Orthoprojection of circle onto a plane

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SUMMARY

The orthoprojection of a circle onto a plane results in an ellipse, as demonstrated through geometric transformations. Given a circle ##{\cal C}## with center ##O## and radius ##R## on plane ##P##, the projection onto another plane ##P'## involves translating the circle and utilizing an orthonormal basis. The relationship between the coordinates of the circle and the resulting ellipse is established through the equation ##\frac{x^2}{R^2} + \frac{(y\cos(\theta))^2}{R^2 \cos(\theta)^2} = 1##, confirming the elliptical shape in plane ##P''##.

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Homework Statement


Show that the orthoprojection of a circle onto a plane is an ellipse

Homework Equations



Let's say the circle ##{\cal C}## of center ##O## and radius ##R## lies on plane ##P## and we want to orthoproject ##{\cal C}## onto ##P'##

The Attempt at a Solution



We can say that up to a translation of vector ## \vec{O\pi_{P'} (O)}## , we can orthoproject ##{\cal C}## onto plane ##P''## such that ##P## and ##P''## intersect over line ##D##, and its center ##O \in D##.

Assume that ##D## is directed by unit vector ##\vec u##. We can find an orthonormal basis of ##\vec P##, say ##(\vec u, \vec v)##, such that ##(\vec u, \vec w = \text{rot}_{\vec u, \theta} (\vec v))## is an orthonormal basis of ##\vec {P''}##, where ## \theta = \angle (P,P'')##.

Then ##M(x,y) \in P \Rightarrow \pi_{P''}(M) = (x \vec u,\vec u) \vec u + (y\vec v,\vec {w}) \vec w = x \vec u + y \cos (\theta) \vec w ##

And then ## M \in {\cal C} \iff x^2 + y^2 = R^2 \iff \frac{x^2}{R^2} + \frac{(y\cos(\theta))^2}{R^2 \cos(\theta)^2} = 1 \iff \pi_{P''}(M) ## is located on an ellipse of ##P''##

Is this working for you ?
 
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Looks right.
I would explicitely replace y cos(θ) by w for the last step.
 
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Ok, thanks !
 

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