Three-circle intersection for circles of unbounded integer radius

Question

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.

Answer 1

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)

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

Answer 2

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.