Circle-Circle intersection centroid in a 3D space

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I am trying to find the centroid of a circle-circle intersection (the shaded region in the image) in a 3D space. The circles are defined by their center points, their radii, and their plane's normal vector. It is known that the circles lie within the same plane. I know the centroid of a circular segment is $ \bar{y} = \frac{4Rsin(\frac{\theta}{2})^3}{3(\theta-sin(\theta)}$ if it is in an x-y plane and the circular segment is in the top area of the circle. From getting the two circular segment centroids I should be able to find the centroid of the total area. I am unsure of how to find the centroid coordinates of the individual segments, or total area, in the 3D space. Would translating the axes be best or is there a way to find the centroid in x,y,z coordinates? I don't know which is best, and I don't know how to do either. I do not know the circles intersection points, the central angle, or cord length.

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You want to convert this to a 2D problem. Let $(\mathbf n_1, \mathbf c_1, r_1)$ and $(\mathbf n_2, \mathbf c_2, r_2)$ be the normal vector, center point, and radius of the two circles respectively, where the normal vector is normalized (ie, $\|\mathbf n\| = 1$).

First, check if $(\mathbf n_1 = \mathbf n_2 \text{ and } \mathbf n_1 \cdot \mathbf c_1 = \mathbf n_2 \cdot \mathbf c_2)$ or $(\mathbf n_1 = -\mathbf n_2 \text{ and } \mathbf n_1 \cdot \mathbf c_1 = -\mathbf n_2 \cdot \mathbf c_2)$. If not, the circles lie in different planes, and so do not have a centroid for their intersection.

Next, let $c = \|\mathbf c_2 - \mathbf c_1\|$ and $\mathbf u = \frac{\mathbf c_2 - \mathbf c_1}{c}$, and let $\mathbf v = \mathbf n_1 \times \mathbf u$, which is orthogonal to both $\mathbf n_1$ and $\mathbf u$. We can coordinatize your plane by mapping $(x, y)$ to $\mathbf c_1 + x \mathbf u + y\mathbf v$. Then your problem becomes one of finding the centroid of the intersection of two circles: 1 of radius $r_1$ about the origin, the other of radius $r_2$ about the point $(c, 0)$.