Hello,
I have found a problem in College Geometry by N.A. Court, Page 29.
and making very little progress in solving it. Can someone help ?

The problem :
In a given circle to draw a diameter such that it shall subtend a given angle at a given point. ( I placed the point outside of the circle )

>The problem :
>In a given circle to draw a diameter such that it shall
>subtend a given angle at a given point. ( I placed the point
>outside of the circle )

Have a look at

http://www.cut-the-knot.org/Curriculum/Geometry/Inversion2.shtml

The page is about inversion with negative power.

All circles that pass through a given point A and two diamterically opposite points on a given circle, always pass through another point A'. You can get A' by inverting A in the given circle and then reflecting the result in the its center.

The circle you are looking for will pass through both A and A'. On the other hand you can find its radius or diameter from the condition that an inscribed angle subtends a chord of a given length (= the diameter of the given circle.)

You must be able to construct a circle of the required size through two available points.

why did you put the point outside the circle?

>why did you put the point outside the circle?

I did not, tvarhegyi did, probably with a reference to a diagram. The fact is not essential for the construction.

Alex, For the problem from Court you used a property of inversion
that Tvarhegyi didn't know. It wasn't necessary. Let A be the point
and S_{0} the circle with centre O. Take any diameter BC not assing
through A and draw the circle S_{1} through A,B,C. Any circle that
goes through A and the intersection of S_{1} with AO produced will
meet S_{0} in diametrically opposite points because the three radical axes meet at O.

As a retired engineer I am going to supply the solution alike one of my draftmen will probably would have given to me.

1) Chose any diameter D in your circunference c and center O draw an isosceles triangle, being D the base and others two equal sides, forming the given angle, let us call it W

2) Draw a new circunference q passing for the three vertexs of the isosceles triangle. So any given point of the circunference will form an angle W with the ends points of the D which has became a chord of circunference q.

3) Draw another circunference r with the same center O that the original circunference c and passing for the given point P. The circunference r the radius OP will intersect the circunference q in two, zero points an exceptionally in one.

4) If there is zero point there is no real solution, if there are two point lets call them M1 and M2 there are two solutions. Lets take one of the two points, M1 draw a radius OM1, which will form and angle X with OP.

5 ) Now all you have to do it with center in O rotate the circunference q including the diameter D (a chord of circunference q.) an angle X, so M1 will transform into P, and the original diameter D will rotated to D1 which is one of the solutions. The same procedure with the other M2.

Of course you can attack the problem using vectors, is an elegant concise equation using the scalar product. However to solve the radical that will appears in the denominator ends in a cumbersome quartic equation with two or four non real solutions.

Same things will happen using complex numbers

There is at least one more beautiful approach using stereographic projection somewhat linked to AlexB response but for me with English as a second language it is very difficult to express. But I believe the above one is real simple.

Thank you for giving the solution to the N.A.Court problem I posted.
After several attempts I gave in and used brute force, that is trigonometry.

I implemented your construction using Geometer's Sketchpad. It is an elegant and in retrospect a "simple" solution, reminds me the problem sets found in Yaglom's Geometric Transformations.

I could not quite figure out the solution suggested using circle inversion, but I will keep trying.

Tamas Varhegyi

My browser settings do not allow applets. So I haven't understood the inversion method given by Alex. Maybe i should begin studying the matter contained in this website (I am a new member and this is only my second visit to this site)
Here is a simple method to construct the diameter - (Refer to the attached file)

Plot another point on the other side of the center of the circle (mirror image of the point - about the center of the circle - at which the diameter subtends the given angle). Using the point and its mirror image, draw an arc that inscribes the supplementary angle of the angle that the diameter is supposed to subtend at the given point. The point of intersection of the arc and the circle gives one end of the required diameter.