The Problem:
Basically, just what the title says. I have two concentric circles, and a third circle with its center vertically aligned with their center and positioned below it. An irregular quadrilateral is defined by these two centers, and the two intersection points of the three circles. Technically, there are four intersection points, but they're mirrored pairs.
Known:
- The radii of the two concentric circles.
- The distance between the two intersection points.
- The upper-left angle of the quadrilateral defined by the two circle centers & the two intersection points.
Unknown:
- The radius of the third circle that intersects them
- The vertical distance between its center and the center of the two concentric circles.
I have uploaded an image to help visualize this problem, which I created with Inkscape:
What is this for/from?
I didn't get this out of any kind of math textbook. I created this problem myself to try to define geometric primitives in Blender 3D, made via Geometry Nodes, for something I'm trying to model. Basically, I'm trying to carve an indentation into a torus slice along its length, a virtual internal ditch, defined by the intersection of these three circles. The smaller of the two concentric circles represents the torus cross-section, the larger one defines the base of the ditch, and the third circle that intersects them both defines the side slopes that connect the ditch's base with its edges.
Is this problem even solvable with the current pre-defined measurements? I'm not sure. If it's unsolvable, what additional measurement(s) would I need to provide to make it solvable?
It has been quite a while since I've solved any kind of trigonometric or geometric problem, so I'm definitely rusty. If this is solvable, I could probably solve this on my own, given enough time to re-familiarize myself with the subject, but I figured I'd just post it here in the meantime to see if someone is willing & able to solve it faster. What I'm hoping is that this problem is not only solvable in its current state, but mere child's play to someone who solves geometry and/or trigonometry problems all the time.
I have relabeled your image and rotated it, so that we can identify the points on a coordinate plane. Let $O = (0,0)$ be the origin, and $D = (d,0)$ be the distance between the centers of the concentric circles and the offset circle.
We are given the lengths $r_1$, $r_2$, and angles $\alpha$, $\beta$. This is sufficient to identify the circle $D$. The points $A$ and $B$ have coordinates
$$A = (x_A, y_A) = (r_1 \cos \alpha, r_1 \sin \alpha)$, \\ B = (x_B, y_B) = (r_2 \cos(\alpha + \beta), r_2 \sin (\alpha + \beta)).$$ The midpoint $M$ of $AB$ is simply $$M = \frac{A+B}{2} = (x_M, y_M) = \left(\frac{r_1 \cos \alpha + r_2 \cos (\alpha+\beta)}{2}, \frac{r_1 \sin \alpha + r_2 \sin (\alpha+\beta)}{2}\right).$$ The $x$-intercept of the perpendicular bisector of $AB$ is $D$, so we write the equation of the line through $M$ and perpendicular to $AB$. The slope of $AB$ is $$m_{AB} = \frac{y_B - y_A}{x_B - x_A} = \frac{r_2 \sin(\alpha+\beta) - r_1 \sin \alpha}{r_2 \cos(\alpha+\beta) - r_1 \cos \alpha}.$$ So the point-slope form of the line is $$y - y_M = -\frac{1}{m} (x - x_M).$$ The $x$-intercept $D = (d,0)$ is the value of $x$ for which $y = 0$; i.e., $d$ satisfies $$-m(0 - y_M) = d - x_M,$$ or $$d = x_M + mY_M = \frac{r_1 \cos \alpha + r_2 \cos (\alpha+\beta)}{2} + \frac{r_2 \sin(\alpha+\beta) - r_1 \sin \alpha}{r_2 \cos(\alpha+\beta) - r_1 \cos \alpha} \cdot \frac{r_1 \sin \alpha + r_2 \sin (\alpha+\beta)}{2}. \tag{1}$$ This fixes all circles in the figure. The radius $r$ is solved via the law of cosines applied to $\triangle OBD$:
$$r^2 = d^2 + r_2^2 - 2 d r_2 \cos(\alpha + \beta). \tag{2}$$
Although you state that $AB$ is given, it too can be derived via the law of cosines applied to $\triangle AOB$:
$$AB^2 = r_1^2 + r_2^2 - 2r_1 r_2 \cos \beta. \tag{3}$$
As for the remaining lengths and angles, these can be solved via further applications of the law of sines and cosines; e.g., $$\frac{\sin \angle OBD}{d} = \frac{\sin (\alpha+\beta)}{r}, \tag{4}$$ hence $$\frac{\sin \beta}{BC} = \frac{\sin (\pi - \beta - \angle OBD)}{r_2}, \tag{5}$$ and so on.