# Calculate the distance of any point on the arc from the center of circle

This is my first time posting so I hope my formatting is correct.

Consider this, I have two circles, one big one small with radius $$r_1$$ and $$r_2$$. The borders of both circles are touching. See image:

Correct me if I'm wrong, I believe the angle from the center of the big circle is $$2\arcsin\left(\dfrac{r_2}{r_1-r_2}\right)$$

What I am actually interested in is subtracting the smaller circle from the larger circle, making a small channel like this:

Is there an expression where I can find the radius of the bigger circle to any point of the arc of the channel? For the shortest distance is easy, basically just $$r_1-2r_2$$.

But what about all the other points? How do I go about calculating the distance to any point on the arc of the channel? I can approximate it from the middle and approximate triangles within small steps but if there is a mathematical expression for it, that would be great. The ideal expression would have $$r_1,r_2,\theta$$

Thank you

• Is $\theta$ the angle from the horizontal line in the big circle or in the small one? – Andrei Aug 13 '20 at 13:45
• The angle from the center is not $2\arctan$ but $2\arcsin$. The hypotenuse is $r_2-r_1$ and the opposite side is $r_2$ – Andrei Aug 13 '20 at 15:37
• Thanks for clearing my mistake. I will edit it. To answer your other question, $\theta$ would be from the horizontal line in the big circle – hazziqueeee Aug 13 '20 at 15:52
• @hazziqueeee, I think you have to find distance of any point on the arc from the center. – SarGe Aug 13 '20 at 16:09
• @SarGe edited the post to reflect the changes you suggested. Thanks! – hazziqueeee Aug 13 '20 at 16:14

In reference to this image

and complementing other answers, the points on the green arc are represented by the equation $$r = (r_1-r_2)\cos\theta-\sqrt{r_2^2-(r_1-r_2)^2\sin^2\theta},\qquad|\theta|\leq\arcsin\left(\frac{r_2}{r_1-r_2}\right),$$ while the points on the red arc are represented by the equation $$r = (r_1-r_2)\cos\theta+\sqrt{r_2^2-(r_1-r_2)^2\sin^2\theta},\qquad|\theta|\leq\arcsin\left(\frac{r_2}{r_1-r_2}\right).$$ In particular, the points on the arc from $$A$$ to $$B$$ are represented by the second of previous equations with $$\arctan\left(\frac{r_2}{r_1-r_2}\right)\leq\theta\leq\arcsin\left(\frac{r_2}{r_1-r_2}\right).$$

• -@enzotib, thanks for the correction. Also, nice explanation, good one. – SarGe Aug 13 '20 at 17:32
• Thanks for this solution this is exactly what I needed. A quick calculation on excel and I can verify that the green arc expression works like a charm. It is exactly what I needed. I will put the equation to test in some real-life application and if it works, I will put another update to show you! Thanks again! – hazziqueeee Aug 13 '20 at 17:33
• hazziqueeee: thanks to you, but this was intended as a long comment to @SarGe, the explanation of derivation of equation remain that of Andrei – enzotib Aug 13 '20 at 17:36
• The condition on the right side of the equations for $\theta$ are the same. Are they not supposed to condition which equation to use ? Am I missing something ? – Malcolm Jan 20 at 18:45

Draw the lines from the center of the big circle and the center of the small circle to the point on the arc. Draw the perpendicular from the point on the arc to the line connecting the circles. You will form two right angle triangles. In the left one (the way your figure is drawn) the hypotenuse is $$r$$, the component along the horizontal is $$r\cos\theta$$, and the vertical segment is $$r\sin\theta$$. For the right side triangle, the hypotenuse is $$r_2$$, the vertical line is $$r\sin\theta$$, and the horizontal is $$(r_1-r_2)-r\cos\theta$$. Now write Pythagoras' theorem in the triangle on the right: $$r^2\sin^2\theta+[(r_1-r_2)-r\cos\theta]^2=r_2^2$$ Expanding the square you get a quadratic equation in $$r$$: $$r^2-2r(r_1-r_2)\cos\theta+r_1^2-2r_1r_2=0$$ The smaller solution is the distance to the front of the arc. Notice that you don't get a real solution if the angle $$|\theta|<\arccos\frac{\sqrt{(r_1-r_2)^2-r_2^2}}{r_1-r_2}=\arcsin\frac{r_2}{r_1-r_2}=\arctan\frac{r_2}{\sqrt{(r_1-r_2)^2-r_2^2}}$$.

• I think the limit for $\theta$ is wrong, it should be $\arctan\frac{r_2}{\sqrt{r_1(r_1-2r_2)}}$ – enzotib Aug 13 '20 at 15:15
• @enzotib You are right. I meant $\arcsin$, but your formula is also valid. I'll fix my answer – Andrei Aug 13 '20 at 15:31
• Thank you for taking your time to get me an answer @Andrei. I've followed your instructions and I also come to the same expression as you. I have not solved the quadratic equation just yet since I'm off work but I don't see any reason this expression wouldn't work. Thanks for solving it! – hazziqueeee Aug 13 '20 at 15:57

You have to find the distance $$(r)$$ of the point $$P$$ from the origin.

The point $$P$$ will move on the green arc. The equation of the circle in polar coordinates is given by $$r = (r_1-r_2)\cos\theta+\sqrt{r_2^2-(r_1-r_2)^2\sin^2\theta}$$ where $$r$$ is distance from origin and $$\theta$$ is the angle made by polar axis (or $$x$$-axis) with the line joining $$P$$ and origin. You have calculated that $$-\sin^{-1}\left(\frac{r_2}{r_1-r_2}\right)\le \theta \le\sin^{-1}\left(\frac{r_2}{r_1-r_2}\right)$$

• The green arc in the image should be reduced, because it does not represent the limitation on $\theta$ shown in the answer. – enzotib Aug 13 '20 at 16:16
• @enzotib, when $P$ will be topmost point, $\theta$ will be less than the maximum. Everything seems perfect to me. BTW, what do you suggest, what it should be? – SarGe Aug 13 '20 at 16:23
• – enzotib Aug 13 '20 at 16:30
• @enzotib, according to my best, my diagram seems to be correct. At topmost point $\theta$ will be less than the maximum. Consider a line coinciding with $y$-axis and now start rotating it w.r.t. origin in clockwise direction. It will first touch the circle at point with $\theta_{max}$ and then it will pass through the topmost point where $\theta<\theta_{max}$. – SarGe Aug 13 '20 at 16:35
• See my answer, please. In particular, you plus sign in front of the square root is wrong. – enzotib Aug 13 '20 at 16:52