 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2024/paper-2/3/d/solution

Let

s = ∣ r^{'} ∣ > 0

and

p = r^{'} / s^{2}

. For

∣ r ∣ = 1

, expanding both squared Euclidean distances gives

s^{2} ∣ r - p ∣^{2} = s^{2} - 2 r \cdot r^{'} + 1 = ∣ r - r^{'} ∣^{2} .

Taking nonnegative square roots proves

∣ r - r^{'} ∣ = ∣ r^{'} ∣ ∣ r - p ∣.

This is the distance identity behind inversion in a circle and the method of images for a circular boundary.

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