Derive formula for coordinates of internal and external centers of similitude.

Question

Given 2 circles $(x - x_1)^2 + (y - y_1)^2 = r^2$ and $(x - x_2)^2 + (y - y_2)^2 = r'^2$

(with radii $r, r'$) coordinates of the internal and external centers of similitude $C_i, C_e$ are given by

$C_i = (\frac{x_1 \cdot r' + x_2 \cdot r}{r + r'}, \frac{y_1 \cdot r' + y_2 \cdot r}{r + r'})$ and $C_e = (\frac{x_1 \cdot r' - x_2 \cdot r}{r' - r}, \frac{y_1 \cdot r' - y_2 \cdot r}{r' - r})$

The diagram below shows that this is true, but I was wondering if anyone knows how to derive the equations?

Answer 1

Consider the two pairs of similar right triangles (red and blue) in the diagram.

Writing $C$ for either $C_{e}$ or $C_{i}$, we have $$\frac{|\overline{AC}|}{a} = \frac{|\overline{BC}|}{b} \qquad\to\qquad a\;|\overline{BC}| = b\;|\overline{AC}|$$

Since $A$, $B$, and $C$ are collinear, we can write coordinate-vector equations $$a\;(C-B) = \pm\;b\;(C-A)$$ (where "$\pm$" is "$+$" when $C=C_{e}$, and "$-$" when $C=C_{i}$, reflecting the fact that the vectors $AC$ and $BC$ point in the same, or in opposite, directions), so that $$C = \frac{aB\mp bA}{a\mp b} \qquad\to\qquad C_{e} = \frac{aB - bA}{a - b}\qquad C_{i} = \frac{aB + bA}{a + b}$$

Your formulas follow from substituting the circle centers $A =(x_1, y_1)$ and $B = (x_2,y_2)$ and radii $a=r$ and $b=r^\prime$.

Answer 2

If you have two points, $A$ and $B$, then you can consider the linear combination

$$P=A+\lambda(B-A)=(1-\lambda)A+\lambda B$$

It will result in a point on the line $AB$ which satisfies

$$\frac{\lvert A,P\rvert}{\lvert A,B\rvert}=\lambda\quad\implies\quad \frac{\lvert A,P\rvert}{\lvert B,P\rvert}=\frac{\lambda}{1-\lambda}$$

You can also reverse this, and obtain a linear combination for a given length ratio:

$$\frac{\lvert A,P\rvert}{\lvert B,P\rvert}=\frac{\alpha}{\beta} \quad\implies\quad P=\frac{\beta}{\alpha+\beta}A+\frac{\alpha}{\alpha+\beta}B$$

Now if your circles have centers $M$ and $M'$, then the center of similarity will, due to the intercept theory, have distances from these centers whose ratio equals that of the radii. So you get

$$C_i=\frac{r'}{r+r'}M+\frac{r}{r+r'}M'$$

For the outer center, you treat one radius as negative, e.g. $r'$, and you get

$$C_e=\frac{-r'}{r-r'}M+\frac{r}{r-r'}M'$$

Now all that remains is extracting the coordinates of the centers from your circle equations, simply by completing the square.

Your formulas seem to have mixed up their signs, as I had when I first wrote this answer.