how to draw a circle passing from given 2 points and tangent to a given line

[Tyke]

Apologies for my previous answer. it was due to the fact that my brain skipped a couple of steps before my fingers hit the keyboard.

 

If A and B are parallel to the tangent line, then the median bisector of AB would intersect the tangent line at right angles, i.e. the intersection point would be the tangent point. This is a special case, although if you took the mathematics, you could probably convince yourself that CD was the average of CA and CB

 

Then the circle could be constructed by using median bisectors, or the fact that the angle at the circumference is twice the angle at the centre

 

Right gotcha I think my brain was a bit fogged up over the weekend, it's back to normal now that I' back in the office and the caffine levels are back up.

[NAUTILUS]

I was expecting that you could create a graphical solution for the term CA x CB = CD², and that is why I put the (probably) at the end.

 

You have a rectangle with sides CA and CB. All you have to do is to find the square of equal area which will give you CD.

 

Over to you now.

 

Im definitely not that kind of brain

 

had a look at your link, seemed to be the solution, but i guess I would use 3P (point, point, tan) of AutoCAD

 

thanks again

[irneb]

Hi guys, sorry to revive this thread but I've got a similar situation, only I'm trying to draw an arc between 2 fixed points extending tangentially to a circle. See this approximation (though the white arc is intersecting the green circle twice so it's not exactly tangential).

 

I've tried this and googled for other solutions, but this thread is the closest I could find. Perhaps someone could point me into some direction?

[nestly]

You may have to use CIRCLE with the 3Point option (Node,Node,Tan) and then TRIM it to get the desired result.

[JD Mather]

Since you are using 2013 you could also use geometry constraints on the Parameter tab.

[irneb]

You may have to use CIRCLE with the 3Point option (Node,Node,Tan) and then TRIM it to get the desired result.

Thanks yes! That actually works. Strange that it does on circles but not arcs. It's as if when drawing a circle the tan snap calculates a tan'ed curve, while the arc only does a tan line from last point.

 

Since you are using 2013 you could also use geometry constraints on the Parameter tab.

I suppose that's another option, yes. Thanks.

 

I'm just beating my head around this ... there should be some way to construct this shouldn't there? I mean, straight forward geometry? The circle idea does show that there is a way to calculate it - what it uses internally I don't know (either temporary / phantom geometry or algebraic Cartesian calculation).

[nestly]

Yeah, the 3Point arc method using TAN appears to place the 3rd point tangent to the last pick point, not the arc being drawn. I can see an argument either way.

 

[irneb]

Finally found it: http://apollonius.wikispaces.com/PPC

 

Though it's a very convoluted way of going about it. Especially those inverse geometries