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## Homework Statement

let f: J --> R^2 be a unit speed curve curve and define it's tangentially equidistant campanion by g(u) = f(t) + r*f'(t) for a fixed r>0. Show that the centre of the osculating circle of f at some u in J is the intersection of the line normal to f'(u) through f(u) and the normal line to g'(u) through g(u).

## Homework Equations

T(u) = f'(u) as f is unit speed (ie. parametrized by arclength)

N(u) = vector normal to f'(u) = R(T(u)) where R is a rotation in the plane by pi/2 in the counterclockwise direction.

The centre of the osculating circle is f(u) + [1/k]*(N(u)), where k is the signed curvature of f.

k := f''(u) dot product with R(f'(u)) for a unit speed curve.

The Frenet-Serret equations also tell us that: N'(u) = -k*T(u)

## The Attempt at a Solution

My attempts have thusfar consisted of applying the defition to find the normal line to f'(u) and g'(u) and setting them equal to eachother then trying to reduce with seriously no luck in proving the claim no matter how I try to manipulate them. I've tried realizing that:

g'(u) = f'(u) + r*(f''(u)) ==> R(g'(u)) = vector pointing in direction normal to g'(u) = R(f'(u)) + r*R(f''(u)) = N(u) + r*(R(f'(u))' = N(u) + r*N'(u) = N(u) + r*(-k*f'(u))

The last few steps can be justified by the fact that R is linear ==> its derivative is itself so that we can manipulate the term R(f''(u)) as shown (I hope) which at least brings the curvature into the fold and intuitevly makes sense in terms of how it is expressed.

No matter what I've tried to do (even going so far as to seemingly uselessly express the desired point in terms of the orthonormal basis (T, N)) has got me no where. Any ideas I'm missing that could get me on the right track?

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