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Representation of Circle as Conjugation of a Dyadic by Radial Vector

Theorem
Define a dyadicn  
\[\mathbf{\lambda} =\mathbf{i} \mathbf{i} + \mathbf{j} \mathbf{j} + \mathbf{k} \mathbf{k}\]
. Then  
\[\mathbf{r} \cdot (\mathbf{i} \mathbf{i} + \mathbf{j} \mathbf{j} + \mathbf{k} \mathbf{k}) \cdot \mathbf{r}=4\]
  descriibes a circle radius 2.
Proof
\[\begin{equation} \begin{aligned} \mathbf{r} \cdot \mathbf{\lambda} \cdot \mathbf{r} &=(x \mathbf{i} +y \mathbf{j} + z \mathbf{k}) \cdot (\mathbf{i} \mathbf{i} + \mathbf{j} \mathbf{j} + \mathbf{k} \mathbf{k}) \cdot (x \mathbf{i} +y \mathbf{j} + z \mathbf{k})= \\ &=(x \mathbf{i} +y \mathbf{j} + z \mathbf{k}) \cdot (\mathbf{i} \mathbf{i} + \mathbf{j} \mathbf{j} + \mathbf{k} \mathbf{k}) \cdot (x \mathbf{i} +y \mathbf{j} + z \mathbf{k}) \\ &= (x \mathbf{i} +y \mathbf{j} + z \mathbf{k}) \cdot (x \mathbf{i} (\mathbf{i} \cdot \mathbf{i}) + y \mathbf{j} ( \mathbf{j} \cdot \mathbf{j})+ z \mathbf{k} ( \mathbf{k} \cdot \mathbf{k}) ) \\ &= (x \mathbf{i} +y \mathbf{j} + z \mathbf{k}) \cdot (x \mathbf{i} + y \mathbf{j} + z \mathbf{k} ) \\ &= x^2 \mathbf{i} \cdot \mathbf{i} + y^2 \mathbf{j} \cdot \mathbf{j} + z^2 \mathbf{k} \cdot \mathbf{k} =4 \end{aligned} \end{equation}\]
%nbsp; which is the equation of a circle.
Note that  
\[\mathbf{i} \cdot \mathbf{j} = \mathbf{i} \cdot \mathbf{k}=\mathbf{j} \cdot \mathbf{k}=\mathbf{j} \cdot \mathbf{i}=\mathbf{k} \cdot \mathbf{i}=\mathbf{k} \cdot \mathbf{j}=0\]

\[\mathbf{i} \cdot \mathbf{i} = \mathbf{j} \cdot \mathbf{j}=\mathbf{k} \cdot \mathbf{k}=1 \]

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