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 March 15th, 2015, 04:31 PM #1 Newbie   Joined: Mar 2015 From: uk Posts: 13 Thanks: 0 infinite straight line to circle transformation Consider the line x = 1/2, can I algebrically transform this line into a unit circle centered around some point . March 16th, 2015, 03:50 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,675 Thanks: 2655 Math Focus: Mainly analysis and algebra Yes. The centre may be at $(0, -\frac12)$. The line passing through $(0, -\frac12)$ - the point on the circle farthest from the line - and $(x, \frac12)$ intersects the circle at one point, which is the image of $(x,y)$. You can do this with a circle above the line instead. Thanks from topsquark March 16th, 2015, 06:56 AM #3 Global Moderator   Joined: Dec 2006 Posts: 20,937 Thanks: 2210 Did you mean the point nearest the line x = 1/2 where the line through (0, -1/2) and (1/2, y) intersects the circle is the image of (1/2, y)? I don't see how this can work, as there will be one point on the circle that is not in the image of the line. March 16th, 2015, 07:00 AM #4 Global Moderator   Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms The projective line x = 1/2 (or any other projective line!) is homeomorphic to any given circle on the affine plane. But the affine line x = 1/2 is only homeomorphic to a circle minus a point. Thanks from topsquark March 16th, 2015, 12:21 PM #5 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,675 Thanks: 2655 Math Focus: Mainly analysis and algebra Here's a diagram. Our point $C(a,y)$ is to be projected onto the unit circle centred at $A(a-1,b)$. It's image is at $E(x',y')$. Now $|\vec{BC}| = y-b$ and $|\vec{DB}| = 2$. These are the opposite and adjacent (respectively) of the angle $\alpha$ in the right-angled triangle $\triangle{BCD}$. Thus we have $$\tan \alpha = {y-b \over 2} = t$$ Now, the triangle $\triangle{ADE}$ is isosceles since $|\vec{AD}| = |\vec{AE}| = 1$ so $\angle{AED} = \alpha$ and thus $\angle{DAE} = \pi - 2\alpha$ and $\angle{BAE} = 2\alpha$. Thus we have $$x' = a - 1 + \cos 2\alpha \qquad y' = b + \sin 2\alpha$$ Now we notice that $$\cos 2A = \cos^2 A - \sin^2 A = {1 - \tan^2 A \over sec^2 A} = {1 - \tan^2 A \over 1+\tan^2 A} \\ \sin 2A = 2\sin A \cos A = { 2\tan A \over sec^2 A} = {2 \tan A \over 1+\tan^2 A}$$ And thus we have $$x' = a - 1 + {1 - t^2 \over 1+t^2} \qquad y' = b + { 2t \over 1+t^2}$$ or $$x' = a - 1 + {4 - (y-b)^2 \over 4 + (y-b)^2} \qquad y' = b + {2(y-b) \over 4 + (y-b)^2}$$ Note that $b$ is arbitrary in terms of the validity of the projection, so we would normally choose $b=0$. Thanks from topsquark March 16th, 2015, 02:15 PM #6 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,675 Thanks: 2655 Math Focus: Mainly analysis and algebra $$y' = {4(y-b) \over 4 + (y-b)^2}$$ Thanks from topsquark Tags circle, infinite, line, straight, transformation Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post shalikadm Algebra 3 January 24th, 2012 10:48 PM colinbeaton1 Algebra 1 January 15th, 2011 03:11 PM outsos Algebra 7 December 23rd, 2010 07:15 PM Kiranpreet Algebra 3 August 10th, 2008 07:45 AM Kiranpreet Algebra 2 August 9th, 2008 09:29 AM

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