= Solution
Write $z=x+iy$ and square the <modulus> equation. It becomes
$$
x^2+(y+b)^2=P^2[x^2+(y-b)^2].
$$
For $P\ne1$, completing the square gives the <Apollonius circle>
$$
x^2+\left(y-\frac{b(P^2+1)}{P^2-1}\right)^2=\frac{4b^2P^2}{(P^2-1)^2}.
$$
Hence
$$
\boxed{\text{center }\left(0,\frac{b(P^2+1)}{P^2-1}\right),\qquad\text{radius }\frac{2bP}{|P^2-1|}.}
$$
The printed statement needs the exception $\boxed{P=1\ \Longrightarrow\ y=0}$: equal distances from $ib$ and $-ib$ give their perpendicular bisector, a straight line rather than a finite-radius <circle>. For $P>1$ the center lies in the upper half-plane; this is the case needed in part (ii). The original PDF confirms the numerator $z+ib$, correcting the duplicated, corrupted OCR formula.
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