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Please Help finding proper formula for Tangent point of two arcs. enter image description here

1st arc R = .030; 2nd R = 0.015, need to get mathematical explanation on how to get (0.02494, 0.01333) point. Assume that starting point of the 1st arc is (0,0)

user117911
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  • Please describe the problem more properly. Also specify what you yourself have tried. – user2369284 Dec 27 '13 at 17:56
  • I believe that the full question should be: I've got a circle arc, tangent to the $x$-axis at the origin with radius 0.30; I've got another circle arc, tangent to the line $y = 0.020, with radius 0.15, as shown in the diagram. The right-hand arc's center is far enough from the origin that at the point where the two meet, their tangent lines are identical. How can I determine (a) the location of the center of the second arc, and (b) (more important) the location of the joining-point?

    Is that about right?

     – John Hughes Dec 27 '13 at 18:06
  • Yes, that is exactly what I need to know : "the location of the joining-point", specifically a formula or a way of finding out that point. ps: R1=.030; R2=.015 – user117911 Dec 27 '13 at 18:14

2 Answers2

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enter image description here

My image is scaled up ($q=1000$). Tangent point: $P=(|EH|,|GH|)$

$|AG|=|AE|=30, |GC|=15, |CI|=20-15=5$

$CIJ$ and $AJE$ are similar $=> \frac{|AE|}{|AJ|}=\frac{|CI|}{|CJ|} => |CJ|=9$

$CIJ$ and $GHJ$ are similar $=> \frac{|GH|}{|GJ|}=\frac{|CI|}{|CJ|} => \underline{|GH|=13.333...}$

From Pythagoras theorem: $|EJ|=\sqrt{|AJ|^2-|AE|^2}=44.8998886413, |HJ|=\sqrt{|GJ|^2-|GH|^2}=19.9555060628$

$|EJ|-|HJ|=\underline{|EH|=24.9443825785}$

Szabolcs Sipos
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Here is a solution. I will use $R_1$ for 0.03, $R_2$ for 0.015, and $h$ for $0.020$

Clearly the center of the two circles are $$C_1=(0,R_1),~~ C_2=(c_x, h-R_2)$$ where $c_x$ is unknown.

Distance between the centers is $$ \sqrt{c_x^2+(R_1+R_2-h)^2}$$ and for the two circles to have a common tangent, we need $$ \sqrt{c_x^2+(R_1+R_2-h)^2} = R_1+R_2 ~~\Rightarrow~~c_x = \pm \sqrt{2\,h\,R2+2\,h\,R1-{h}^{2}}$$ From the figure we pick the positive root.

Now the line connecting the two points have to be divided in the ratio of $R_1:R_2$. Hence the common point of contact is $$ P = \frac{R_2 C_1 + R_1 C_2}{R_1+R_2} =\left(\frac{R_1 c_x}{R_1+R_2}, \frac{h R_1}{R_1+R_2}\right)$$

Now substitute the values to get the answer.

If this is a homework problem, you should show some work when you ask a question. Since you did not show any work, I decided to wait half an hour before answering!

user44197
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