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

[khoshravan]

This is a geometry question, but I think folks here can help me to find a solution to draw it with CAD capabilites

 

I have 2 points and a line given (both points on one side of line)

I want to draw a circle so that it passes through two points and tangent to the given line.

I tried a little bit but seams not be so easy (at least for me)

Any idea.

I have to say that still I couldn't solve the geometry of the question but there might be a solution in ACAD.

[Organic]

Draw -> Circle -> 3 Points

 

Click the two points then the line (which the circle will then be tangential to).

[khoshravan]

Draw -> Circle -> 3 Points

 

(which the circle will then be tangential to).

clicking the line, doesn't insure that it will be tangent.

[Organic]

Sorry, forgot the key point: Only have the 'Tangent' Osnap on when selecting the line (the 3rd point).

[Michaels]

Follow the steps ,

 

circle :

3p

1st point

2nd point

Right-Click with the mouse after holding down the Shift button and choose Tangent

Select the Line.

 

 

Congratulations .

 

Michaels

[Jack_O'neill]

Follow the steps ,

 

Right-Click with the mouse after holding down the Shift button and choose Tangent

 

Michaels

 

OR, simply type TAN at this point if you've changed your right click to do something besides displaying the osnap menu.

[khoshravan]

Follow the steps ,

 

circle :

3p

1st point

2nd point

Right-Click with the mouse after holding down the Shift button and choose Tangent

Select the Line.

 

 

Congratulations .

 

Michaels

 

Dear All

 

Thanks. I learned that it is possible to input new snap command "tan" in the middle of another command. Very nice solution.

[NAUTILUS]

amazing solution, just out of curiosity, does anyone knows the geometrical solution? ( without using CAD options)

[JD Mather]

amazing solution, just out of curiosity, does anyone knows the geometrical solution? ( without using CAD options)

 

Draw a circle with center at point 1 to point 2.

Draw a circle with center at point 2 to point 1.

Draw line from intersection of circle 1 and 2 perpendicular to the line. The third point is now known.

[eldon]

Draw a circle with center at point 1 to point 2.

Draw a circle with center at point 2 to point 1.

Draw line from intersection of circle 1 and 2 perpendicular to the line. The third point is now known.

 

I just tried that and it did not work for me. Could you please post a screen shot to show your construction? Thank you.

[Organic]

I just tried that and it did not work for me. Could you please post a screen shot to show your construction? Thank you.

 

I tried it and for the case I tried, it worked on one intersection point where the two original circles overlapped (the intersection point closest to the line) although failed on the other intersection point.

[eldon]

I am afraid that JD Mather mis-read the latest request that was for a Geometrical solution, and his solution must be said to be incorrect.

 

After a bit of delving, I think that the Tangent-Secant Theorem is the geometrical solution.

 

Given two points A and B and a tangent line, then proceed thusly:-

 

produce the line AB to meet the tangent line at C. Then the length CD is given by the equation

 

CB x CA = CD²

 

A calculator is needed (probably), but there are graphical solutions.

Edited May 13, 2013 by eldon
corrected expression

[Tyke]

I am afraid that JD Mather mis-read the latest request that was for a Geometrical solution, and his solution must be said to be incorrect.

 

After a bit of delving, I think that the Tangent-Secant Theorem is the geometrical solution.

 

Given two points A and B and a tangent line, then proceed thusly:-

 

produce the line AB to meet the tangent line at C. Then the length CD is given by the equation

 

AB x AC = CD²

 

A calculator is needed (probably)

 

eldon, how would that work if the line AB was parallel to the tangent line CD? You would never get an intersection. Honestly though, I can't see how to do it.

[eldon]

I was giving a general solution.

 

The case if the two points were parallel to the tangent line would be simplicity itself, because the angle at the circumference is half the angle at the centre.

[NAUTILUS]

i'm afraid its not the right solution.

[NAUTILUS]

.... A calculator is needed (probably)

 

then its not the geometric solution.

 

what I mean by geometric solution is pure drawing, no calculations.

 

thanks anyway

Edited May 13, 2013 by NAUTILUS

[eldon]

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.

Edited May 13, 2013 by eldon
corrected expression - oops

[eldon]

In fact, here is a link to do exactly that

 

O ye of little faith

[eldon]

In fact, I have just spotted my deliberate mistake.

 

In the posted picture, the equation should be CA x CB = CD²

[eldon]

eldon, how would that work if the line AB was parallel to the tangent line CD? You would never get an intersection. Honestly though, I can't see how to do it.

 

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