testing

[imatfaal]

[math] m\left(\sqrt{(x - j)^2 + (y - k)^2}\right) = n\left(\sqrt{(x - r)^2 + (y - s)^2}\right) [/math]

get rid of roots

[math] m^2\left((x - j)^2 + (y - k)^2\right) = n^2\left((x - r)^2 + (y - s)^2\right) [/math]

multiply out brackets

[math] m^2\left((x^2-2jx+j^2) + (y^2-2ky+k^2)\right) = n^2\left((x^2-2rx+r^2) + (y^2-2sy+s^2)\right) [/math]

multiply through

[math] m^2x^2-2m^2jx+m^2j^2 + m^2y^2-2m^2ky+m^2k^2 = n^2x^2-2n^2rx+n^2r^2 + n^2y^2-2n^2sy+n^2s^2 [/math]

gather like terms

[math] m^2x^2 - n^2x^2 + m^2y^2 - n^2y^2 -2m^2jx +2n^2rx -2m^2ky +2n^2sy = +n^2r^2 +n^2s^2 - m^2k^2 - m^2j^2[/math]

factorise a bit

[math] (m^2 - n^2)x^2 + (m^2 - n^2)y^2 -x(2m^2j -2n^2r) -y(2m^2k -2n^2s) = n^2r^2 +n^2s^2 - m^2k^2 - m^2j^2[/math]

this is an equation for a circle - the more general form is this

[math] (x-x_{centre})^2 +(y-y_{centre})^2 = radius^2[/math]

I am convinced I should be able to get one into the other - but at the moment I cannot.

This is why this is all in the sandbox - cos I could be up a gum tree

And thanks for the protip SchrHat - appreciated and implemented.

Edited November 28, 2013 by imatfaal

[Schrödinger's hat]

\left and \right are handy.

so is the preview button

[math] m\left(\sqrt{(x - x_1)^2 + (y - y_1)^2}\right) = n\left(\sqrt{(x - x_2)^2 + (y - y_2)^2}\right) [/math]#

[Tres Juicy]

How can I use this function?

 

If, on the odd occaision I have posted equations I do them using Outlook and copy and paste them

[Schrödinger's hat]

How can I use this function?

 

If, on the odd occaision I have posted equations I do them using Outlook and copy and paste them

 

use the

[math] and [/math]

tags.

Most latex is accepted. There is a tutorial in the maths section.

You can also click on someone's equation and it will show you what is typed to make it (I have noticed it isn't always 100% accurate, but the mistakes I did notice were infrequent/some time ago).

The reply/multiquote buttons will also get you a copy of what someone else typed (eg. for correcting someone else's post), strip out the quote tags and surrounding text and modify the eqn to suit.

 

[math] (m^2 - n^2)x^2 + (m^2 - n^2)y^2 -x(2m^2j -2n^2r) -y(2m^2k -2n^2s) = n^2r^2 +n^2s^2 - m^2k^2 - m^2j^2[/math]

 

this is an equation for a circle - the more general form is this

 

[math] (x-x_{centre})^2 +(y-y_{centre})^2 = radius^2[/math]

 

I am convinced I should be able to get one into the other - but at the moment I cannot.

 

Almost there. Just complete the square(s).

It'll probably get even messier, either live with it for a couple more lines or play around with redefining some variables.

[imatfaal]

 

Almost there. Just complete the square(s).

It'll probably get even messier, either live with it for a couple more lines or play around with redefining some variables.

 

I keep trying to complete the square and get too messed up - I know what it should come out as and just cannot get there

[Schrödinger's hat]

Because I'm bored/curious/out of practise with algebra:

[math] (m^2 - n^2)x^2 + (m^2 - n^2)y^2 -x(2m^2j -2n^2r) -y(2m^2k -2n^2s) = n^2r^2 +n^2s^2 - m^2k^2 - m^2j^2[/math]

Tidy up a bit. Get some stuff we're not interested in on the rhs.

[math] (m -n)(m + n)x^2 + (m -n)(m + n)y^2 -x2j(m^2j - n^2r/j) -y2k(m^2 -n^2s/k) = n^2r^2 +n^2s^2 - m^2k^2 - m^2j^2[/math]

 

[math] (m - n)(m + n)x^2 + (m - n)(m + n)y^2 -x2j(m -n)(m + nr/j) -y2k(m - n)(m + ns/k) = n^2r^2 +n^2s^2 - m^2k^2 - m^2j^2[/math]

 

[math] (m + n)x^2 + (m + n)y^2 -x2(km + nr) -y2(km + ns) = \frac{n^2r^2 +n^2s^2 - m^2k^2 - m^2j^2}{m-n}[/math]

^Edit: Mistake. Have fun.

Now get the quadratics in a nice form:

[math] x^2 + y^2 -x\frac{2(km + nr)}{(m + n)} -y\frac{2(km + ns)}{(m + n)} = \frac{n^2r^2 +n^2s^2 - m^2k^2 - m^2j^2}{m^2-n^2}[/math]

For convenience, set:

[math]a = \frac{-(km + nr)}{(m + n)}[/math]

[math]b = \frac{-(km + ns)}{(m + n)}[/math]

 

Now:

[math] x^2 + y^2 +2ax + 2by = \frac{n^2r^2 +n^2s^2 - m^2k^2 - m^2j^2}{m^2-n^2}[/math]

 

Note that: [math] (x+a)^2 = x^2+2ax + a^2 \rightarrow x^2 + 2ax = (x+a)^2 - a^2[/math]

 

[math] \left(x - \frac{(km + nr)}{(m + n)}\right)^2 + \left(y - \frac{km + ns)}{(m + n)}\right)^2 = \frac{n^2r^2 +n^2s^2 - m^2k^2 - m^2j^2}{m^2-n^2} + \left(\frac{(km + nr)}{(m + n)}\right)^2 + \left(\frac{(km + nr)}{(m + n)}\right)^2 [/math]

 

And you're left with a horrific mess on the RHS which can be simplified, and a k I substituted instead of a j on line 4.

Edited March 8, 2012 by Schrödinger's hat

[Schrödinger's hat]

Lots of that junk on the RHS will cancel down or turn into neatly bracketed multiplication of stuff if you care. Turning it into [math](nr\pm mj)[/math] type of thing before combining the brackets should help

[imatfaal]

damn - I just combined the brackets

[Schrödinger's hat]

damn - I just combined the brackets

 

I meant after that last step you wrote, but at any rate, I feel it is now time for a DON'T CROSS THE STREAMS

Too lazy to find the image, I'll let your imagination do the work.

[imatfaal]

I think I just got slimed.

[Schrödinger's hat]

Also, now that I actually turn my brain on, this seems a lot simpler.

[math] m\left(\sqrt{(x - j)^2 + (y - k)^2}\right) = n\left(\sqrt{(x - r)^2 + (y - s)^2}\right) [/math]

Set some variables:

[math] x' = x-j,\;y' = y-k,\; a = \left(\frac{m}{n}\right)^2\;b=j-r\;c = k-s[/math]

Now:

[math] a\left((x')^2 + (y')^2\right) = (x' + j - r)^2 + (y' + k - s)^2 [/math]

 

[math] a\left((x')^2 + (y')^2\right) = (x' + b)^2 + (y' + c)^2 [/math]

Set:

[math]d=a-1[/math]

And:

[math] d(x')^2 - 2bx' + d(y')^2 - 2cy' = b^2 + c^2 [/math]

[math] (x')^2 + \frac{-2b}{d}x' + (y')^2 + \frac{2c}{d}y' = \frac{b^2 + c^2}{d} [/math]

[math] (x' - \frac{b}{d})^2 + (y' - \frac{c}{d})^2 = \frac{b^2 + c^2}{d} + \left(\frac{b}{d}\right)^2 + \left(\frac{c}{d}\right)^2 [/math]

[math] (dx' - b)^2 + (dy' - c)^2 = db^2 + dc^2 + b^2 + c^2 = a(b^2 + c^2) [/math]

Finally:

[math] \left(x - j - \frac{b}{d}\right)^2 + \left(y - k - \frac{c}{d}\right)^2 = \frac{a}{d^2}(b^2 + c^2) [/math]

Resubbing in variables could be done before dividing by d to simplify things somewhat further.

And finally realise one is doing algebra for entertainment and go to find something productive to do...

Edited March 8, 2012 by Schrödinger's hat

[imatfaal]

I feel very productive - OK so the people who pay for me to sit at this PC might disagree....

 

Thanks all your help and guidance.

[Schrödinger's hat]

May I enquire as to what your profession is such that it entailed doing this calculation by hand?

Or is this merely your version of goofing off?

[Tres Juicy]

[math]x^2=y/3[/math]

 

 

Cool but how do I use a large operator?

Edited March 8, 2012 by Tres Juicy

[Schrödinger's hat]

[math]x^2=y/3[/math]

 

 

Cool but how do I use a large operator?

 

fractions are \frac{}{} with the top in the first parentheses, and the bottom in the second.

Click on the images of the equations to see the source:

[math]
\oint_{s}\left(\iiint\limits_{\mbox{ham}}^{\mbox{fish}^a}  
\psi^{
\left[\begin{array}{ccc}
   	\frac{a}{b}    &\partial_s&\tfrac{\partial}{\partial  \pi}\\
  	1	&2    	&3\\
       4	&5    	&6
\end{array}\right]} \;  
d\mathcal{V}\right) d\mathbf{R}\rightarrow \pm \infty = 
\sqrt{3}[/math]  

Produces

[math] \oint_{s}\left(\iiint\limits_{\mbox{ham}}^{\mbox{fish}^a} \psi^{\left[\begin{array}{ccc}\frac{a}{b}&\partial_s&\tfrac{\partial}{\partial \pi}\\1&2&3\\4&5&6\end{array}\right]} \; d\mathcal{V}\right) d\mathbf{R}\rightarrow \pm \infty = \sqrt{3}[/math]

 

Most things in LaTeX are fairly straightforward. If I don't know how to do something, 4/5 times my first guess will be correct.

There's a tutorial in the maths section, as well as a bunch of examples in the sandbox and maths section (remember, click the images).

The forum often eats your formatting on the raw input. So if it's really complicated, paste it elsewhere (or get used to writing horrible unformatted messy latex ).

[Schrödinger's hat]

Ah, in that case, you can totally get away with changing to a new coordinate system (centred on one of the circles and re-scaled by:

[math]\left(\tfrac{m}{n}\right)^2 - 1[/math]

so it's nice and pretty).

[math]x'=d(x-j),\;y'=d(y-k)[/math]

[math] \left(x' - b \right)^2 + \left(y' - c \right)^2 = a(b^2 + c^2) [/math]

 

Feel free to copy anything I did over at the appropriate time.

Edited March 8, 2012 by Schrödinger's hat

[imatfaal]

Ah, in that case, you can totally get away with changing to a new coordinate system (centred on one of the circles and re-scaled by:

[math]\left(\tfrac{m}{n}\right)^2 - 1[/math]

so it's nice and pretty).

[math]x'=d(x-j),\;y'=d(y-k)[/math]

[math] \left(x' - b \right)^2 + \left(y' - c \right)^2 = a(b^2 + c^2) [/math]

 

Feel free to copy anything I did over at the appropriate time.

 

Yeah - I had almost done that to prove it to myself. Not with scaling but setting origin on one of the points and rotating till both point on an axis - got rid of a load of work.

[Tres Juicy]

fractions are \frac{}{} with the top in the first parentheses, and the bottom in the second.

Click on the images of the equations to see the source:

[math] \oint_{s}\left(\iiint\limits_{\mbox{ham}}^{\mbox{fish}^a} \psi^{ \left[\begin{array}{ccc} \frac{a}{b} &\partial_s&\tfrac{\partial}{\partial \pi}\\ 1 &2 &3\\ 4 &5 &6 \end{array}\right]} \; d\mathcal{V}\right) d\mathbf{R}\rightarrow \pm \infty = \sqrt{3}[/math]

 

[math]6\frac{17}{19}= infinity * \sqrt{3.1}/{9}[/math]

 

[math]21^a}[/math]

Edited March 8, 2012 by Tres Juicy

[Schrödinger's hat]

[math]\infty \mbox{infty} \div \mbox{div} \times \mbox{times} \star \mbox{star} \cdot \mbox{cdot}[/math]

[math]S p\,a\;c\quad e\qquad s[/math]

[math]21^a[/math]

 

It's very finnicky about parentheses, and rarely gives intelligible error messages when you misplace one (I don't think thhe forum gives intelligible error messages at all)

Edited March 8, 2012 by Schrödinger's hat

[Tres Juicy]

[math]9*{21/22}{3}*\sqrt5=Friday![/math]