Associated Topics || Dr. Math Home || Search Dr. Math

### Constructing a Tangent to a Circle, Continued

```Date: 03/26/2013 at 21:24:37
From: Zach
Subject: Constructing tangents to circles

At the following page, there is a construction of a tangent to a circle:

http://mathforum.org/library/drmath/view/55103.html

In referencing points, I will refer to its diagram:

puzzled over why the midpoint of segment PC is necessary for the

At first, my wife and I tried drawing a circle with center C and radius
CP. We called R and S the intersections of this circle with circle CQ'.
Drawing the segments CR and CS as stated means that the segments are
congruent, but CP is also congruent to CR and CS. This somehow breaks down
in my mind.

Any ideas? Does my confusion even make sense?

```

```
Date: 03/26/2013 at 22:45:15
From: Zach
Subject: Constructing tangents to circles

It seems that the point of tangency to the circle is also unique. How come
the midpoint construction creates the point of tangency (from an exterior
point) to a circle?

```

```
Date: 03/27/2013 at 20:00:17
From: Doctor Peterson
Subject: Re: Constructing tangents to circles

Hi, Zach.

You're essentially asking for a proof (or justification) of the
construction. Let's see what we can do.

Here is the reasoning behind the construction, as stated on that page:

Let C be the midpoint [center] of the circle, and P the point
outside the circle, through which point we are given the task of
constructing a tangent to the circle. Suppose that Q will be the
point where this tangent will meet the circle. Then angle(PQC)
will be 90 degrees, so Q must be on the circle with diameter PC.

What Doctor Floor is saying is that the goal of the construction is to
make angle PQC -- between the tangent-to-be PQ and the radius CQ -- a
right angle, since that is necessary for any tangent. We have a theorem
that an angle inscribed in a semicircle is a right angle; that is, if an
angle the vertex of which is on a circle has its sides passing through the
endpoints of a diameter, then the angle itself is a right angle. So in
order to make PQC be a right angle, we want Q to lie on a circle of which
PC is a diameter.

That is why we find the midpoint M of PC: When we draw the circle with its
center at M, points P and C will be the endpoints of a diameter of that
circle, and PQC will be a right angle.

A proof would go something like this: By construction, Q is on circle M,
so angle PQC is a right angle; since CQ is a radius, PQ is tangent to
circle C, as desired.

By the way, we have another page that refers to that one, and gives two
other constructions for the tangents. The one at the end of the page is
particularly interesting (but I'd have to do some thinking to reconstruct
how I proved it):

Constructing Tangents to Circles
http://mathforum.org/library/drmath/view/60678.html

Based on your last paragraph, I think you misstated what you did. I
believe you meant to say that you drew a circle with center P, not C,
having radius CP, so that it passed through C; and hoped that the
intersections of this with the given circle, R and S, would give the
desired tangents PR and PS. They don't. PR will be perpendicular to your
circle (arc RC) -- not to radius RC, as needed.

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/

```

```
Date: 04/08/2013 at 18:40:53
From: Zach
Subject: Constructing tangents to circles

Thank you very much, I appreciate this. I need to ponder over it a little,
but the basic idea makes sense.

Thank you! :)
```
Associated Topics:
High School Conic Sections/Circles
High School Constructions

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search