= Solution
Rotate the disc so that $P=r\in(0,1)$ lies on the positive real axis. From part (b), the perpendicular hyperbolic line $\ell$ is a Euclidean circle orthogonal to the unit circle, with centre $c>1$ on the real axis and radius $R=c-r$. Orthogonality of the two circles gives
$$
c^2=1+R^2.
$$
Combining the two equations yields
$$
c=\frac{1+r^2}{2r},
\qquad
R=\frac{1-r^2}{2r}.
$$
The radial distance formula from part (b) says
$$
\rho=2\operatorname{artanh}r,
\qquad r=\tanh(\rho/2).
$$
The hyperbolic double-angle identities now give
$$
\sinh\rho=\frac{2r}{1-r^2},
\qquad
\tanh\rho=\frac{2r}{1+r^2}.
$$
Therefore
$$
\boxed{R=\frac1{\sinh\rho},
\qquad c=\frac1{\tanh\rho}}.
$$
This is the <Euclidean circle representing a perpendicular hyperbolic line>.
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