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I tried to make a C++ program to make a hyperbolic line.

This is how I started:

I have two points, $P=(x_1, y_1)$ and $Q=(x_2, y_2)$, in the Poincaré disk. And the radius is 1.

$$\begin{align} \alpha &= \frac{1}{x_1^2 + y_1^2} \qquad P^{-1} = (\alpha x_1, \alpha y_1) \\[4pt] \beta &= \frac{1}{x_2^2 + y_2^2} \qquad Q^{-1} = (\beta x_2, \beta y_2) \end{align}$$

Then calculate the center of the circle as the center of the two inverted points and the radius.

But the new circle does not intersect $P$ and $Q$.

What I doing wrong?

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  • $\begingroup$ I took the liberty of changing some notation to be more mathematical. If you have strong objections, you may reverse those changes. $\endgroup$
    – Blue
    Commented Mar 2, 2020 at 21:53
  • $\begingroup$ Thank you! Sorry for my bad formatting. $\endgroup$
    – ricsi
    Commented Mar 2, 2020 at 21:54
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    $\begingroup$ The center of the target circle isn't "the center of the two inverted points". It's the center of the circle determined by (any three of) $P$, $Q$, $P^{-1}$, $Q^{-1}$. You can find this as the intersection of the perpendicular bisectors of two pairs of those points; you can standardize this by taking the pair $P$ and $P^{-1}$ and the pair $Q$ and $Q^{-1}$. $\endgroup$
    – Blue
    Commented Mar 2, 2020 at 22:07
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    $\begingroup$ Thank you very much, it's worked. $\endgroup$
    – ricsi
    Commented Mar 3, 2020 at 20:21
  • $\begingroup$ You should post your working solution as an answer, so that this question won't get auto-reposted in the future. (Plus, we can upvote your success.) $\endgroup$
    – Blue
    Commented Mar 3, 2020 at 20:34

1 Answer 1

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Based on that site, I can make it.

http://www.ambrsoft.com/TrigoCalc/Circle3D.htm

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