How to find the intersection points of a 3d plane and 2d circle

Question

I've tried multiple searches but have been unable to find answer or guidance on this question (that I can follow). I also apologize in advance if my terminology is not quite correct.

Consider a right had coordinate system where y is up (i.e. x forward, y up, z out / towards you).

Now for example, consider a plane that is rotated about the x axis by 30 degrees and the z axis by -15 degrees and defined by the point P0(x0,y0,z0) as (5000, 3500, -500). Given these rotations, a normal N(a,b,c) to this plane can be defined as (-48.3, 83.65, -25.88). So in point / normal notation we can define this plane as:

$ a * (x-x0) + b * (y-y0) + c * (z-z0) = 0 $

$ -48.3(x-5000) + 83.65(y-3500) - 25.88(z+500) = 0 $

Now consider a 2d circle (r/theta notation) that lies on the xz plane at a known height y(3000) with a know radius r(5000). You can find the x / z points that lie on the circle as follows:

$ x = r * cos(\theta) = 5000 * cos(\theta)$

$ z = -r * sin(\theta) = -5000 * sin(\theta)$

Substituting the height/y of the xz plane circle and the formulas for x and z into the formula for the 3d plane you now have:

$ -48.3((5000 * cos(\theta))-5000) + 83.65(3000-3500) - 25.88((-5000 * sin(\theta))+500) = 0 $

Now, if the circle intersects the plane, there should be two values of $\theta$ that cause the above equation to equal $0$. Using an iterative solver to two decimal places, the values for this example are $\theta(18.85, 284.78)$.

In other situations there could be no solution if the circle does not intersect, one if it just touches and I suppose infinite if it lies exactly on the 3d plane.

My hope is to arrive at a series of equations that allow me to solve for $\theta$ rather than using the iterative approach.

Answer 1

Substituting circle into the plane we can get equation for $\theta$:
$ a (r\cos \theta-x_0) + b (h_y-y_0) + c (r\sin \theta -z_0) = 0 $ or $a\cos \theta + c \sin \theta =\frac{ax_0+cz_0-b (h_y-y_0)}{r}$
This can be rewritten as $$\sin(\theta+\alpha)=\frac{ax_0+cz_0-b (h_y-y_0)}{r\cdot\sqrt{a^2+c^2}}$$ where $\alpha=\tan^{-1}\frac{a}{c}$. The last equation has one or two solutions when right hand side is between $-1$ and $1$.

Answer 2

If you expand your equation for $\theta$, you'll get $$A \cos \theta + B \sin \theta = C$$ with some values of $A$, $B$, $C$ that are known from your input data.

Such type of equation can be solved analytically.

I believe $A^2+B^2 > 0$ otherwise equation is $0=C$ which has eiher no solutions or infinitely mane solutions depending on $C$

First, divide both parts by $\sqrt{A^2+B^2}$: $$\frac{A}{\sqrt{A^2+B^2}} \cos \theta + \frac{B}{\sqrt{A^2+B^2}} \sin \theta = \frac{C}{\sqrt{A^2+B^2}}$$

Then we can find $\beta \in [-\pi/2;3\pi/2)$ such that $\sin\beta=\frac{A}{\sqrt{A^2+B^2}}$, $\cos\beta=\frac{B}{\sqrt{A^2+B^2}}$: $$\beta=\frac{\pi}{2}+\left(\arcsin \frac{A}{\sqrt{A^2+B^2}}-\frac{\pi} {2}\right)\cdot k,$$ where $k=-1$ if $B<0$ otherwise $k=1$

Then equation transforms to $$\sin \beta \cos \theta + \cos \beta \sin \theta = \frac{C}{\sqrt{A^2+B^2}}$$ $$\sin (\beta+\theta) = \frac{C}{\sqrt{A^2+B^2}}$$ $$\beta+\theta=\frac{\pi}{2}\pm\left(\arcsin \frac{C}{\sqrt{A^2+B^2}}-\frac{\pi}{2}\right)$$ $$\theta=\frac{\pi}{2}-\beta\pm\left(\arcsin \frac{C}{\sqrt{A^2+B^2}}-\frac{\pi}{2}\right)$$