Given a random point in a funnel area in 2d I'm trying to find the radius of the circle that has center in the bisector of that funnel.

This is for a geometry program so I can extract a lot of information. I think I'm just missing a small detail but not sure what.

enter image description here

This is the situation. I'm given the center of a funnel and there is a circle that can goes along the bisector of that funnel, expanding or contracting until the edges of the funnel depending on the distance. I'm given a point P as the image shows. I'm trying to find the radius of circle that this point belongs to. Obviously there are 2 circles that contain that point however I want the one showed in the image, to the right of the point P. I have the distance D to that point from the center of the funnel and the angle alpha it makes with the bisector. All the letters A, B, C, E are easily obtainable however I can't seem to find the a way to use them to calculate the radius of the circle. I also want to calculate the distance from the circle center to the center of the funnel but that is easy if I have the radius. Any ideas or hints?

  • $\begingroup$ Do you know the angle of the funnel? If yes: Consider the radius from the center to the point of tangency. You have there a right angle. Compute the radius using the sine function. $\endgroup$ – user376343 Oct 22 '18 at 19:32
  • $\begingroup$ @user376343 I do have the angle of the funnel. I could use the sine function but I don't know the distance from the funnel center to the point of tangency. $\endgroup$ – Dozed12 Oct 22 '18 at 19:51
  • $\begingroup$ Can you list all the information you have? Angles, distances etc. $\endgroup$ – Seyed Oct 22 '18 at 19:59
  • $\begingroup$ @Seyed Sure. I have the angle of the funnel, angle from the bisector to the point P and distance to P from the funnel center. I think that should be enough. At least with just this information you can only define 1 point in the funnel although there are 2 circles that contain it but I do want the one closest to the funnel center. $\endgroup$ – Dozed12 Oct 22 '18 at 20:05
  • $\begingroup$ Do you need to know the radius? Or do you rather need to find the center of the circle? $\endgroup$ – user376343 Oct 22 '18 at 20:07

CONSTRUCTION based on a homothety:

Denote $\mathcal{K}$ the circle you want to construct, and H its center. Denote V the vertex of the funnel. Construct an arbitrary circle $\mathcal{L}$ tangent to the arms of the tunnel, with center S at the bissector of the funnel and such that |SV|<|HV|.
The half-line VP cuts this small circle at two points. That one which is closer to P (denote it R) is an image of P through a homothety. The small circle $\mathcal{L}$ is image of $\mathcal{K}$ through this same homothety. Thus, the segments RS and PH are parallel. We have R, S, P, thus we can easily construct H.

  • $\begingroup$ Thanks this seems to work just fine. And I assume if we had |SV|>|HV| we will get the other circle that contains P. $\endgroup$ – Dozed12 Oct 22 '18 at 20:40
  • $\begingroup$ Yes, it is so. I am glad my solution was useful. $\endgroup$ – user376343 Oct 22 '18 at 20:47

Denote with $\gamma$ the acute angle that radius R (shown in your picture) creates with the funnel bisector. Denote the half angle of the funnel with $\beta$.



Square these two equations and add them. Unknown angle $\gamma$ will vanish and you will get a simple quadratic equation with R being the only unknown.

  • $\begingroup$ I'm following your idea, I even saw the relation between those 2 triangles earlier but wasn't sure how to solve it. However I'm not really understanding your solution when you say "Square these two equations and add them". $\endgroup$ – Dozed12 Oct 22 '18 at 21:09
  • $\begingroup$ @Dozed12 Pretty much exactly what he said: square both sides of each equation, then add one to the other. The resulting left-hand side will be $R^2\cos^2\gamma+R^2\sin^2\gamma$, which is just $R^2$. $\endgroup$ – amd Oct 22 '18 at 23:17
  • $\begingroup$ Oh just add them like that. Got it. $\endgroup$ – Dozed12 Oct 22 '18 at 23:36

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