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Constructing Tangents to Circles

Date: 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 

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: 

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 

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 
Associated Topics:
High School Constructions
High School Coordinate Plane Geometry

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