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I need to solve for variables $u$ and $v$ in this system of equations:

$(x+u)^2+(y+v)^2=1$

$u^2+v^2=k$

How do I isolate $u$ and $v$ to get them both in terms of $x$, $y$, and $k$?

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    $\begingroup$ $u =\sqrt{ k - v^2}$. From there, substitution is your best bet. $\endgroup$
    – Kaynex
    May 16, 2017 at 2:43
  • $\begingroup$ Why did you delete then repost what's essentially the same question? The hint in my previous comment still stands as written: subtract the two equations and you get a linear equation in $u,v\,$. Use it to eliminate one of the variables, then substitute in either equation and solve the quadratic in the other variable. $\endgroup$
    – dxiv
    May 16, 2017 at 2:45
  • $\begingroup$ @dxiv. I didn't know what I was asking before. I cleaned it up a little. I'll try doing some substitution methods but I'm having difficulty so far. $\endgroup$
    – name
    May 16, 2017 at 2:48
  • $\begingroup$ @dxiv: my answer to this one sounds very close to what you suggested, though a slightly different way to express it. Great minds think alike. $\endgroup$ May 16, 2017 at 2:49
  • $\begingroup$ @RossMillikan Can only hope your answer works out better than my hint ;-) $\endgroup$
    – dxiv
    May 16, 2017 at 2:56

1 Answer 1

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If you expand the squares in the first equation, you can use the second to eliminate the $u^2,v^2$ terms. That leaves you with one linear equation and one quadratic. Solve the first for $u$ and substitute into the second. That gives you a quadratic in $v$ which you can solve, getting two roots. Plug them into the first and get two solutions for $u$. Check them both and you are done.

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  • $\begingroup$ What about the 2xu, 2yv terms? I've even tried to use SymboLab to solve it for me, but no luck $\endgroup$
    – name
    May 16, 2017 at 2:54
  • $\begingroup$ Those are the linear terms left after the quadratic ones disappear. You essentially have $au+bv=c$ or $u=\frac 1a(c-bv)$. Plug that into the second and you have the promised quadratic in $v$. $\endgroup$ May 16, 2017 at 2:59
  • $\begingroup$ It gets more complicated as I do these things. Are you absolutely sure it is as simple as that? $\endgroup$
    – name
    May 16, 2017 at 3:16
  • $\begingroup$ Write it out in detail and see how it goes. Yes, it works like that. $\endgroup$ May 16, 2017 at 3:43

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