Constructing Tangents to CirclesDate: 05/08/2002 at 23:32:23 From: Little Subject: tangent lines How do you construct a line tangent to a circle through a point outside the circle? And how do you construct a line tangent to a circle through a point outside the circle using only a straightedge? Date: 05/09/2002 at 05:13:36 From: Doctor Mitteldorf Subject: Re: tangent lines Let's call the circle center O and the point outside P. You want to find the point T on the circle such that OTP is a right angle. One way to do this is to find the correct length PT for a leg of a right triangle. Then you can swing an arc of this length from P and note where it intersects the circle. So our strategy is to construct this length, equal to sqrt(OP^2 - radius^2), by constructing a right triangle with leg equal to radius and hypotenuse equal to OP. To construct such a right triangle, first construct a right angle, then measure the distance [radius] along one leg, and from that measured point swing an arc equal in length to OP, marking where it intersects the other leg. Now you have the length of the leg that you need in order to swing the arc from P. There's a different solution to this problem in our archives: http://mathforum.org/library/drmath/view/55103.html The only way to do this without a compass is to hold one end of the straightedge on the point P and rotate the other end until it just touches the circle. This works pretty well in practice, but does not conform to standard rules about what constitutes a geometric construction. - Doctor Mitteldorf, The Math Forum http://mathforum.org/dr.math/ Date: 05/12/2002 at 20:58:48 From: Little Subject: tangent lines Thank you very much for your time and for answering my questions. Date: 06/23/2002 at 23:36:22 From: Doctor Peterson Subject: Re: tangent lines Hi, Little. You asked this over a month ago, but I just saw it recently and realized there is more that could be said about the second part. It is an amazing fact that any construction that can be done with compass and straightedge can be done with only a straightedge, as long as there is also a circle, with its center, available for use. That fits this case exactly, so I know there is a solution to the second part of your question. But ... is it an easy solution? Since you were asked to do it, it seemed worth trying, and I found a nice way to solve the problem! First construct the diameter OP through P, meeting the circle at A and B. Then construct any secant PCD of the circle through P. Now construct lines AC, AD, BC, and BD, and find the intersections E (of AC and BD) and F (of AD and BC). The two points where line EF cuts the circle, T and T', are the two points of tangency! - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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