Scale intersecting circles fixed at pivots so that they have only one point in common

Question

Given two points, A and B; Given two circles, having 2 points in common, I1 and I2:

• one circle at center C1, with radius r1, with the point A on to it
• and another circle at center C2, with radius r2, with the point B on it.

Find the centers and radii of the two circles which still touch the points A and B in the same manner but only have one point in common located "mid-way" between the original circles.

Edit: Im not sure if the phrasing of the question is entirely correct, also im not sure if im missing some corner cases. Basically the question is revolving around detecting the collision of two circles "glued" to the points A and B. In the solution, center of the first circle should belong to the line (A, C1), while center of the second to (B, C2)

Answer 1

We find new configurations by extensions, not by scaling.

By enlarging first circle radius by an arbitrary amount $\delta_1$ and contracting the second circle radius by $\delta_2$ to maintain tangential contact, a set of centers and their radii of the two circles can always be found. There is no unique result.

To find $\delta_2$ as a function of $\delta_1$ the sum of vector projections ( found by vector dot products) of two radii, and projection of line of centers is always constant equal to distance between given two fixed points of contacting circles.

Before applying the $\delta s$

$$ r_1 \cos \alpha + (r_1+r_2) \cos \gamma + r_2 \cos \beta = d. $$

Answer 2

The equations for $C_1, C_2 $ are

$ (x - x_1)^2 + (y - y_1)^2 = r_1^2 $

$ (x - x_2)^2 + (y - y_2)^2 = r_2^2 $

which has $6$ degrees of freedom (i.e. $6$ free parameters). However, by the conditions in the problem we must have

$ (x_A - x_1)^2 + (y_A - y_1)^2 = r_1^2 $

$ (x_B - x_2)^2 + (y_B - y_2)^2 = r_2^2 $

$ (x_1 - x_2)^2 + (y_1 - y_2)^2 = (r_1 + r_2)^2 $

The solution of the the first $2$ equations is

$C_1 = (x_1, y_1) = (x_A , y_A) + r_1 (\cos(\phi), \sin(\phi) ) $

$C_2 = (x_2, y_2) = (x_B, y_B) + r_2 ( \cos(\theta), \sin(\theta)) $

Angles $ \theta, \phi $ are selected such that the third equation is satisfied, i.e.

$ \| A - B + r_1 u_1 - r_2 u_2 \| = r_1 + r_2 $

where $u_1 = (\cos(\phi), \sin(\phi) ) , u_2 = (\cos(\theta), \sin(\theta) ) $

Squaring, and simplifying,

$ \| A - B \|^2 + 2 (A - B) \cdot (r_1 u_1 - r_2 u_2) - 2 r_1 r_2 (u_1 \cdot u_2) = 2 r_1 r_2 $

Noting that $u_1 \cdot u_2 = \cos(\phi - \theta) $, the above simplifies to

$ \| A - B \|^2 + 2 (A - B) \cdot (r_1 u_1 - r_2 u_2) + 2 r_1 r_2 (1 - cos(\phi - \theta) ) = 0 $

So to find circles $C_1, C_2$:

Choose $r_1, r_2,$ and $\phi $

and solve the final equation for $ \theta $