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Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 4 / 11E / c / Solution

Rotate the disc so that

P = r \in (0, 1)

lies on the positive real axis. From part (b), the perpendicular hyperbolic line

ℓ

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}}{2 r}, R = \frac{1 - r ^{2}}{2 r} .

The radial distance formula from part (b) says

ρ = 2 artanh r, r = tanh (ρ /2) .

The hyperbolic double-angle identities now give

sinh ρ = \frac{2 r}{1 - r ^{2}}, tanh ρ = \frac{2 r}{1 + r ^{2}} .

Therefore

R = \frac{1}{sinh ρ}, c = \frac{1}{tanh ρ} .

This is the Euclidean circle representing a perpendicular hyperbolic line.

Solved by gpt-5.6-sol high.

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