Dear Joseph, thank you very much for your response and the references. It helps me a lot.
I was not sufficiently precise in my question: my three circles can be intersecting, so I think that the solution should be simpler (?).
For example, i can have the three following circles (from an example on which I work):

C1: $x^2+(y-3)^2 = 9$

C2: $(x^{\prime}+3) ^2+y^{\prime 2} = 9$

C3: $(x^{\prime\prime}+3)^2+ y^{\prime\prime 2}=16$

$(x_a, y_a)=(17/6, 4) \in C1$

$(x_b, y_b)=(-17/6, 4) \in C1$

$p_1= (x,y)$, $(x_a, y_a)$ and $p_2=(x^\prime, y^\prime)$ are aligned

$p_1= (x,y)$, $(x_b, y_b)$ and $p_3=(x^{\prime\prime}, y^{\prime\prime})$ are aligned

$\dfrac{(x-x^{\prime\prime})^2+(y-y^{\prime\prime})^2}{(x^\prime-x^{\prime\prime})^2+(y^\prime-y^{\prime\prime})^2}= \alpha$

$\dfrac{(x-x^{\prime\prime})^2+(y-y^{\prime\prime})^2}{(x-x^{\prime})^2+(y-y^{\prime})^2}= \beta$

$\alpha$, $\beta$, $(x_a, y_a)$ and $(x_b, y_b)$ are known.

The output (the triangle to find) is : $p_1=(x,y)$, $p_2=(x^\prime, y^\prime)$
and $p_3=(x^{\prime\prime}, y^{\prime\prime})$.

One angle of the triangle to find (the one in the corner $p_1=(x,y)$) can be implied from the 1st equation, it is equal to the inscribed angle formed by $(x_a, y_a)$, $(x_b, y_b)$ and $(x,y)$.

I think that since I have some additionnal constraints (the 6th and 7th equations) my problem should be easier to solve. I drawed this in GeoGebra and I could find numerically the solution, but I don't know how to solve it analytically.

Does anyone have any idea ?
Thank you in advance.

exactnumber? (The usual definition of the word "algorithm" is modelled after our familiar digital computers that don't have infinite-precision real numbers.) – Sergei Ivanov Sep 17 '10 at 20:39