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A line is given and two points in one half plane of the line are given. Construct a circle passing through these two points such that the given line is tangent to this circle.

I have no idea how to do this. I have derived the radius of the required circle in the terms of perpendicular distances of the two points from the line and distance between them but it is too much complex and not useful.

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  • $\begingroup$ A quick picture should tell you that this is impossible. $\endgroup$ – JP McCarthy Feb 24 '15 at 14:49
  • $\begingroup$ @Jp McCarthy: What is impossible? $\endgroup$ – Bernard Feb 24 '15 at 15:24
  • $\begingroup$ @Bernard I read the question incorrectly. $\endgroup$ – JP McCarthy Feb 24 '15 at 15:56
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    $\begingroup$ you need this: math.stackexchange.com/questions/793125/… everything else is easy enough $\endgroup$ – Will Jagy Feb 24 '15 at 20:35
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    $\begingroup$ @WillJagy's comment and the link really do answer your question. Do you need more detail? I saw that link a few months ago and I have a construction that is a little more simple. $\endgroup$ – Rory Daulton Feb 27 '15 at 11:30
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Here is a construction. There are actually two solutions.

The construction relies on the notion of power of a point with respect to a circle. It can be proved that, for any point $K$ in the plane and any circle with centre $O$ and radius $R$, if a line through $K$ intersects the circle in two points $A$ and $B$, the product $KA\cdot KB$ is constant (and is called the power of $K$ w.r.t. the circle):

enter image description here

This power is $\enspace KA \cdot KB =KT^2 =KO^2- R^2.$

In particular if the line $(KT)\,$ is tangent to the circle at $T$, $KT$ is the geometric means of $KA$ and $KB$.

Thus in order to solve the problem, we have to take the intersection $K$ of the line $(AB)$ with the given line, and construct geometrically the geometric means of $KA$ and $KB$.

This can be done with the right triangle altitude theorem: let $C$ be the symmetric of $A$ ww.r.t. $K$ and $P$ be an intersection of the line perpendicular to $(AB)$ through $K$ and the circle with diameter $[AB]$:

enter image description here

By the right triangle altitudet theorem, $KP^2=KC\cdot KB=KA\cdot KB$.

There remains to take the intersections $T$ and $U$ of the circle with centre $K$ and radius $KP$ with the given line. These will be the points of contact of the circles with the line.

To finish the construction, the centres of the circles are the points of intersection of the lines perpendicular to the given line through $T$ and $U$, with the perpendicular bisector of $[AB]$:

enter image description here

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