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# Thread: Geometry or Formuls to construct a circle to three entities.

1. Hi,

I need to be able to construct a circle that is tangental to two circles and a staight line. The url below has an expanded explanation of the problem with two diagrams. I can't currently add pictures to this post so I've had to use my question in a photo forum I belong to instead. Sorry for the inconvenience.

https://www.ephotozine.com/forums/to...rmulae--119523

Thank you in advance for any help.  Reply With Quote

2. This one doesn't need much more than Pythagoras to solve it.

Call the distance of the centre of a blue circle from the horizontal line h (note - this is not the centre-centre spacing of the two blue circles)
Call the distance of the centre of a blue circle from the vertical line x
Call the radius of the small circle r

Then the radius of the large circle is: (h^2+x^2-r^2)/(2*(x-r))

I would have shown my working but typing out equations like this is too tedious!

No guarantees, but I've checked the results for a few values against a sketch in Fusion 360 and it gives the same answers. F360 won't let you construct a construct a circle with those constraints either, but you can create a circle in roughly the right position and then apply appropriate constraints afterwards to get the same effect.
Last edited by Neale; 07-07-2019 at 09:37 PM.  Reply With Quote

3. Thank you for your help, much appreciated. I'll try it out at work today.

I presume both the ^ and * symbols in the formula stand for squared?  Reply With Quote

4. Originally Posted by theforester Thank you for your help, much appreciated. I'll try it out at work today.

I presume both the ^ and * symbols in the formula stand for squared?
"^2" means squared; "*" means multiplication. That's one reason why I didn't want to write out the derivation - too much chance of getting it wrong!  Reply With Quote

5. Nice to have an excuse to get the old geometric grey matter working! I've just turned to paper and pencil and come up with the same answer as Neale. I don't fancy trying to work out the option where the two blue circles are not symmetrical about the horizontal line though. That turns it into a two dimensional problem instead of just one.

Kit  Reply With Quote

6. Originally Posted by Neale This one doesn't need much more than Pythagoras to solve it.
Then I would have expected a square root in the answer. Interesting question, but I would have liked to know how the solution was determined (the method, not the equation derivation).

Cheers,

Rob-T  Reply With Quote

7. Originally Posted by cropwell Then I would have expected a square root in the answer. Interesting question, but I would have liked to know how the solution was determined (the method, not the equation derivation).

Cheers,

Rob-T
Well, strictly, I suppose, Pythagoras was a means to an end, and it's one reason why there is a bunch of squared terms in the final answer.

I started by sketching the problem, putting in the dimensions as per my first response (above). You can draw a triangle with vertices at the intersection of horiz and vert lines, centre of small circle, and assumed centre of large circle (radius R). Then drop a perpendicular from centre of small circle to horiz line. You have a right-angled triangle with one side = h, and another = R-x.

Write out the equation for the square on the hypotenuse (length y in my sketch): y^2=h^2+(R-x)^2

But from the sketch you can see that y=R-r, so replace y in the Pythogoras equation with (R-r). Multiply it all out and the R^2 terms magically cancel each other out. Then it's just a matter of rearranging for R, which is where we came in. Technically, we should state some assumptions (like r<=x, r<=h) to make sure that we can actually draw the thing. I am also assuming that the large circle is tangent to the small circles on the side of the circle closer to the vertical line.

I would like to say that it was easy, but the real trick is spotting how to get some kind of relationship between the things we know and the one thing that we don't, and I happened to strike lucky with my "draw a triangle" approach.  Reply With Quote

8. I googled the problem and found this https://planetcalc.com/1421/ you have to scroll down to the 'Segment defined by chord and height' and that will give you the radius of that circle that passes through the centers of the smaller circle. So you just add on the radius of the smaller circle.

Cheers,

Rob-T  Reply With Quote

9. A big Thank you Neale for your solution to the problem. It works a treat:-) This will save me a load of time.

Just seen and read the way you went about solving this. I'll have to a closer look and brush up on my Maths.  Reply With Quote

10. Originally Posted by cropwell I googled the problem and found this https://planetcalc.com/1421/ you have to scroll down to the 'Segment defined by chord and height' and that will give you the radius of that circle that passes through the centers of the smaller circle. So you just add on the radius of the smaller circle.

Cheers,

Rob-T
Rob - unfortunately, it's not quite that simple! I generally try to find some sort of sanity check when I'm doing sums like this - does it give the right answer for the extreme cases? Does the answer move the right way if I change one of the given values? In this case, if you increase the size of the small circle to be equal to the segment height, the answer should be "infinity" for the radius of the big circle as the circle becomes a straight line. With your approach (which will be fairly close if the size of the small circles is small). as the small circle size increases, R increases - but that means that it no longer touches the vertical line. What's needed is the chord length and height at the points of tangency with the small circles - which you don't know at this point.

If it's any consolation, I'm typing this rather than being in the workshop because I've just fouled up a job on the CNC router. I wanted to move the tool up and clear so that I could add an extra fixing screw towards the end of a 1.5hr machining operation. Due to a complete brain fade, I moved Z up using machine rather than work coordinates - which meant that it plunged down rather than going up before moving in X and dragging work and spoil board sideways, leaving a big gouge in the work. The only good thing was that at least it was a thick spoil board so that the cutter didn't reach the machine bed, and I didn't break the 2mm carbide cutter. We all have off days...  Reply With Quote