# Point in Circle, Distance to Edge

I'm standing in a circle (Point P), radius of 2000 with a center point 0/0. Then, I'm looking straight to the edge of the circle on a random point (B1 or B2) of the edge. How can I get the distance (f1 or f2) to the point when my rotation/angle is given?

Advanced: What happens if teta is not given, is it still possible to solve it?

Image of Problem

My first tries: Depending on the angle, I'm using sin, cos, tan of my rotation * (the distance to the center point + radius). I tested some numbers and it looks like it's working. Does that makes sense and is there a better solution to solve my problem? It looks like I need several formulas depending on the angle teta and the Position P.

• making sure here: you don't wish to find the point on the circle that's closest to you, but the one you're looking at in a particular direction? Sep 1, 2016 at 4:43
• What do you mean by "my rotation is given"?
– user345851
Sep 1, 2016 at 4:57
• I recommend drawing a picture. Geogebra is a good tool. It is much easier to understand what you are asking. Sep 1, 2016 at 5:05
• No, this is not a dupe, I updated my question to make the problem more understandable <3. Sep 1, 2016 at 10:24
• You are overcomplicating. Extend your line d so that it forms a diameter of the circle. Your picture will then be the same as that in the answer below, and the answer below will give you either f1 or f2. Sep 1, 2016 at 12:21

If I understand you, the situation is shown below. $g=2000$ is the radius of the circle, $r$ is your distance to the center, $\theta$ is the angle from the radius you are on to the direction you are looking. You want to find $f$. First use the law of sines: $\frac g{\sin \theta}=\frac r{\sin \angle ADC}$ to get $\angle ADC$, then $\angle CAD=\pi-\theta-\angle ADC$ and now use the law of cosines $f^2=r^2+g^2-2rg\cos \angle CAD$