Constructing a Tangent to a Circle, ContinuedDate: 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: My wife is teaching a lesson about this in a geometry class. We have both puzzled over why the midpoint of segment PC is necessary for the construction. What is special about this point M, and not another point? 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! :) |
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