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The theory said: The inverse of a point $P$, with respect to a circle centered at $O$ and has a radius $r$, is the point $P'$ such that

  1. The three points $O$, $P$ and $P'$ are colinear.
  2. $OP \times OP'=r^2$

But I don't figure out how to calculate it. For example, given a point $P=(x,y)$, what is the formula to calculate its inverse $P'=(x',y')$ with respect to the circle centered at $O=(0,0)$ and has the radius $r=1$.


Solution

Thanks to @Cameron Buie 's hint the solution is $x'=\alpha x$ and $y'=\alpha y$ where $\alpha = \frac{r^2}{x^2 + y^2}$.

And for the more general case with the circle of inversion centered at any point $O=(h,k)$ rather than only at the origin, the solution becomes $x'=\alpha (x-h) + h$ and $y'=\alpha (y-k) + k$ where $\alpha = \frac{r^2}{(x-h)^2 + (y-k)^2}$

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  • $\begingroup$ Nice work! +1 for cranking it out on your own. For extra fun, you can generalize this approach to figure out how to sign symmetric points when the circle is centered at a point $A$ other than the origin. $\endgroup$ Jun 28, 2013 at 15:08
  • $\begingroup$ Did you mean when the circle is centered at $A=(h,k)$. Then the formula becomes $\alpha = \frac{r^2}{(x-h)^2 + (y-k)^2}$, right? Thank a lot. :D $\endgroup$ Jun 28, 2013 at 15:34
  • $\begingroup$ That is almost the whole story. What is $x'$? $y'$? $\endgroup$ Jun 28, 2013 at 17:05
  • $\begingroup$ I'm not sure, could it be $x'=\alpha (x-h) + h$ and $y'=\alpha (y-k) + k$. $\endgroup$ Jun 28, 2013 at 20:24
  • $\begingroup$ And done! Good work. $\endgroup$ Jun 28, 2013 at 22:30

1 Answer 1

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Hint: Since you want the points collinear and your circle is centered at the origin, then you need $x'=\alpha x,y'=\alpha y$ for some positive $\alpha$. Can you use requirement number $2$ to solve for $\alpha$ in terms of $x,y,r$?

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