I'm trying to solve this exercise:

Show that the transformation of inversion in the unit circle is given analytically by the equations $$x'=\frac{x}{x^2+y^2}, y'=\frac{y}{x^2+y^2}$$Find the inverse transformation and prove analytically that inversion transforms the totality of lines and circles into lines and circles.

I know that the image of a point $P$ is defined to be the point $P'$ lying on the line $OP$ (where $O$ is the center of a unit circle) on the same side of $O$ and $P$ and such that $OP\cdot OP'=r^2$; which, with our circle, would mean $OP\cdot OP'=1$. Based on this information, how do I figure out what a function might look like? Obviously it has to correspond to $x'$ and $y'$, but I can't see how.

In terms of figuring out the inverse function, all I can get to is:

$$ x^2x'+y^2x' = x \\ y^2x'=x(1-xx') $$

Is there a way of neatly factoring to get $x$ and thus inverting the function?

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    $\begingroup$ For finding the inverse function, note that the criterion for "the image of $P$ is $P'$" is symmetric in $P$ and $P'$ -- that is, inversion in a circle is its own inverse! $\endgroup$ – Henning Makholm Aug 2 '15 at 19:57
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    $\begingroup$ @HenningMakholm so would the inverse function of $x'$ be $x=\frac{x'}{x'^2+y'^2}$? Should it be $y^2$ or $y'^2$? $\endgroup$ – hohner Aug 2 '15 at 20:01
  • $\begingroup$ x @hohner: Yes! $\endgroup$ – Henning Makholm Aug 2 '15 at 20:22
  1. $\vec{OP'}=\lambda \vec{OP}$, as $P$ and $P'$ are aligned. Then $OP\cdot OP'=\sqrt{x^2+y^2}\cdot\lambda\sqrt{x^2+y^2}=1$, giving the value of $\lambda$.

  2. The inversion is its auto-inverse, as


  1. Plugging the transformed coordinates in the implicit equation of a circle, $$\left(\frac x{x^2+y^2}-x_c\right)^2+\left(\frac y{x^2+y^2}-y_c\right)^2=r^2$$ we get $$x^2-2x_cx(x^2+y^2)+x_c^2(x^2+y^2)^2+y^2-2y_cy(x^2+y^2)+y_c^2(x^2+y^2)^2=r^2(x^2+y^2)^2$$ which can be simplified by $x^2+y^2$ $$1-2x_cx+x_c^2(x^2+y^2)-2y_cy+y_c^2(x^2+y^2)=r^2(x^2+y^2),$$

which is also the equation of a circle. In the event that $x_c^2+y_c^2=r^2$, the circle degenerates to a straight line.

  • $\begingroup$ Won't $\lambda$ equal $\frac{1}{x^2+y^2}$ rather than $\frac{x}{x^2+y^2}$? $\endgroup$ – hohner Aug 2 '15 at 20:19
  • $\begingroup$ @hohner: Yes, it will. Remember that $(x',y')=\lambda(x,y)$, so the numerator comes from this multiplication. $\endgroup$ – Henning Makholm Aug 2 '15 at 20:23
  • $\begingroup$ No, my question was not about substituting in the numerator. I still don't understand how you arrive at $x'=\frac{x}{x^2+y^2}$ in the first place. $\endgroup$ – hohner Aug 2 '15 at 20:25
  • $\begingroup$ @hohner: $(x',y')=\lambda(x,y)$ and $\lambda=\frac{1}{x^2+y^2}$, so $x'=\lambda x=\frac{1}{x^2+y^2}x=\frac{x}{x^2+y^2}$. $\endgroup$ – Henning Makholm Aug 2 '15 at 20:28
  • $\begingroup$ @hohner: I think this discussion doesn't belong here. $\endgroup$ – Yves Daoust Aug 2 '15 at 20:33

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