Question

show that the point A(2,-6,0),B(4,-9,6),C(5,0,2),D(7,-3,8) are concyclic.

Answer 1

Answer on Question #52829 - Math – Analytic Geometry

Show that the point $A(2,-6,0)$, $B(4,-9,6)$, $C(5,0,2)$, $D(7,-3,8)$ are concyclic.

Solution

First of all, for A, B, C and D to be concyclic they all should lie in a single plane. It is true if the following condition holds:

$$\overline{AB} \cdot (\overline{AC} \times \overline{AD}) = 0$$

Since $\overline{AB} = (4, -9, 6) - (2, -6, 0) = (2, -3, 6)$, $\overline{AC} = (5, 0, 2) - (2, -6, 0) = (3, 6, 2)$ and

$\overline{AD} = (7, -3, 8) - (2, -6, 0) = (5, 3, 8)$, we obtain

$$\begin{array}{l} \overline{AB} \cdot (\overline{AC} \times \overline{AD}) = \begin{vmatrix} 3 & -3 & 6 \\ 3 & 6 & 2 \\ 5 & 3 & 8 \end{vmatrix} = 3 \begin{vmatrix} 6 & 2 \\ 3 & 8 \end{vmatrix} + 3 \begin{vmatrix} 3 & 2 \\ 5 & 8 \end{vmatrix} + 6 \begin{vmatrix} 3 & 6 \\ 5 & 3 \end{vmatrix} = 3 \cdot 42 + 3 \cdot 14 - 6 \cdot 21 = \\ = 42 \neq 0 \end{array}$$

It means that these 4 points don't lie in a single plane and therefore they can't be concyclic.

**Answer**: not concyclic.

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