow a user automatically? bash: ./program: cannot execute binary file: Exec format error My processor is 64-bit - does that mean I need the amd64 image? My processor is an Intel Core 2 Duo P8600 (2.40GHz). 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