# A couple of basic questions, arc, circle tangency

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I’m pretty new to AutoCAD, but am familiar with TurboCAD, and I’m trying to figure out how to do some pretty simple operations in AutoCAD. It seems to me that some arc and circle tools are missing and I wonder what is the best way to work around this

1 – how can I draw a circle tangent to endpoint #2. I can’t find a circle tool which takes a single tangent line and a radius, only TTR or TTT. Is there any way simpler to do this than adding a perpendicular construction line from the radius point?

2- If I want to draw a tangent circle between the 2 lines I would use TTR, but is there any way to get a ‘dynamic’ circle drawn after you pick the two tangent points like when you draw a 3 point circle but with the 2 points moving tangent the picked lines ?. For instance after I pick the 2 lines I want the circle to be tangent at the mid-point at line 3-4.

3 – Drawing an arc tangent from the last point by hitting enter (space) is a really nice feature, but is there any way to ‘set’ the point back to e.g. point #2, or how can I start an arc tangent to this point?

4 – in general, what is the easiest way to draw a tangent line FROM and arc and draw and arc tangent FROM a line?

Jesper

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oops, just realized I forgot the image

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AutoCAD has all the tools one needs for circle and arc, but as yet has no automatic operations that cover up for a basic lack of knowledge of geometrical properties. Do not be afraid to draw temporary construction lines! They are easily erased afterwards. There are many ways of getting to the same end result, and everyone works out a way that suits them. Try them all.

1 - Offset line by the radius, then drawn the circle centred on the end of the line.

2 - Why do you want a dynamic circle when you know that your tangent is the midpoint of 3-4? Construct the centre of the circle which lies at the intersection of the perpendicular bisector of 3-4 and the angle bisector of the lines 1-2 & 3-4.

3 - AutoCAD only remembers where the end of the last entity was drawn, so there is no way to 'set' the point back to #2. If you have not planned ahead to draw things in a useful order, then draw a temporary line at the end of line 1-2 (first point near the line and the end of the line at #2).

4 - If the arc or line is finished at the correct place, then drawing a line after the arc, or an arc after the line, it is always tangential if, when asked for a starting place, you just hit Enter (or Space bar).

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I appreciate your reply. I do have some basic knowledge of geometric as you mention, but I’m still learning AutoCAD and I'm always looking for the fastest way to do things (as few clicks and keyboard commands as possible), and being used to TurboCAD I was looking for similar ways to solve the problems in AutoCAD.

#1 - thank you for this idea, this is definitely faster than drawing a perpendicular line 'from point - to line' and then move it to the end point.

#2 - Why did I ask this ? – For instance when you use the cad for something like a trackplan sketch (model train layout), you won’t always now your radius, all you know is your to tangent lines and then you 'play' with the circle until it fits. I can see in AutoCAD I would do this by adding the 3rd line indicating my max radius and then user TTT.

A follow-up question for the angle bisector, the only way I can figure out to do this is to fillet the 2 lines with radius 0. Now if those two lines were ‘close’ to parallel, the vertex point would be somewhere way off the drawing area, is there an easier way to find the bisector point of the angle between 2 lines?

Jesper

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My apologies if I seemed a bit harsh, but some people (present company excepted) think that AutoCAD is magic instead of it being just a tool.

For the angle bisector, I would draw a circle centered on the intersection of the lines (start the Intersection and pick the lines one at a time), and then pick a radius point so the circle crossed your two lines. Then trim the external arc (leaving the small arc between your lines) and draw a line from the centre point of the arc to the midpoint of the arc. As long as your two lines are not parallel, the vertex point will be within the drawing (which extends to infinity).

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No need to apologize, I should have spend a little more time playing before I asked the question.

Anyway, in case anyone is interested, these are the steps I found to make an arc tangent to 2 lines when only one point on a line is given.

Problem:

Given, line A, B and point pA, now draw an arc tangent to both lines tangent at point pA – without resizing or panning the drawing area.

Solution:

#1 – create perpendicular line at pA. quickest way I found (someone pointed it out to me, I didn’t find it myself) is to use the ROTATE / COPY command.

Select ROTATE, select line A, set base point to pA, select Copy, rotate 90 – done.

#2 – find the bisecting angle between the 2 lines, here I use the idea given by eldon, but instead of drawing a circle I use a construction line XLINE and chose Bisect. At the ‘specify angle vertex point’, shift –rightclick and selected “apparent intersect” from the OSNAP menu, select both lines. At the ‘angle start point, select one of the lines (anywhere ON the line), ‘specify angle endpoint’, select second line. Now you have a construction line dividing the angle between the 2 lines.

#3 – now draw the arc, using the intersection point at pC as center, then click on pB and finally pA.

Can anyone come up with a quicker way (less clicks and commands) ?

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Here’s different method. I wouldn’t say, necessarily, that it’s quicker or requires fewer clicks – or, for that matter, that it is completely different. It may be of some interest however:

1. Construct perpendicular “A” as previously described.

2. Mirror “B” about “A+ 90"

3. Circle – with 3pt option and running Tangent osnap. Pick A, B, mirrored B

4. Delete construction geometry and Trim circle to Arc

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Here’s different method. I wouldn’t say, necessarily, that it’s quicker or requires fewer clicks – or, for that matter, that it is completely different. It may be of some interest however:

Good one, I would never have thought of that solution. Just shows that the same problem can be solved in many ways.

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Solution:

#1 – create perpendicular line at pA. quickest way I found (someone pointed it out to me, I didn’t find it myself) is to use the ROTATE / COPY command.

Select ROTATE, select line A, set base point to pA, select Copy, rotate 90 – done.

#2 – find the bisecting angle between the 2 lines, here I use the idea given by eldon, but instead of drawing a circle I use a construction line XLINE and chose Bisect. At the ‘specify angle vertex point’, shift –rightclick and selected “apparent intersect” from the OSNAP menu, select both lines. At the ‘angle start point, select one of the lines (anywhere ON the line), ‘specify angle endpoint’, select second line. Now you have a construction line dividing the angle between the 2 lines.

#3 – now draw the arc, using the intersection point at pC as center, then click on pB and finally pA.

Can anyone come up with a quicker way (less clicks and commands) ?

Excuse me, but your geometry does not fit.

In your solution, the lines are drawn nearly parallel, but if you exaggerate the dividing angle, you will see what you have drawn is not exact.

#1. I drew the perpendicular bisector using polar, and drawing a line at right angles to A. When this line crosses B the angle is not a right angle, therefore pB is not a tangent point!

#2. I like the XLINE. It was not available when I started, but I am still learning.

#3. If you draw an arc centred on pC and starting on pB, it does not pass through pA.

#4. However, if you draw a line from pC perpendicular to B, you arrive at the proper position for pB1.

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One comment about drawing arcs. There are so many different set-ups, and you have to remember that it draws in an anticlockwise direction, that I find it easier to draw a circle and trim.

In your problem, you could have drawn an arc (Centre, Start, Angle), centred on pC, starting on pA, with an angle of -180 (to make it draw the right way). Then extend that arc to line B.

P.S. I thought SEANT's solution was very nice

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Excuse me, but your geometry does not fit.

In your solution, the lines are drawn nearly parallel, but if you exaggerate the dividing angle, you will see what you have drawn is not exact.

#1. I drew the perpendicular bisector using polar, and drawing a line at right angles to A. When this line crosses B the angle is not a right angle, therefore pB is not a tangent point!

You are absolutely right, of course pB must be perpendicular to B for the tangent point as you have shown in pB1. - I don't know what I was (or wasn't) thinking :oops:

I think I will just stay quiet for the rest of the day.

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