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How do we find center of a circle passing through three points:

$ A(x_1,y_2,z_3),B(x_1,y_2,z_3),C(x_1,y_2,z_3) $?

Can we minimize $ (d_{OA}+...+... ) $ with condition $ d_{OA}=...=... ,$

or is the volume of tetrahedron $ OABC =0$ adequate to find it with the latter distance equalities ?

Is the result/formula known? Thanks for the indications.

Ken
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Narasimham
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  • Do you know off hand what the values of $A, B$, and $C$ are? or do you want a formula in general terms? – TravisJ May 29 '15 at 16:24
  • @Narasimham: I'd consider perpendicular bisector planes of $AB,,BC,,CA$ intersection (it's a line) and intersection of this line with plane $ABC$. – Alexey Burdin May 29 '15 at 16:26

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Define

$$\begin{cases} a &= |\vec{B}-\vec{C}|,\\ b &= |\vec{C}-\vec{A}|,\\ c &= |\vec{A}-\vec{B}| \end{cases} \quad\text{ and }\quad \begin{cases} \alpha &= a^2(b^2+c^2-a^2)\\ \beta &= b^2(c^2+a^2-b^2)\\ \gamma &= c^2(a^2+b^2-c^2) \end{cases}$$ The center you want is given by the formula

$$\vec{X}_{circum} = \frac{\alpha \vec{A} + \beta \vec{B} + \gamma\vec{C}}{\alpha + \beta + \gamma}$$

What we have done is lifting the formula for circumcenter in barycentric coordinates from $\mathbb{R}^2$ to $\mathbb{R}^3$. The $X_{circum}$ so constructed is the circumcenter of the triangle $\triangle ABC$ in the plane containing it.

achille hui
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