A system of geometric equations I have a question about solving a system of geometric equations. I really hope someone here can help me, it's been several months since I try to solve the problem but without success. As I am not an expert in maths, I count on your help. Thank you in advance.
Problem:
I have three circles $C_1, C_2, C_3$. For example (refer to this figure)
$$\begin{eqnarray*}
C_1 :& (x+1)^2+(y−4)^2=9 \\
C_2 :& (x+4)^2+y^2=25 \\
C_3 :& (x−2)^2+y^2=16
\end{eqnarray*}$$
$N_2$ and $N_3$ are two points:
$N_2$ is located at the intersection of $C_1$ and $C_2$, and is outside $C_3$.
$N_3$ is located at the intersection of $C_1$ and $C_3$, and is outside $C_2$.
$T_1$, $T_2$ and $T_3$ are three points (defining the blue triangle in the figure):
$T_1$ is located at $C_1$ (and inside $C_2$ and $C_3$).
$T_2$ is located at the intersection of $C_2$ and the line that passes through $T_1$ and $N_2$.
$T_3$ is located at the intersection of $C_3$ and the line that passes through $T_1$ and $N_3$.
Question:
Given any three circles $C_1$, $C_2$, and $C_3$, is there a point $T_1$ (defined as above) such that $\text{distance}\space (T_1, T_2)=\alpha\cdot (\text{distance}\space (T_1, T_3))$ ($\alpha$ is some constant, for example equal to $1$)?
I tried Maple, it gives me the solution, but not the way how to calculate it. Any idea?
Edit: What I need is a proof of calculability.
For example, in a system of linear equation with $x$ variables, we know that $x-1$ equations are needed, otherwise there is an infinity of possible solutions. For my system, I know how to prove that if a solution exists, it is necessarily unique. But I don't know an algorithm that can calculate it because the equations are not linear. 
Thank you.
 A: Here's an approach that leverages the geometry of the situation.
Let $K_i$ be the center of circle $C_i$.
Points $N_2$ and $N_3$ on $C_1$ can be considered "known". Hence, the measure of the central angle $N_2 K_1 N_3$ (corresponding to the "top" arc $N_2 N_3$ in your diagram) is also known, and easily computable with dot products; call that $2\theta_1$. (Why the "2"? Read on...) By the Inscribed Angle Theorem, the angle $N_2 T_1 N_3$ --no matter where $T_1$ is on "bottom" arc $N_2 N_3$-- is half the size of the central angle; that is $N_2 T_1 N_3$ has measure $\theta_1$. This must also be the measure of angle $T_2 T_1 T_3$, at the top of your blue triangle.
Let the blue triangle's angle at $T_2$ have (unknown) measure $\theta_2$. Let $P_2$ be the (other) point at which line $T_2 T_3$ meets circle $C_2$. The angle $N_2 T_2 P_2$ has measure $\theta_2$ (because it's the "same angle" as $T_1 T_2 T_3$). By the Inscribed Angle Theorem again, the central angle $N_2 K_2 P_2$ (measured along the "outer" arc $N_2 P_2$) must be twice as big: $2\theta_2$. Therefore, if we know the size of $\theta_2$, then we know how far around circle $C_2$ we have to travel from $N_2$ to get to $P_2$.  Likewise, with (unknown) $\theta_3$ the blue triangle's angle at $T_3$ and $P_3$ the (other) point where line $T_2 T_3$ meets $C_3$, we have that angle $N_3 K_3 P_3$ has measure $2\theta_3$, which tells us how to get to $P_3$ from $N_3$.
So, if we can find formulas for $\theta_2$ and $\theta_3$ in terms of $\theta_1$ and $a$, then we'll effectively know the locations of $P_2$ and $P_3$ relative to $N_2$ and $N_3$. The line joining the $P$s determines the points $T_2$ and $T_3$, which in turn give us the location of $T_1$ (as the intersection of lines $T_2 N_2$ and $T_3 N_3$ with circle $C_1$).
Let's get to it ...
The Law of Sines, and your assumption about the ratio of side lengths, tells us that
$$\frac{\sin{\theta_3}}{\sin{\theta_2}}=\frac{|T_1 T_2|}{|T_1 T_3|}=a$$
So, 
$$a \sin{\theta_2} = \sin{\theta_3}$$
But $\theta_3=\pi-\left(\theta_1+\theta_2\right)$.
$$a\sin{\theta_2} = \sin{\left(\pi-\left(\theta_1+\theta_2\right)\right)}=\sin{\left(\theta_1+\theta_2\right)}=\sin\theta_1 \cos\theta_2+\cos\theta_1 \sin\theta_2$$
$$\sin{\theta_2}\left(a-\cos\theta_1\right)=\sin\theta_1 \cos{\theta_2}$$
$$\sin^2{\theta_2}\left(a-\cos\theta_1\right)^2=\sin^2\theta_1 \cos^2{\theta_2}$$
$$\left( 1 - \cos^2{\theta_2} \right)\left(a- \cos\theta_1\right)^2 = \sin^2\theta_1 \cos^2\theta_2$$
$$\left(a-\cos\theta_1\right)^2 = \cos^2{\theta_2}\left(a^2 - 2 a \cos\theta_1+ \cos^2\theta_1+\sin^2\theta_1\right)= \cos^2{\theta_2}\left(1 +a^2 - 2a \cos\theta_1\right)$$
$$\cos^2{\theta_2}=\frac{\left(a-\cos\theta_1\right)^2}{1 + a^2 - 2 a \cos\theta_1}$$
Therefore, 
$$\frac{1+\cos\left(2\theta_2\right)}{2}=\frac{\left(a-\cos\theta_1\right)^2}{1 + a^2 - 2 a \cos\theta_1}$$
$$\cos\left(2\theta_2\right)=\frac{a^2-1-2a\cos\theta_1+2\cos^2\theta_1}{1 + a^2 - 2 a \cos\theta_1}$$
Similarly, swapping $\theta_2 \leftrightarrow \theta_3$ and $a \leftrightarrow \frac{1}{a}$ ...
$$\cos\left(2\theta_3\right)=\frac{1-a^2-2a\cos\theta_1+2a^2\cos^2\theta_1}{1 + a^2 - 2 a \cos\theta_1}$$
Now, use $N_2$ and $N_3$ and $2\theta_2$ and $2\theta_3$ to locate $P_2$ and $P_3$, then $T_2$ and $T_3$, and finally $T_1$. I doubt the formulas are particularly elegant, but each step in finding them should be relatively straightforward.
