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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

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