# How do I solve this system of 2 equations?

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 $k$?

• $u =\sqrt{ k - v^2}$. From there, substitution is your best bet. – Kaynex May 16 '17 at 2:43
• 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. – dxiv May 16 '17 at 2:45
• @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. – name May 16 '17 at 2:48
• @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. – Ross Millikan May 16 '17 at 2:49
• @RossMillikan Can only hope your answer works out better than my hint ;-) – dxiv May 16 '17 at 2:56

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.
• 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 $v$. – Ross Millikan May 16 '17 at 2:59