# Three-circle intersection for circles of unbounded integer radius

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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Are you looking for (a) the point where all three of them intersect if it exists? (b) the (up to 6) points where pairs of them intersect? (c) the area where all three overlap? – Henry Feb 17 '12 at 10:10
@Henry: a. I guess I'll try to make the question clearer. – DeadMG Feb 17 '12 at 10:10

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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Yeah, I think your conclusion is kind of fishy. That would seem to indicate that I can substitute in any kind of values I wanted for n, m, and o, and compute a collision, whereas I used a simple graphing program to show that none exist where 1 <= n, m, o <= 4. – DeadMG Feb 17 '12 at 11:06
You used the details provided by (1), (2) and (1),(3). But there is one more constraint, if you substract (2) and (3) – Beni Bogosel Feb 17 '12 at 11:24
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. – anthus Feb 17 '12 at 11:28

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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I got that far on my own. But I have no idea how to solve them. If I could solve it on my own, I wouldn't be asking the question. – DeadMG Feb 17 '12 at 10:34
@DeadMG,In each system expand the second equation and then subtract the first equation from the second one.... – pedja Feb 17 '12 at 10:38
@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 exactly what you were hoping for. – Gerry Myerson Feb 17 '12 at 12:08