= Solution
Because $A$ is symmetric, translating $\mathbf r=\mathbf x+\mathbf a$ gives
$$
\mathbf x^{\mathsf T}A\mathbf x+(2A\mathbf a+\mathbf b)^{\mathsf T}\mathbf x
+\mathbf a^{\mathsf T}A\mathbf a+\mathbf b^{\mathsf T}\mathbf a=1.
$$
The linear term vanishes when $\boxed{2A\mathbf a+\mathbf b=0}$. Solving gives $\boxed{\mathbf a=(-1,0,0)^{\mathsf T}}$. Since $\mathbf b=-2A\mathbf a$, the remaining constant is
$$
\boxed{C=1+\mathbf a^{\mathsf T}A\mathbf a=3,\qquad
\mathbf x^{\mathsf T}A\mathbf x=3.}
$$
This is <completing the square> for a <quadratic form>.
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