# 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.

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This site is Mathematics - Stack Exchange, not MathOverflow. –  Américo Tavares Nov 20 '10 at 20:36
Also, they'll dismiss this question as "not research level" on MathOverflow. I'm not sure what you're asking here, since you got a solution with Maple, you clearly know how to set up the equations. The rest is just lots of algebra. You could show some partial work and someone would help you from there. –  trutheality Nov 20 '10 at 20:44
@trutheality: actually, the question has already been dismissed as "not research level" on MathOverflow. The user just copied the text directly over to here, hence the funny bit about MO in the first paragraph. –  Willie Wong Nov 20 '10 at 20:57
Also, the solution is probably not going to have nice forms. While $N_3 = (2,1)$, the $x$ coordinate of $N_2$ is $-1.54 - 0.24 \sqrt{91}$. So computation by hand can get tedious. –  Willie Wong Nov 20 '10 at 21:34
Look up "resultants" and/or "Sylvester's Eliminant". (In Mathematica, see Resultant[].) The resultant of two polynomials is a polynomial with one variable eliminated. With a larger system, proceed with a kind of enhanced Gaussian Elimination strategy: pick your favorite equation, then pair it with each remaining equation, in each case computing a resultant that eliminates (say) $x$. Then choose a favorite among those resultants, pair it with the others to eliminate (say) $y$. Etc. You end up with a (high-degree) polynomial in the remaining variable. (The "why it works" is rather fascinating.) –  Blue Nov 20 '10 at 22:16

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.

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Waw ! Many Thanks Day Late Don. Very elegant solution. Thank you very much. –  user3749 Nov 21 '10 at 3:13