How to calculate the inverse of a point with respect to a circle?
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
- The three points $O$, $P$ and $P'$ are colinear.
- $OP \times OP'=r^2$
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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$\begingroup$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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