# Radius of circle inside a funnel

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. 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?

• 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. – user376343 Oct 22 '18 at 19:32
• @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. – Dozed12 Oct 22 '18 at 19:51
• Can you list all the information you have? Angles, distances etc. – Seyed Oct 22 '18 at 19:59
• @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. – Dozed12 Oct 22 '18 at 20:05
• Do you need to know the radius? Or do you rather need to find the center of the circle? – 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.

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

$$R\cos\gamma=D\cos\alpha-\frac{R}{\sin\beta}$$

$$R\sin\gamma=D\sin\alpha$$

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.

• 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". – Dozed12 Oct 22 '18 at 21:09
• @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 $R^2$. – amd Oct 22 '18 at 23:17
• Oh just add them like that. Got it. – Dozed12 Oct 22 '18 at 23:36