The lifted determinant records a power-sphere side

Put $h_i=\mathrm{\parallel }{p}_i{\mathrm{\parallel }}^2-w_i$. Let $A$ be the $d\times d$ matrix whose rows are $p_i-p_0$ for $i=1,\dots ,d$. Subtract row $p_0$ from each other row of the orientation matrix defining $O_d$, then expand along its final column. Since the base points are affinely independent, $\det A\mathrm{\ne }0$, and the cofactor sign gives $\det A=(-1{)}^d{O}_d$.

Expanding the equal power-sphere equations in [A1] gives $h_i-h_0=2c\cdot (p_i-p_0)$ for $1\le i\le d$. For the query row, the same expansion gives $h_q-h_0=2c\cdot (q-p_0)+\mathrm{\Delta }(q)$.

In the lifted determinant defining $P_d$, subtract row $p_0$ from all other rows and expand along the final column. The cofactor sign is $(-1{)}^{d+3}$. Subtract $2c_j$ times coordinate column $j$ from the lifted-height column for every $j$; by [step 1.2] the first $d$ height entries become zero and the query height entry becomes $\mathrm{\Delta }(q)$. The remaining determinant is $\det A\mathrm{\Delta }(q)$, so $P_d=(-1{)}^{d+3}\det A\mathrm{\Delta }(q)=-O_d\mathrm{\Delta }(q)$ by [step 1.1].

Multiplying the identity in [step 2.1] by $O_d$ gives $O_d{P}_d=-O_d^2\mathrm{\Delta }(q)$. Since affine independence gives $O_d\mathrm{\ne }0$, the factor $O_d^2$ is positive; hence the product has the claimed inside/on/outside sign.