I'm trying to programmatically find the intersection points of two circles with different radii. Solving their equotations would be an option, but I thought of using vectors to do so.
Assuming I have two circles A and B (A, B as center, $r_1$, $r_2$ as radii), $\overrightarrow{AB}$ as direction vector of their straight, $E_A$ as endpoint of circle A, $E_B$ as endpoint of circle B. I can get the length $l$ between the two endpoints. The intersection points are on the normale of $V_{direction}$, starting from point X. $$ A = \bigg(\begin{array}(u\\v\end{array}\bigg); B = \bigg(\begin{array}(x\\y\end{array}\bigg); r_1 = k; r_2 = m \\ \overrightarrow{AB} = \overrightarrow{B} - \overrightarrow{A} \\ \overrightarrow{V_{direction}} = \overrightarrow{AB^0} \\ \overrightarrow{E_{A}} = \overrightarrow{A} + r_1\cdot\overrightarrow{V_{direction}} \\ \overrightarrow{E_{B}} = \overrightarrow{B} - r_2\cdot\overrightarrow{V_{direction}} \\ \overrightarrow{D} = \overrightarrow{E_{B}} - \overrightarrow{E_{A}} \\ l = |\overrightarrow{D}| \\ \overrightarrow{X} = \overrightarrow{E_A} + l \cdot factor \cdot \overrightarrow{V_{direction}} $$
If both circles had the same radius, X would be located in the center between Ea and Eb. In my case, I used $A = (350 ; 350), B = (150 ; 200), r_1 = 175, r_2 = 125$. X was about X = (226.326 ; 257.245). After thinking about it, I came up with the idea, that X has to be located at (r2/r1)*l.
But $(r_2/r_1) = (125/175) = 0.7142857142857143$, I measured a relation of $0.6896173016811695$.
Red cross = $\overrightarrow{E_B} + \overrightarrow{V_{direction}}*(r_2/r_1)*l$, green cross = correct point of intersection
My question:
How do the radii correlate to splitting the straight $[E_AE_B]$?
Rotate the coordinates so that $A$ is located at the origin and $B$ is located at $(\ell, 0)$. This simplifies the system of equations to
$$\left\{\begin{array}{l}x^2 + y^2 = r_1^2 \\ (x-\ell)^2 + y^2 = r_2^2\end{array}\right.$$
Subtracting, we obtain
$$2 x \ell - \ell^2 = r_1^2 - r_2^2;$$
solving,
$$x = \frac{r_1^2 - r_2^2 + \ell^2}{2 \ell}.$$
Since you asked about thinking of this as a fraction of the distance along the line segment $\overline{AB}$, that fraction is not $(r_2/r_1)$ but rather
$$\rho := \frac{x}{\ell} = \frac{r_1^2 - r_2^2 + \ell^2}{2 \ell^2} = \frac{1}{2}\left[\left(\frac{r_1}{\ell}\right)^2 - \left(\frac{r_2}{\ell}\right)^2 + 1\right]$$