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Given two points of a circumference in $\mathbb{R}^3$ space $P_1 = (x_1,y_1,z_1)$ and $P_2 = (x_2,y_2,z_2)$, and two coordinates of its center $x_c$, $y_c$, is it possible to obtain the missing $z_c$?

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  • $\begingroup$ Infinitely many circumferences go through two given points. All point of the straight bisector of the line segment determined by the two points is a center of one of these uncountable circumferences. $\endgroup$ – Piquito Nov 10 '17 at 12:44
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Note that $$(x_1-x_c)^2+(y_1-y_c)^2+(z_1-z_c)^2=(x_2-x_c)^2+(y_2-y_c)^2+(z_2-z_c)^2.$$ Expand, simplify and find $z_c$, given that $z_1\not=z_2$, \begin{align}z_c&=\frac{(x_1-x_c)^2-(x_2-x_c)^2+(y_1-y_c)^2-(y_2-y_c)^2+z_1^2-z_2^2}{2(z_1-z_2)}\\ &=\frac{(x_1-x_2)(\frac{x_1+x_2}{2}-x_c)+(y_1-y_2)(\frac{y_1+y_2}{2}-y_c)+\frac{z_1+z_2}{2}}{z_1-z_2}. \end{align} When $z_1=z_2$ if
$$(x_1-x_c)^2+(y_1-y_c)^2=(x_2-x_c)^2+(y_2-y_c)^2$$ then any $z_c$ is fine, otherwise the problem of finding $z_c$ has no solution.

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  • $\begingroup$ That was really helpful, now I know what to do. If z1=z2 then z1=z2=zc right? $\endgroup$ – Daniel Viaño Nov 10 '17 at 11:20
  • $\begingroup$ See my edited answer. Have you found $z_c$ when $z_1\not=z_2$? $\endgroup$ – Robert Z Nov 10 '17 at 11:27
  • $\begingroup$ zc = (-(x1-xc)² - (y1-yc)² + (x2-xc)²+(y2-yc)²-z1²+z2²) / (-2z1 + 2z2) $\endgroup$ – Daniel Viaño Nov 10 '17 at 11:33
  • $\begingroup$ Yes, compare with my answer. $\endgroup$ – Robert Z Nov 10 '17 at 11:51

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