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I have three circles. One is at $(0,0)$ and has radius $n$, another has is at $(1,0)$ and has a radius $m$, and the third is at $(0.5, \sqrt{0.75}))$ and has a radius of $o$. All of the radius values are integers.

How can I compute where these three circles intersect, for any value of $n$, $m$, and $o$? Edit: The values here are outputs, not inputs- that is, I want to determine possible values for $n$, $m$, $o$, not give them. For example, there are obviously no solutions with $n = m = o = 1$.

Bonus points for any solution that can solve for more than three circles under the same constraints.

Sorry: I want to know how to find the point(s) at which the three circles meet, if there are any.

Edit: My mistake! The initial positions for the circles actually violate the original constraint. The second circle is positioned at $(1, 0)$.

I appear to have asked a question of significantly higher difficulty than I had first imagined. My apologies.

Edit:

$x^2 + y^2 = n^2$ $x^2 + y^2 - 2x + 1 = m^2$ $1 - x - \sqrt3y + y^2 + x^2= o^2$

Therefore,

$-n^2 = 1 - 2x - m^2 = 1 - \sqrt3y - x - o^2 = -x^2 - y^2$ $ n^2 = m^2 + 2x - 1 = o^2 + \sqrt3y + x = x^2 + y^2$

This places a number of simultaneous conditions if $m$, $n$, and $o$ are to be integral. I don't see how any x and y could satisfy this equation for any integral m, n, o.

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    @Henry: $a$. I guess I'll try to m$a$ke the question clearer.2012-02-17

2 Answers 2

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The equations for your circles are:

$x^2 + y^2 = n^2 $ $(x-1)^2 + y^2 = x^2 + y^2 - 2x + 1 = m^2 $ $\left(x-\frac{1}{2}\right)^2 + \left(y-\sqrt{0.75}\right)^2 = x^2 +y^2 -\sqrt{3}y - x+ 1 = o^2 $

Combining the first and second we observe that $n^2 - m^2 = 2x-1$

Likewise, the first and third give that $n^2 - o^2 = \sqrt{3}y+x-1$

Rearranging, $x = n^2 - o^2 - \sqrt{3}y + 1$ and substituting back gives:

$ n^2 - m^2 = 2(n^2-o^2-\sqrt{3}y+1)-1 $

So we can solve for $y$ in term of $n,m$ and $o$:

$ -2\sqrt{3}y = n^2+m^2-2o^2+1$ $ y = \frac{-n^2-m^2+2o^2-1}{2\sqrt{3}}$

Solving for $x$:

$x = n^2 - o^2 +1 - \frac{(-n^2-m^2+2o^2-1)}{2}=\frac{3n^2+m^2-4o^2+3}{2}$

This is assuming that a solution exists in the first place. Also, if the solution lies along $y=0$ it follows that $x^2 = n^2$, $(x-1)^2 = m^2$ and $(x-\frac{1}{2})^2 = o^2$, which is impossible.

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    Indeed, I have implicitly assumed that a solution exists. If, for instance, $n^2-m^2$ does not lie on the line $2(x+y-1)$ no solution will exist.2012-02-17
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To find intersection points you should solve following systems of equations :

A)

$\begin{cases} x^2+y^2=n^2 \\ (x-1)^2+(y-1)^2=m^2 \end{cases}$

B)

$\begin{cases} x^2+y^2=n^2 \\ (x-0.5)^2+(y-\sqrt{0.75})^2=o^2 \end{cases}$

C)

$\begin{cases} (x-1)^2+(y-1)^2=m^2 \\ (x-0.5)^2+(y-\sqrt{0.75})^2=o^2 \end{cases}$

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    @DeadMG, on this website, when someone tries to help you, the done thing is to thank the person, not to get sarcastic. Did pedja have any way of knowing you got that far on your own? Had you showed so much of your work that people could see you had gotten that far? You come here asking for help - be gracious, not rude, when you are offered some, even if it isn't e$x$actl$y$ what you were hoping for.2012-02-17