Draw a tangent line to a circle

[Johnny5]

In another thread, I asked how would you locate the center of a given circle, using only a compass and straightedge, and I got absolutely wonderful answers.

 

During that thread, many solutions involved already knowing how to construct a tangent line to one of the points on the circumference of the given circle.

 

I must confess, I don't know how to do it.

 

So here is a related question.

 

Using only a compass, and a straightedge, how do you construct a line which is tangent to a given circle, at a given point on the circumference?

 

Regards

[5614]

Just look at the circle and do it!

 

Select a point on the circle and draw a straight line so it touches that point.

 

Due to the definition of "circle" and "straight line" if the straight line touches the circle at one (on its outter side) then it cannot touch any other part of the circle, thus its a tangent.

[Johnny5]

Just look at the circle and do it!

 

Select a point on the circle and draw a straight line so it touches that point.

 

Due to the definition of "circle" and "straight line" if the straight line touches the circle at one (on its outter side) then it cannot touch any other part of the circle' date=' thus its a tangent.[/quote']

 

Think about the question the following way. One of the postulates of Euclid, was that you can draw a straight line from any point, to any other point.

 

So, in order to use that postulate, you need two points. A point on the circumference of the circle is given to you. BUT, and here is the whole point of the excersise (and I myself don't know the easiest way to do it), the goal is to find a second point, somewhere in the plane of the given circle, which lies on the tangent line. Then you can use postulate one of Euclid, to draw the straight line from the point on the circle to this other point which you found, and then you can use posulate two of Euclid to elongate the straight line from either endpoint. Then you have to logically demonstrate that the straight line just constructed is actually tangent to the circle.

 

It's a very hard problem.

[5614]

I still can't see what's so hard about it.

 

Use your eyes, chose a point on the circle and a second point which if you drew a line between these two points the line would only touch the circle at the original point and would not cross through the circle, join up the two points and you've got a tangent.

[syntax252]

 

I see that I should have said that when you draw the line from the center of the circle to the arc on the outside of the circle, it needs to pass through the spot that you marked on the circumfrence.

[Johnny5]

 

I see that I should have said that when you draw the line from the center of the circle to the arc on the outside of the circle' date=' it needs to pass through the spot that you marked on the circumfrence.[/quote']

 

I followed your construction!

 

How did you just know that you had to adjust the compass to more than 1/2 the distance, and you would succeed though?

 

I wouldn't have figured that out anytime soon.

 

Also, your comments about trial and error confused me a bit. Initially you just wanted to inscribe a regular hexagon inside the given circle right?

[syntax252]

I followed your construction!

 

How did you just know that you had to adjust the compass to more than 1/2 the distance' date=' and you would succeed though?

 

I wouldn't have figured that out anytime soon.

 

Also, your comments about trial and error confused me a bit. Initially you just wanted to inscribe a regular hexagon inside the given circle right?[/quote']

 

Starting with a circle already drawn and only a compass and a straightedge, I first had to find the center of the circle. As we discussed yesterday, there are a number of ways to do that, but the side of a hexegon that is just inscribed by a circle, is equal to the radius of the circle so I used that method. Once the compass was set to the redius, it is simple to mark the center of the center by drawing two arcs from different places on the circumfrence.

 

Then mark the point at which you want the lind to be tangent, and from that marked spot, draw an arc opposite the center of the circle that is in line with the marked point and the center.

 

Now, if you draw a line from the center to that outer arc, and it passes through the place you picked as the tangent spot, you will have a line that is exactly 2 radius long that passes through the point in question.

 

All you need to do is construce a perpendicular line that divides the 2 radius line exactly in half which is what you get if you draw arcs from each end that are more than 1/2 the distance from each end.

[Johnny5]

 

Now' date=' if you draw a line from the center to that outer arc, and it passes through the place you picked as the tangent spot, you will have a line that is exactly 2 radius long that passes through the point in question.

 

All you need to do is construce a perpendicular line that divides the 2 radius line exactly in half which is what you get if you draw arcs from each end that are more than 1/2 the distance from each end.[/quote']

 

I see it. After you increased the radius of your compass, you were able to find two points, where arcs intersect. You then could construct a parallelogram, with one of its diagonals equal to 2R, where R is the radius of the given circle, and the other diagonal is tangent to the given circle at the given point. Marvelous.

 

I'm not exactly sure how you would logically prove that the line drawn is indeed tangent to the given circle at the given point, but i can see that your construction works anyhow.

[syntax252]

I see it. After you increased the radius of your compass' date=' you were able to find two points, where arcs intersect. You then could construct a parallelogram, with one of its diagonals equal to 2R, where R is the radius of the given circle, and the other diagonal is [i']tangent [/i] to the given circle at the given point. Marvelous.

 

I'm not exactly sure how you would logically prove that the line drawn is indeed tangent to the given circle at the given point, but i can see that your construction works anyhow.

 

The reason that it works is because the circle divides the R2 line at the exact center. Now, say that you have a line that is 4" long. If you adjust your compass to 3" and put the point at one end of that line and draw a circle and then go the other end of that line and draw another circle, you will find that those 2 circles intersect in two places, once above the line, and once below the line. A line drawn between those two points of intersection will be perpendicular to the line, and in the center of the line.

 

That is how you know that the second line drawn in my diagram is both tangent to the circle and tangent at the point that you chose.

[Johnny5]

You know, this was one of the first things I thought about this morning. I was running through your construction again.

 

I was also thinking about how to prove (in the Euclidean spirit), that the perpendicular line which you finally drew is tangent to the circle at the given point.

 

You have a given circle and a given point on it. You draw a radius, from the center of the circle to the given point. Now, all you have to do, is construct a line which is perpendicular to the radius, and you're done, you have the tangent.

 

But how would Euclid have concluded that the line you finally constructed is a tangent. That's what I was thinking about this morning. I know he would label everything, but exactly how he would draw the conclusion i don't know.

 

So now, I'm going to see where in Euclid, such a construction is given, and put a link here. It just seems appropriate.

 

Here it is, book three, proposition 17.

 

Euclid's Elements, Book III, Proposition 17

 

In the proposition above, the given point was outside the given circle, which wasn't this problem, and I don't see this exact problem in Book three.

[syntax252]

You know' date=' this was one of the first things I thought about this morning. I was running through your construction again.

 

I was also thinking about how to prove (in the Euclidean spirit), that the perpendicular line which you finally drew is tangent to the circle at the given point.

 

You have a given circle and a given point on it. You draw a radius, from the center of the circle to the given point. Now, all you have to do, is construct a line which is perpendicular to the radius, and you're done, you have the tangent.

 

But how would Euclid have concluded that the line you finally constructed is a tangent. That's what I was thinking about this morning. I know he would label everything, but exactly how he would draw the conclusion i don't know.

 

So now, I'm going to see where in Euclid, such a construction is given, and put a link here. It just seems appropriate.

 

Here it is, book three, proposition 17.

 

Euclid's Elements, Book III, Proposition 17

 

In the proposition above, the given point was outside the given circle, which wasn't this problem, and I don't see this exact problem in Book three.

 

 

OK, I can see that we are missing wach other here, sooo.......

 

The key to the final line being tangent to the circle is that when I drew the line from the center through the circumfrence and out to the are that arc I had drawn, that line was exactly 2R in length.

 

So knowing that one can devide any line precisely between two points by the method that I discribed, and that that line will be perpendicular to the other line, it is a simple matter to establish that perpendicular line precisely at the 1R point.

 

I cannot tell you where you will find the law regarding deviding the line, because it is something I had to learn about 40 years ago to be a machinist, but I think it was part of what I was taught about trig. This method of establishing perpendicular lines and deviding a line is something that is used every day in machine shop work at the layout bench.

 

EDITED TO ADD:

 

This guy explains it better than I do.

http://mathforum.org/~sarah/hamilton/ham.fractions.ruler.html

[Asimov Pupil]

if you can come up with the equation of the circle at any point. then the tangent is just the derivative. so just pick a point.

[Primarygun]

I'm lazy to read the replies. I don't know whether the problem has been solved.

I tell you my solution.

1.Centre can be easily found. Join the given point to the centre. Now, we got a line passing through the given pt..

2.Use the given point as the centre of a circle, draw its circumference.

So we got two points on the line drawn.

3. Then, the problem can be easily solved as we just need to find out the perpendicular bisector of the line. And it's the tangent.

[Asimov Pupil]

that's what i said but i meant using calculus