"Circle-Circle Intersection" in vector notation

[jxu]

Article: Circle-Circle Intersection

Problem:
If the problem is only determining how many intersections there are (not including same center circles), there is a simple geometric interpretation: check if the distance between both centers is no more than the sum of radii. If equal, there is exactly one intersection, if less then 2 intersections, if more then 0 intersections.

To actually find the intersection, in vector notation (which generalizes to higher dimensions)

WLOG circle 1 is centered at origin and circle 2 is centered at $P$.

By definition of circle, circle 1 is points $x : |x| = r_1$ and circle 2 is points $x: |x-P| = r_2$. Squaring both sides and substituting $|x|^2$ gives the radical line described by $r_1^2 - 2 x \cdot P + |P|^2 = r_2^2$.

[adamant-pwn]

Yeah, I also don't like direct work with coordinates. Would you like to make a pull request on that?

[adamant-pwn]

So, essentially with circles $(A, r_1)$ and $(B, r_2)$ it boils down to

$$ (x-A) \cdot (B - A) = \frac{r_1^2 - r_2^2 + (B-A)^2}{2} $$

Geometrically, it defines the line CD on the picture below:

I think, in higher dimensions intersection would be a hypersphere defined on a space formed by the equation above. So, I assume we would need another equation or formula to get the radius of the hypersphere.

[jxu]

In higher dimensions I think there are still only at most two intersections of the sphere and hyperplane and all the equations are the same: $|x|=r$ for sphere and $n \cdot x = c$ for hyperplane

[adamant-pwn]

I think, the intersection of an $n$-sphere and a hyperplane is an $(n-1)$-sphere... For example, in 3D the intersection of a sphere and a plane is a circle.

[jxu]

oh yeah you're right. I was thinking of the same vector equations applying to line-sphere intersections, such as in ray tracing. there is also a cool barycentric coordinates trick for ray-triangle intersection which is particularly common in computer graphics

http://www.graphics.cornell.edu/pubs/1997/MT97.html

[jxu]

I feel like there must be a nicer solution than boiling it down to circle-line intersection, but I can't think of one. Maybe something with angles...

[jxu]

Oh the geometric solution is simple. Draw two triangles with sides the radii and vector between centers. Use law of cosines to find one if the angles and apply a rotation matrix and scale by radius length

here's an even nicer solution https://math.stackexchange.com/a/1033561