Help solving equations

**quicksilver** · June 1st 2009, 07:03 AM

I have these equations I was working on that although seemed simple, turned out to be deceptively hard.

$ r^2/s_1^4 = (x-x_1)^2 + (y-y_1)^2 $
$ r^2/s_2^4 = (x-x_2)^2 + (y-y_2)^2 $
$ r^2/s_3^4 = (x-x_3)^2 + (y-y_3)^2 $

I need to solve for r, x, and y in terms of s1, s2, s3, x1, x2, x3, y1, y2, y3.

I tried solving by hand and was able to get it down to just one equation with one unknown but the equation was so complicated I could not figure out how to solve it. Perhaps my approach was not correct, could anyone give me some pointers?

Thanks!

**VonNemo19** · June 1st 2009, 07:57 AM

Originally Posted by quicksilver

I have these equations I was working on that although seemed simple, turned out to be deceptively hard.

r^2/s1^4 = (x-x1)^2 + (y-y1)^2
r^2/s2^4 = (x-x2)^2 + (y-y2)^2
r^2/s3^4 = (x-x3)^2 + (y-y3)^2

I need to solve for r, x, and y in terms of s1, s2, s3, x1, x2, x3, y1, y2, y3.

I tried solving by hand and was able to get it down to just one equation with one unknown but the equation was so complicated I could not figure out how to solve it. Perhaps my approach was not correct, could anyone give me some pointers?

Thanks!

Yeah. If all you need to do is solve for r, x, and y it's easy:
$r=\pm{\sqrt{s_1^4[(x-x_1)^2+(y-y_1)^2]}}$

I think you can do the rest. Hint:

When solving for y, subtrat $(y-y_1)^2+\frac{r^2}{s_1^4}$ from both sides. Now take roots....

**quicksilver** · June 1st 2009, 08:51 AM

We can easily remove the r from the first equation and simplify it to two equations.

$ s_1^4*((x-x_1)^2+(y-y_1)^2) = s_2^4*((x-x_2)^2+(y-y_2)^2) $
$ s_1^4*((x-x_1)^2+(y-y_1)^2) = s_3^4*((x-x_3)^2+(y-y_3)^2) $

We can take the first equation above and solve for Y:
$ s_1^4*(x^2-2x*x_1+x_1^2+y^2-2y*y_1+y_1^2) = s_2^4*(x^2-2x*x_2+x_2^2+y^2-2y*y_2+y_2^2) $

s1^4*x^2-s1^4*2x*x1+s1^4*x1^2+s1^4*y^2-s1^4*2y*y1+s1^4*y1^2 = s2^4*x^2-s2^4*2x*x2+s2^4*x2^2+s2^4*y^2-s2^4*2y*y2+s2^4*y2^2
s1^4*y^2-s1^4*2y*y1-s2^4*y^2+s2^4*2y*y2 = s2^4*x^2-s2^4*2x*x2+s2^4*x2^2+s2^4*y2^2-s1^4*x^2+s1^4*2x*x1-s1^4*x1^2+s1^4*2y*y1-s1^4*y1^2
(s1^4-s2^4)*y^2+(2*s2^4*y1-2*s1^4*y2)*y = s2^4*(x^2-2x*x2+x2^2+y2^2) - s1^4(x^2-2x*x1+x1^2+y1^2)
(s1^4-s2^4)*y^2+(2*s2^4*y1-2*s1^4*y2)*y + (s1^4(x^2-2x*x1+x1^2+y1^2) - s2^4*(x^2-2x*x2+x2^2+y2^2)) = 0
y = (2*s1^4*y2-2*s2^4*y1 \pm{\sqrt{(2*s2^4*y1-2*s1^4*y2)^2-4*(s1^4-s2^4)*s1^4(x^2-2x*x1+x1^2+y1^2) - s2^4*(x^2-2x*x2+x2^2+y2^2)}})/(2s1^4-2s2^4)

We can solve for Y of the second equation the same way giving us:

y = (2*s1^4*y3-2*s3^4*y1 \pm{\sqrt{(2*s3^4*y1-2*s1^4*y3)^2-4*(s1^4-s3^4)*s1^4(x^2-2x*x1+x1^2+y1^2) - s3^4*(x^2-2x*x3+x3^2+y3^2)}})/(2s1^4-2s3^4)

By combining these two equations we get:

(2*s1^4*y2-2*s2^4*y1 \pm{\sqrt{(2*s2^4*y1-2*s1^4*y2)^2-4*(s1^4-s2^4)*s1^4(x^2-2x*x1+x1^2+y1^2) - s2^4*(x^2-2x*x2+x2^2+y2^2)}})/(2s1^4-2s2^4) = (2*s1^4*y3-2*s3^4*y1 \pm{\sqrt{(2*s3^4*y1-2*s1^4*y3)^2-4*(s1^4-s3^4)*s1^4(x^2-2x*x1+x1^2+y1^2) - s3^4*(x^2-2x*x3+x3^2+y3^2)}})/(2s1^4-2s3^4)

This is one equation with only one unknown (x). However solving this for x eludes me.

edit:sorry the equations looks so bad, I tried putting it in the pretty math viewer you have but the equations were too big to fit.

edit: here is as far as I can get in solving that equation:

I can make this more readable by replacing some sections with shorthand variables. For instance:

$ a_1 = s_1^4-s_2^4 $
$ b_1 = 2*s_2^4*y_1-2*s_1^4*y_2 $
$ c_1 = s_1^4(x^2-2x*x_1+x_1^2+y_1^2) - s_2^4*(x^2-2x*x_2+x_2^2+y_2^2) $

$ a_2 = s_1^4-s_3^4 $
$ b_2 = 2*s_3^4*y_1-2*s_1^4*y_3 $
$ c_2 = s_1^4(x^2-2x*x_1+x_1^2+y_1^2) - s_3^4*(x^2-2x*x_3+x_3^2+y_3^2) $

And we have:
$ (-b_1 \pm{ \sqrt{ b_1^2 - 4*a_1*c_1} })/(2a_1) = (-b_2 \pm{ \sqrt{ b_2^2 - 4*a_2*c_2} })/(2a_2) $

Since the c's are the only ones that contains the unknown, we can seperate that.
$ a_2*(-b_1 \pm{\sqrt{b_1^2-4*a_1*c_1}} = a_1*(-b_2 \pm{ \sqrt{ b_2^2 - 4*a_2*c_2} }) $
$ -a_2*b_1 \pm{a_2*\sqrt{b_1^2-4*a_1*c_1}} = -a_1*b_2 \pm{a_1*\sqrt{b_2^2-4*a_2*c_2}} $

$ a_1*b_2-a_2*b_1 = \pm{a_1*\sqrt{b_2^2-4*a_2*c_2}} - \pm{a_2*\sqrt{b_1^2-4*a_1*c_1}} $

so basically I don't know how to combine those two sqrt's to solve for the variable in the c's.

**quicksilver** · June 2nd 2009, 12:59 PM

Well i got a little bit farther, I got a very large equation and put that in a ti-89 emulator set to run super fast, and it's been working for nearly 24 hours now and I don't have confidance it will give me an answer. I am thinking there must be somethign wrong in my approach of this problem, but I cannot think of a way to solve it.

And I do thank VonNemo for his reply, but I do need this solved in terms of variables excluding x, y, and r.

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